Talk:Monty Hall problem/Arguments/Archive 1

Strategy or reality? (Random host variant)
Behaviour creates reality. If reality is bound by presumptions (no car revealed), a strategy that goes beyond such reality (random door opening) should not be accounted into any valid probability theory. Actually the point I'm making underneath is that there is a fundamental difference between an experiment in which doors are randomly opened, thus not reducing the sample space by experiment, and a single situation in which the sample space has been reduced definitely. If reality is only that single situation (B), without being part of a greater reality A (total sample space), it makes no sense to define P(B) against A. Heptalogos (talk) 12:37, 13 February 2009 (UTC)

I will try to prove that the version in which the host opens doors randomly, but never picks a door with a car behind it, does not decrease the player’s chance to win the car when switching doors, compared to the original situation.

The original situation, in which a player picks door 1, has six theoretical possibilities:

a = auto; g = goat; empty spaces are unopened doors with goats behind them

Explanation: if door1 hides a goat (player’s chance = 2/3), the host has two options: reveal a goat behind door2 or reveal a goat behind door3. Since the goats are placed randomly, these have even chances.

The version in which a host randomly opens door 2 or 3 has the same possibilities with different chances:

* In 4/6 = 2/3 of the possibilities the host does not reveal the car. If we would have to statistically test this, we could do the game e.g. 999 times and create about 666 situations in which no car was revealed. Then we take a look into only those 666 cases, and pick one of those. Now there are four options that could have created this situation:

Indeed, they add up to a total chance of 4/6 = 2/3, which is exactly the part we are in now. But these chances are old chances that created this situation. What is the chance that any of these four possibilities is now true? Of course it is 1. The sum of chances changed from 2/3 to 1, because we stepped out of the 999 options into 666, carefully selected by one variable. In fact, the ‘knowledge’ of the host in the original problem has been replaced by the ‘knowledge’ of the creator of this new situation. By removing all ‘cars-revealed-by-host-possibilities’ afterwards, we create the same situation as when ‘not-reveal-cars-in-the-first-place’.

If the sum of four possibilities is 1, what are the individual chances? Now the player finds himself in a new situation in which one of those four possibilities is true. This is obvious because no car but a goat has been revealed. It doesn’t matter to him why the host opened a door, random or by knowledge; his behaviour is in both cases the same, leading to the same new knowledge of the player. Could the host have changed the chances of the two possibilities in which the car is behind door1? No, he could only prefer one ‘goat’ above another, but they share no relevant difference. So he should have changed the chances of the other possibilities. Is that possible? Of course, revealing a goat behind doorX disables the other option which has the car behind doorX, so these chances become one, in number. These are the new chances:

Which are the same as the chances in the original problem. Could I have explained this easier? Probably. Test it yourself by playing three cards of which two are identical and one is unique. Let a player pick one and turn around another one yourself. Every time you turn around the unique card, stop the game and start again. It is not different from knowing the cards and turn around the twin card in the first place. Heptalogos (talk) 23:23, 31 January 2009 (UTC)

"The conclusions of this article have been confirmed by experiment" Does anyone have information about experiments on the 'random host variant'? Heptalogos (talk) 16:48, 1 February 2009 (UTC)


 * The primary purpose of the bold disclaimer at the top of this page is to encourage folks who are unconvinced switching wins 2/3 of the time in the regular version to think about it before posting comments here doubting the solution. Many experiments and simulations have shown the 2/3 result.  I am not aware of similar experiments regarding the random door variant.


 * However, I think you may be slightly misunderstanding this variant. If the host reveals the car accidentally we don't keep the same initial player pick and have the host pick another door (until he manages to not reveal the car), we ignore this case completely.  In your playing card version if you reveal the unique card, stop the game and start again and don't count this time, i.e. re-randomize the cards and have the player pick one.  I think you understand that if you do this 999 times, roughly 333 times you'll reveal the unique card and roughly 333 of the other 666 times the player will have picked the unique card (right?).  In this variant we're only counting those 666 times (not the entire 999), so the chance of winning by switching is 1/2.  Your tables above are confusing because of the "player chance" and "host chance" columns. If you remove these columns and leave the equally possible 6 cases (your second table), the total chance of each is 1/6 as you indicate.  We can count the cases where the player picks the car and see the chance of this is 2/6.  Similarly, counting the cases where the player picks the goat indicate a chance of this of 4/6 (2/6 plus 4/6 better be 1, since the player either picks the car or a goat!).  What happens if we're given that the host didn't reveal the car is that there aren't 6 cases, but only 4.  In 2 of these the player picked the car.  The probability of the player having picked the car changes from 1/3 to 1/2 because we've eliminated 2 of the possible 6 cases (and in both of them the player initially picked a goat).  -- Rick Block (talk) 18:54, 1 February 2009 (UTC)


 * I understand. The 333 cases which we threw away are of course all car picks which now indirectly influence the outcome of the remaining group, as a whole. Now I understand why I cannot accept the issue: it seems from the description of the situation that Monty is playing his usual game, remembering the door he should open, but then in one case forgets. In this situation there is no way to define the chances, just because it's a single event. Actually, I would say that in case he never in his career accidentally picks a car, nothing changes the odds because there is no behaviour change. In other words: the chances only change to 50/50 if the game changes statistically, which would mean that Monty always picks randomly, revealing cars 1/3 of the time. Only then it would be true.


 * Now in the original problem description it says somewhere that "This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car." Only this is true, and it is actually not true, as stated above, that the player should switch (in any individual case). So it can be a good advise to all players of the game generally, to follow consequently, in a consequent game. The statement about a clueless host is correct if this is either a consequent, repeating situation, or if it's defined as generally the best choice, but in the latter case a car should be revealed a significant number of times. I suggest to change the idea of Monty having a memory leak on a particular day, into a consequent situation from which chances can be correctly analyzed. Heptalogos (talk) 20:13, 1 February 2009 (UTC)


 * I still stand with my attempted proof of "the version in which the host opens doors randomly, but never picks a door with a car behind it", because this is something else than my 'easier explanation' below with the card game. The card game is statistically realistic, revealing the 'wrong' card and stopping the game repeatedly. But playing random and yet never open the 'wrong' door is a small chance on it's own! (Which becomes smaller when more reps are done.) Hence this is a different world, starting with the assumption that the right door always opens. The 333 expected options do not exist, are not excluded, and do not change the chances of the remaning part. Heptalogos (talk) 20:34, 1 February 2009 (UTC)


 * Even if the situation is a single game where Monty forgets we can talk about the chances, although in this case in a strict sense it would be a Bayesian probability rather than a frequency probability. And, although there is a small chance playing randomly that a wrong door is not opened, this doesn't mean much about the frequency probability which by definition is the limit over a large number of trials.  If the host hasn't revealed the car roughly 1/3 of the time you simply haven't run enough trials.  -- Rick Block (talk) 00:07, 2 February 2009 (UTC)


 * Actually we can never define chances in any single specific case. We can only say that in indentical situations chances are..., which means that the outcome is... x% of the time. In case Monty only sometimes forgets, I see two possibilities:
 * 1. Monty forgets once every x shows. (We even need to know x.)
 * 2. In all series of identical shows, the host will forget once. (number of shows in a series is irrelevant)
 * We need this population of possibilities to define chances within, one of which could be a random single event's chance, but not a specific single event's chance. How many times Monty forgets? What is the chance he forgets? We only don't need to know this when he always forgets, or in other words, when the host acts randomly, consequently.


 * I think we are reversing the essentials here. To define chances, we basically need a significant group of events, from which we know the overall outcomes. Then we can group similar outcomes and calculate the chance of one of those. Furthermore we may try to understand what causes such groups. A probable cause can be the behaviour of a host randomly opening one of two doors. Now if we presume this behaviour, it is only that we can define the overall chances because we predict the outcome of this behaviour, which is the given group. We assume that the behaviour will create a significant group of outcomes, but it should really reach this number of events before we can be sure that the overall reality looks like it. If this behaviour only creates one single event we cannot say anything about it!


 * Not only do we miss a big enough control group to define solid chances; even if we define such chances, it makes no sense assuming that it will predict a single specific event. The fun part is that one cannot reject this by experiment, because in those cases a control group will already be created. We can only prove that behaviour is random when we create a big enough reality with it. But it is more than that: without this reality there actually is no random behaviour. I apologize for the generality of this discussion, but I only started this because in this case even a description of such a reality is missing! Random behaviour is assumed as a single exception! This is totally unreal and cannot even be closely proved experimentally. As soon as you do, you create reality. Most realities which are big enough to be near-significant, will confirm the chances, just because of the 'size' of the reality. How many small realities deviate? We never know, but if when we can group them, we can create the bigger reality that we need. That's why we need a bigger control group like one of the two possibilities presented above.


 * This is not just a formal filosophical argumentation. There is no such thing as random behaviour in single specific events. Every single event has it's own unique variables which may or may not be known by anybody. That is why we never can predict, by experiment, a single specific outcome of random behaviour. It is only in a significant group of events that all of the uncontrolled variables wipe out each other's relevant influence on the given situation. Heptalogos (talk) 22:17, 2 February 2009 (UTC)


 * If I put 17 marbles in an opaque urn, 16 of which are white and one black but otherwise identical, you put your hand in and grab one, I think it's common to say you have a 1 in 17 chance of picking the black one (even if you only do this once). We don't have to do this a large number of times to analyze what the chances are using probability theory.  Similarly, we can say one individual player in the Monty Hall problem has a 1/3 chance of initially picking the door hiding the car (which we assert is placed "randomly" behind one of the 3 doors).  If the host forgets where the car is and opens a door "randomly" (and if the car was actually placed randomly but the host forgets where it is, opening either door is equivalent to opening a door randomly) and fortuitously doesn't reveal the car, we can (using probability theory) say the player has a 1/2 chance of winning by switching because there are 4 equally likely scenarios with 2 where switching wins.  Could we design a test to show this experimentally?  Sure.  Do we have to in order to believe it's true?  Well, I don't.  I trust probability theory and the law of large numbers. -- Rick Block (talk) 03:40, 3 February 2009 (UTC)


 * It's common to say that you have a 1/17 chance to pick a black marble if you try once, but it's not true. Actually there is no chance directly related to a single specific event. The event is related to (as part of) a group of similar events to which chances are related. These chances are generic and cannot be linked to a unique element of the group. That's why we strictly can only advise a group of players, or a player who will repeat a process. The first spacial similarity focuses on similar players in similar shows, while the timely similarity addresses a certain repeating frequency of the same spacial situation.


 * What happens if we're in a situation in which we (once) seem to have a chance to pick one out of many, is that we make this event part of a group of (seemingly) similar events, and divide the group in two subgroups: the ones and the manies. Now we want to predict in which subgroup we arrive by event. We have no idea (random) what causes access to one of these subgroups, but we only know that one subgroup is bigger. Is it more likely to arrive in a bigger subgroup, just because it’s bigger? If we try, by process, we will see that most events end up in the bigger group. But it only becomes (generally) true because it becomes real, in a big enough group. Fortunately we don't have to start the whole process ourselves; we can make our event part of a group of events which happened in the past. Or we can make it part of similar single events in similar situations at the same time, assuming that they're all random.


 * Could we design a test to show this experimentally? Do we have to in order to believe it's true? My answer is: we don't have to do the experiment, but we should indeed do the design! We should describe a theoretical reality to make it credible. One cannot 'once' act random, we really need the bigger picture. Please design your test, as you say you can, and you will see that no forgetting Monty can exist in such an experiment. You will likely replace him with 'another' random event, which will even be repeated, but that's really something else. The situation is presented as random, but it's not. Apart from that, there is no way to define chances in a single specific event. Heptalogos (talk) 21:18, 4 February 2009 (UTC)

Logic and math as different disciplines
We should consider to distinguish logic from maths. We might even agree then. Logic tells us that if we have x doors, from which a player randomly picks one (door 1) after which a host picks another (door 2), the initial chance of the player choosing the one and only winning door is 1/x. If we assume that the host picks the winning door if possible, and otherwise chooses randomly, the chance that he chooses the winning door is 1-(1/x). In this scenario there are two possible situations:
 * 1. Doors 1 and 2 are numbered after chosen.
 * 2. Doors 1 and 2 are numbered before any choice is made.

In the second situation we will afterwards restrict the experiment to the outcomes mentioned. Logic tells us that such a restriction does not change the second chance. This chance is 1-(1/x), for all x, which is logically valid and can be proved by experiment.

Mathematically, this may be wrong. I am not even interested if this is the fundamental discussion which it seems to be (about strict policies). But we have to respect the fact that mathematics are a product of logic, as a tool to make logic more efficient and effective. It cannot outclass logic itself. If we distinguish logic from maths, we can provide the simple, logical solution, and afterwards present the mathematical solution. The last solution is supervised by experts and needs academical sources, but what about the logical? Does it belong to philosophy? Heptalogos (talk) 08:36, 11 February 2009 (UTC)

Btw, if doors 1 and 2 are 'picked', the host might open all other doors (3-x), or not. The player might switch to the host's choice. Actually, the knowledge of the player doesn't matter, because we decide the chances. That's why it doesn't matter if doors are opened or not, or if players are blindfolded etc. Heptalogos (talk) 09:13, 11 February 2009 (UTC)

Maybe I still need to explain why "Logic tells us that such a restriction does not change the second chance". Suppose we know that in every country of the world men and women exist in equal amounts. Now we have to pick a country after which we randomly have to pick a person in that country. What is the chance this person is a woman? In this scenario there are two possible situations:
 * 1. We have all countries to choose from.
 * 2. We can only choose African countries.

Logically in both situations chances are the same, because the restriction does not affect the probability. Heptalogos (talk) 09:44, 11 February 2009 (UTC)

As you may have noticed, I am an expert in logic. Heptalogos (talk) 09:45, 11 February 2009 (UTC)

I think I understand why maths has to follow a straight line; it cannot make logical jumps. The same occurs writing computer code. The binary code is one-dimensional, while the human brain has more dimensions. Humans can use several time-space coordinate systems. On one hand we may all agree to the validity of certain logical jumps, but on the other hand it may be hard or even impossible to prove it logically. I agree that the mathematical approach, which has solid rules for such proof, is a better proof, or even the only exact proof. On the other hand, maths is a product of the same logic and we need to be pragmatic in some way. It's an interesting idea that we actually only understand things in our conscienceness in a logical way, but we can only make it an objective truth by externalizing and even materializing it (write it down in strict 'code'). To make this less easy: scientists are experts in understanding their own systems, checking if any logic complies to that, without necessarily understanding any of the logic itself.

I call these 'operational' scientists. But think of all the great scientists who had their moments of 'eureka', logically, after which they had to work out new code. They even had to search intensively for the 'right' code to make it fit to their logic! What is good logic and what is bad logic? Besides 'code', the best proof of good logic is 'reality'. I would like to turn that around and state that logic is a recognition of a specific expression of the laws of nature. One recognizes the practice of natural law and tries to make it commonly accessible. The keyword is 'patterns', to be recognized, to be registered, to be tested and to be executed by ourselves.

REALITY	->	LOGIC	->	SCIENCE	->	REALITY

It's a circle of construction. In the famous Monty Hall paradox, we make a logical jump, which really seems realistic, but some scientists make objections. The logical jumps do not fit into their systems. Is it bad logic, does science need an extra dimension, or are these exceptions rather useful to understand the limits of our disciplines? Heptalogos (talk) 11:33, 11 February 2009 (UTC)

Probabalistic analysis
Notation:


 * A=number of the chosen door
 * C=number of door with car
 * D=number of the opened door

Given:


 * P(A=a)=1/3 (a=1,2,3)
 * P(C=c)=1/3 (c=1,2,3)
 * A and C are independent
 * P(D=2|A=1 and C=1) = 1/2
 * P(D=3|A=1 and C=1) = 1/2
 * P(D=3|A=1 and C=2) = 1
 * P(D=3|A=1 and C=3) = 0

Similar for other combinations. For simplicity we write A1 instead of A=1, etc.

Given {A=1 and D=3} what is the conditional probability of {C=2}?
 * $$\!\,P(C_2|A_1 D_3) = \frac{P(C_2 A_1 D_3)}{P(A_1 D_3)}$$
 * $$\frac{P(D_3|A_1 C_2)P(A_1 C_2)}{P(D_3|A_1 C_1)P(A_1 C_1) + P(D_3|A_1 C_2)P(A_1 C_2) + P(D_3|A_1 C_3)P(A_1 C_3)}$$
 * $$\frac{1 \frac 19}{\frac 12 \frac 19 + 1\frac 19 + 0\frac 19}=\tfrac 23$$

Nijdam (talk) 22:30, 13 February 2009 (UTC)

A simple experiment
Prizes were placed randomly. It is unknown whether the player chooses randomly. He might always choose the same door, but we don't even know if he can (distinguish doors). Even if he would always pick the same door (i.e. door No. 1 is always the same physical door), the results are still random because the prizes are placed randomly. Simulation of this event will relate the car to door No. 1 in 1/3 of the times, in random order. Instead of relating the car to door No. 1, it may relate the car to 'C', which means 'player's Choice'.

Computer simulation looks like this:


 * Ca
 * Cg
 * Cg
 * Cg
 * Ca
 * Cg
 * Ca
 * Cg
 * Cg

P(Ca)=1/3, P(Cg)=2/3

Now two prices are left, from which a goat (No. 3) is excluded from the experiment. This is equivalent to offering Switch-door No. 2 ('S') with two possible outcomes:


 * Sa
 * Sg

We know that (C,S) is (a,g) with these possible combined outcomes (only a car and one goat are left in the game):


 * Ca	Sg
 * Cg	Sa

We can simply add it to the simulation above:


 * Ca	Sg
 * Cg	Sa
 * Cg	Sa
 * Cg	Sa
 * Ca	Sg
 * Cg	Sa
 * Ca	Sg
 * Cg	Sa
 * Cg	Sa

Switching leads to the car (Sa) in 2/3 of the times. P(Sa)=2/3, P(Ca)=1/3. Heptalogos (talk) 13:25, 15 February 2009 (UTC)

Why conditional?
Can someone please explain why the problem should be treated as a conditional problem, as P(A|B)? I would like a general answer, even a maths law if possible. I think there are mainly two reasons to see it as conditional:


 * 1. Event B is affecting the sample space.
 * 2. Event B is affecting the probability of event A.

I think the first is no argument for making it obligatory conditional, but it's still used by many people in the Monty Hall problem, without explicitely saying so, or even being aware of it. The second reason is, but it's not the case in the MHP.

One other question, because it's related: isn't it true that the only definite way to identify unique doors, is by one of these:


 * a. It has been picked;
 * b. It has been opened;
 * c. It is related to a car;
 * d. It is related to a goat (only after another door has been opened or another door related to a goat has been picked);
 * e. It is the only one not identified yet, and therefore identified?

Event a. may be called door No.1, event b. may be called door No.3, but it's at least a lable on the event and not for sure a physical identification of a door on beforehand. Heptalogos (talk) 21:23, 15 February 2009 (UTC)

To put the second question differently: I see no information in the number, at least not to reason 2. It might be to reason 1, as a reduction of sample space, but what information does the number give? It is a precondition that the sample space will be reduced by one door (with a goat), so this is no new information. What new information is in a number? Why this numbering? It seems that the only reason to number the doors on beforehand (like Krauss and Wang (2003:10) seem to do, but even that is not sure), is to be able to 'prove' afterwards that a conditional approach is needed. Heptalogos (talk) 22:47, 15 February 2009 (UTC)


 * The way the problem is phrased ("Suppose you're on a game show") gives it a context in which I think it's clear we're not talking about abstract conceptual "doors" that may or may not have any physical reality but doors labeled 1, 2 , and 3 on a physical stage. There's a photo of the actual "Let's Make a Deal" studio at http://www.letsmakeadeal.com/.  The problem is perhaps not about this specific game show, but we're told it's about a game show which we should assume is at least similar.    The decision point (switch or not) is after an initial door has been picked following which the host has opened a door to reveal a goat.  Here's a direct quote from the Gillman paper referenced in the article:


 * Game 1 [the Parade version of the problem] is more complicated: What is the probability P that you win if you switch, given that the host has opened door #3? This is a conditional probability, which takes account of this extra condition. [italics in the original]


 * Regardless of what you call the door the player initially picked (and we can renumber the doors so that we always call this door 1) and what you call the door the host has opened (which we can similarly always call door 3) the question remains should you switch from one door (which we'll call door 1) to another (which we'll call door 2) given the host has opened a door revealing a goat (which we'll call door 3). Given a choice between representing this question as


 * $$P(\text{win by switching})\,$$


 * or


 * $$P(\text{win by switching} | \text{player picks door 1 and host opens door 3})\,$$


 * isn't the conditional question obviously a more appropriate mathematical model? It may turn out that the answer to these two questions are the same, but we can't know this until we answer the second question.  -- Rick Block (talk) 23:34, 15 February 2009 (UTC)

Rick, thank you for your ever continuing effort to try to make other people understand, which I honoustly appreciate. Yet, it's not that easy. Your explanation is very clear as from 'Regardless'. (Let's forget about abstract doors, because of course I agree that they're physical.) You try to make things obvious, but the explicit explanation is still missing. I can do the same:


 * $$P(\text{win by switching})\,$$


 * or


 * $$P(\text{win by switching} | \text{one car and two goats are randomly placed behind the doors})\,$$

Isn't the conditional question obviously a more appropriate mathematical model?

Since we already agree/assume that door 1 is always picked, door 3 is always opened, and a goat is always revealed, these are preconditions rather than new information. Do you get my point? Heptalogos (talk) 11:36, 16 February 2009 (UTC)

As far as I know, only one person tried to explicitely explain, which is Nijdam, stating that when event B is reducing the possible outcomes, P(A|B) should be the calculation method, which is the conditional one. This is at least a mathematical rule being applied here. What we need is rules! How can we just 'show' something and assume that it's obvious? Where is the proof? Isn't this all about statistical dependency? Heptalogos (talk) 11:47, 16 February 2009 (UTC)


 * Perhaps you're questioning what probabilistic universe we're starting from, which I think means deciding which parts of the problem description should be treated as "rules" of this universe rather than conditions on a problem posed within this universe. What we're basically talking about here is what should be considered the Bayesian prior.  If we're fully rigorous, we never write
 * $$P(\text{win by switching})\,$$
 * but rather
 * $$P(\text{win by switching} | I)\,$$
 * where $$I\,$$ is the background information that is known. The Bayesian analysis section of the article follows this approach and makes it very clear what is background.  In a formal Bayesian sense, all problems are conditional problems.  What we're calling the "unconditional" probability here (well, at least what I'm calling "unconditional") is
 * $$P(\text{win by switching} | \text{the game rules})\,$$
 * where "the game rules" is everything before the "Imagine ..." sentence in the K&R description (car/goats randomly placed, player initial pick, host must open a door revealing a goat, host opens random door if player initially picks the car). Does this help? -- Rick Block (talk) 14:53, 16 February 2009 (UTC)

Well it surely helps if it enables you to answer the following question, in general: "which conditions need Bayesian method?" I ask "in general", because you are still not answering my question about the common rules. It may be the same question as "which conditions need conditional probability method?"

What about this: "If events A and B are statistically independent, then P(B|A) = P(B)" Heptalogos (talk) 16:39, 16 February 2009 (UTC)

Heptalogos suggested that we continue a conversation here. It was about what makes a particular event a condition in a probability problem. Let us consider four events that happen (or might happen) between the player's original choice of door and his decisiom to swap or not. Which of these must be taken as a condition in the solution of the MH problem and why?

—Preceding unsigned comment added by Martin Hogbin (talk • contribs)
 * 1) Monty opens a door to reveal a goat
 * 2) Monty uses the word 'pick'
 * 3) A member of the audience coughs
 * 4) It starts to rain outside


 * I'm not trying to be difficult, but what do both of you mean by "need"? It's certainly true that if A and B are independent, then P(B|A) = P(B) - but you've just moved the question to how you know whether A and B are independent (in the case of the MH problem, this is fundamentally the argument that several folks have mentioned is missing from the "unconditional" solution).  The point of Bayesian priors is that it eliminates the need to ask the question you're asking.  It is a more precise formalism that (among other advantages) neatly avoids the exact issue we're talking about.


 * (Maybe I should learn about Bayesian statistics.) Martin Hogbin (talk) 20:54, 16 February 2009 (UTC)


 * The answer to Martin's question is the first one, and only the first one, because it is exactly what the problem asks. If the problem statement were "what is the chance of winning by switching if a member of the audience coughs" we'd have to include this as a condition.  Or am I completely missing what you're asking? -- Rick Block (talk) 19:46, 16 February 2009 (UTC)


 * What about 2? Monty does use the word 'pick' when he asks if the player wants to swap. Martin Hogbin (talk) 19:51, 16 February 2009 (UTC)


 * Martin - I honestly don't have a clue what point you're trying to make. Care to explain? -- Rick Block (talk) 02:00, 17 February 2009 (UTC)


 * It is about what criteria we need to apply to determine whether an event that occurs between the initial choice and the decision to swap must be treated as a condition. You have given one good criterion, that it must be mentioned in the question.  But in the parade statement Monty says, 'Do you want to pick...?'.  Why is it not necessary to formulate the problem with this statement as a condition.  Why do we not need to say, after the initial choice we need to determine the probabilities given that Monty uses the word 'pick' ? —Preceding unsigned comment added by Martin Hogbin (talk • contribs)


 * Is this merely a pedantic point, or are you truly confused and can't see the difference between using the word "pick" and opening a door? Like any word problem, there's a mapping that has to be done from the natural language version to a more precise mathematical version.  I'm saying the natural language version of the problem, as commonly understood, maps to a conditional probability problem, and is specifically asking about:


 * $$P(\text{win by switching} | \text{player picks door 1 and host opens door 3})\,$$


 * where the "unconditional" probability


 * $$P(\text{win by switching})\,$$


 * refers to the probability of winning by switching as seen by all players.


 * What are you saying? -- Rick Block (talk) 19:45, 17 February 2009 (UTC)

When you do the mapping from natural language to a mathematical description, why do you not do it like this?


 * $$P(\text{win by switching} | \text{player picks door 1 and host opens door 3 and Monty says the word pick})\,$$

I think it much the same question that Heptalogos is asking. Martin Hogbin (talk) 22:26, 18 February 2009 (UTC)

To answer you question above Rick, I cannot see any difference in principle regarding saying the word 'pick' and opening a door in relation to the MH problem, if I am missing the obvious please point it out to me. Martin Hogbin (talk) 22:30, 18 February 2009 (UTC)


 * The difference is that it's obvious that the user's initial pick is the car with probability 1/3. The entire question is what impact, if any, the host opening a door has on this probability.  To rephrase slightly, the fundamental question here is the difference between


 * $$P(\text{player has picked the car})\,$$


 * i.e. the probability of initially picking the car (regardless of what the host does), and


 * $$P(\text{player has picked the car} | \text{host has opened a door given the rules of the game})\,$$


 * If we assume there's no impact on the player's initial probability the problem is trivial - in effect, we've assumed the solution. As it turns out, there is no impact, but the only actual way we can figure this out is to figure out the conditional probability. -- Rick Block (talk) 23:02, 18 February 2009 (UTC)


 * You seem to have missed my point which is, why do we not include the fact that the host uses the word 'pick' as a condition. Is it because we assume that using this word cannot change the probability that the player has chosen a car?  Martin Hogbin (talk) 00:01, 19 February 2009 (UTC)


 * I don't know if there are hard and fast rules for how to map a natural language problem to a math problem, which is what it sounds like you're really looking for here (are you trying to force me to admit this?). I've tried to get C S to respond to this (and have recently asked Glopk as well).  They might have a more formal answer, but I think this basically boils down to common sense.  If something is obviously irrelevant with regard to the mathematical model of the problem, we shouldn't include it.  We can choose to include it, and conclude it does not change the probability that the player has chosen a car, but if it's something that's obviously extraneous why should we bother?


 * This might come across condescendingly (not the intent - I just want to use an example that is obviously clear) a word problem like: "If Johnny has 2 apples and 1 orange, and Sally has 3 apples and 4 jelly beans, then how many apples do they have altogether" is obviously (to any adult) the math problem "what is 2 + 3". Similarly, the MH problem is obviously (to a mathematician) a conditional probability question involving the door the host opens (and not whether he uses the word "pick").  -- Rick Block (talk) 02:48, 19 February 2009 (UTC)


 * Martin, as the main editor of the current "Bayesian Analysis" section, I'd like to reply to your question as stated at the beginning of this thread:
 * When you do the mapping from natural language to a mathematical description, why do you not do it like this?
 * $$P(\text{win by switching} | \text{player picks door 1 and host opens door 3 and Monty says the word pick})\,$$
 * What I do not like about this formulation is that it is uneconomical: it is very verbose, but identical in meaning to the one currently in the article. After careful reading of the "Bayesian Analysis" section of the article, you ought to agree that (using the symbols therein):


 * $$\text{win by switching}\quad==\quad C_2\qquad$$, since the player wins by switching if and only if the car is behind Door 2.


 * $$\text{player picks door 1 and host opens door 3}\quad ==\quad H_{13}\qquad$$, by the definition of the symbol $$H_{ij}\,$$.


 * $$\text{Monty says the word pick}\quad\in\quad I\qquad$$, as it specified/implied in the rules of the game.


 * Therefore your probability expression above is the same as $$P(C_2|H_{13}, I)\,$$, which is exactly the one whose value is computed in the Bayesian proof, and found to be 2/3 in the standard MH formulation.


 * So, to repeat again Rich Block's question: what are you saying?glopk (talk) 04:56, 19 February 2009 (UTC)

Allthough I'm still involved in a private discussion with Martin, I might as well try to explain what I think is the essence of Martin's question, which I rephrase as: What events lead to conditioning? I said this before, but no harm in repeating: all events that put limiting constraints on the sample space. So the cough of a member in the audience might be a conditioning factor if "coughing of the audience" was primarily taking part in the sample space. But in general one doesn't take this into account. However if one would do so, a cough would not limit the possible outcomes. But distributing a car and opening a door plays an essential role in the sample space. I gave several times representations of it (without taking coughing and the use of the word "pick" into account!). And I showed, and it is not difficult to understand, that opening a specific door, limits the outcomes. That's why. Nijdam (talk) 13:04, 19 February 2009 (UTC)
 * If your rendition of Martin's question is correct, the question is ill-posed: there is no such thing as "events" that lead to conditioning, particularly not in the standard formulation of the MHP, where the background information $$I\,$$ is stationary (the rules do not change while the game is in progress). Rather, the conditioning among the terms simply reflects logical structure of the problem. However, I believe that Martin is asking a different question, namely what makes the opening of a particular door following the player's selection a relevant conditioning term, whereas a sneeze in the crowd is not. Or, to use the notation of the "Bayesian analysis" section, why it is that as a function of $$k\,$$, $$P(C_k|H_{ij}, \text{Sneeze}, I) == P(C_k|H_{ij}, I) \ne P(C_k|I)\,$$ ? The answer, which Rick Block has already given above, and I restate here formally, is that the conditioning of $$C_k\,$$ on $$H_{ij}\,$$ is due precisely to $$I\,$$. This is because (in the standard formulation) Monty follows a well-defined algorithm to decide which door to open, and the outcome of the algorithm depends on where the car is. Hence $$H_{ij}\,$$ is not independent of $$C_k\,$$ under $$I\,$$, or $$P(H_{ij}|C_k, I) \ne P(H_{ij}|I)\,$$, and therefore (by Bayes's theorem) $$P(C_k|H_{ij}, I) \ne P(C_k|I)\,$$. On the other hand, the statement of the problem does not mention anything about sneezes among the public, phases of the moon, or opening prices of sweet crude at the NYMEX, therefore (by Ockam's razor) we must assume the independence of $$C_k\,$$ from any of them.glopk (talk) 17:18, 19 February 2009 (UTC)
 * I hope Martin will speak for himself, but in the mean time I may comment on your analysis. As the rules I doesn't change during the play (why call this stationary?), I skip mentioning them all the time. In my opinion the main discussion is only about the difference between the unconditional $$P(C_k)$$ and the conditional $$P(C_k|H_{ij})$$, and the necessity of the use of the latter. Nijdam (talk) 17:33, 19 February 2009 (UTC)
 * Nijdam, at this point I really don't understand what you talking about - I suspect you are attaching two different meanings to each of the symbols above. Please, spell out the proposition $$C_k$$ whose probability you wrote above as $$P(C_k)$$. Just the proposition, with no "conditional" or "unconditional" qualifiers. Then we can be sure we're talking about the same thing. Thanks. glopk (talk) 07:19, 20 February 2009 (UTC)
 * I have accpeted Rick's criterion that only things mentioned in the question should be taken into account but Monty does say the word 'pick' in the Parade statement. Why is this not taken into account?  Can we ignore anything that we  deem independent of $$C_k\,$$? Martin Hogbin (talk) 19:45, 19 February 2009 (UTC)

For the less mathematical skilled, I would say that a lot of these problems have the following form. A situation is given and the probability os some event is calculated. Then something happens and one is asked about "the probability" of the mentioned event. I put quotation marks around "the probability" because it is a different probability than initial. Why else should we bother? What is the difference? Well, because something has happened, a new situation has arised. And it is in this new situation the new(!) probability has to be calculated. It may or may not differ from the old one, and it's common to phrase this as: the probability has or has not changed. These problems are of course only of interest if the thing that happens essentially changes the situation. But anyhow has a new probability to be calculated in the new situation. Probabalist call this new situation a condition to the origanal one; and the new probability the conditional probability given the new situation. Nijdam (talk) 13:32, 19 February 2009 (UTC)
 * Firstly as you all have do doubt worked out, I am not an expert of probability, but your statement above seems to me to be pretty much what I thought. This brings be back to my original question, how do we decide which events, of those mentioned in the question, we take into account in calculating the probability? Martin Hogbin (talk) 19:45, 19 February 2009 (UTC)
 * For example we might suppose that the host could always say something like, 'Do you want to pick another door?', when he knows that the player has chosen a car but he might say 'Do you want to choose another door?' when the player has chosen a goat, as a way of giving a hidden clue to the contestants.Martin Hogbin (talk) 23:40, 19 February 2009 (UTC)

Ok. If some event occurs, it usual limits the possible outcomes of the experiment. I.e. in throwing a dice, someone may tell you, the result is an even number. This implies, you only have to consider the possibilities 2, 4 and 6. On forehand the prob. getting 6 was 1/6, now it is 1/3. "Now" means: under the condition the outcome is even. The limiting event in the MHP is the choice of the door by the player and the opening of a door by the host. This event really limits the possible outcomes. And hence the conditional probability is a different function than the unconditional one, even for the car behind the picked door, allthough the numerical value is the same for both functions. Nijdam (talk) 20:28, 19 February 2009 (UTC)

And to be complete: anything that doesn't limit the possible outcomes may be discarded. Normally this will be events that are not an essential part of the problem formulation. Nijdam (talk) 20:31, 19 February 2009 (UTC)
 * But what is the way to decide what is an essential part of the problem formulation? Martin Hogbin (talk) 23:40, 19 February 2009 (UTC)


 * Martin, have you read my reply above? What do you find unsatisfactory about it? It is a standard application of Bayes theorem. In plain language, it says that the probability that car is behind any given door is dependent on what the host does because, due to the standard game rules, Monty must select which door to open taking into account where the car is. On the other hand, the game rules do not mention anything about the word "pick" (or the name "Monty") as having any bearing on which door is selected for opening, and therefore the car's location cannot depend on them. I really don't know how to make it easier than this. glopk (talk) 07:18, 20 February 2009 (UTC)


 * The Parade statement does not make clear what the game rules are, but what it does say is, '...the host, ... opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" '


 * There is nothing in the statement of the question itself which says that opening a door to reveal a goat must be taken as a condition but saying the words "Do you want to pick door No. 2?" need not be considered a condition. My question is, what logical process do you use to determine what is a condition and what is not?


 * Martin, I'll try one last time. In the "standard interpretation" of the Parade statement, the opening of a door to reveal a goat (proposition $$H_{ij}$$ in the article) must be taken as a conditioning term for the location of the car ($$C_k$$), because the selection of which door to open is directly affected by the very location of the car: Monty must reveal a goat and not the car, and he must flip a fair coin to decide which door to open when two are available. Therefore we have no freedom left: the rules say that $$H_{i,j}$$ is not independent of $$C_k$$ as a function of the index k, and from this very fact, thanks to Bayes' theorem, it necessarily follows that $$C_k$$ is not independent of $$H_{ij}$$. Any formulation of the "standard interpretation" that asserted the independence of $$C_k$$ from $$H_{ij}$$ would be either self-contradictory or a violation of Bayes' theorem, hence be useless for reasoning toward a solution.


 * On the other hand, within the "standard interpretation", we are free to ignore the words "Do you want to pick door No. 2?", because this sentence is not affected by where the car is, save for the particular door number - which has already been covered above. The offering of a choice to switch only affects the player's gain/loss function, which is the mathematical formulation of what it means to "win" the game, not the player's judgment of where the car may be. Therefore we are free to ignore it while remaining consistent with the interpretation.


 * Now, you are free to allege that the "standard interpretation" is unsatisfactory, and may come up with any number of increasingly baroque variants (subject only to those minimal requirements of consistency that are needed to make the problem well-posed). You may also add as many irrelevant conditioning terms as you wish. By doing so you'll only run the risk to make your version more and more uninteresting, to the point that a certain gray-bearded monk wielding a razor may start taking an interest in you :-) glopk (talk) 05:23, 21 February 2009 (UTC)

Change of perspective by Morgan
Morgan et al say that there are two distinct probabilities that must be summed to get the unconditional probability, they say: P(Ws) = P(Ws | D3) P(D3) + P(Ws | D2) P(D2).

My assertion is that even when the host does not choose a goat door randomly P(Ws | D3) and P(Ws | D2) must be equal. My reasoning is based on this statement, 'Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence'. This agrees with my intuitive understanding off the subject. The important question to ask ourselves is which individual should we consider. The Parade statement clearly refers to the viewpoint of a player, not that of the host or a third party observer. Now, even if the host always opens the leftmost door whenever possible, the player is extremely unlikely to know this so, even though the hosts action is not random, the player does not know this. To the player the action still is random.

What Morgan et al do is formulate the problem from the point of view of a mysterious third party observer, who somehow knows of the hosts tactics. Their answer may be correct from this point of view but it does not answer the question actually asked. Martin Hogbin (talk) 23:25, 19 February 2009 (UTC)

To give an example, suppose somebody tells you that their friend has tossed a fair coin and asks you what are the chances that it will be heads. You could give the fatuous answer that it will be either 1 or 0 depending on which their friend actually got. This is indeed the correct answer from some perspectives, but not from the point of view of the questioner, where the answer is 0.5. Martin Hogbin (talk) 23:48, 19 February 2009 (UTC)


 * The probabilistic analysis needs to take into consideration everything from the problem statement (well, everything that's relevant to the probability). The analysis can then be used to answer the question.  In the case of the Parade version, the question is "is it to your advantage to switch".  The answer is yes (assuming we clarify that the host always offers the chance to switch and always reveals a goat, but say nothing about how the host picks between two goats).  If the question were "what should the player think her chances are of winning by switching" we'd then have to consider whether information presented in the problem is known to the player (which simply adds another conditional problem to address).  The answer to this question, for the Parade version (clarified for the player in the same way as above), is "my chances are something in the range of 0.5 to 1 depending on how the host picks between two goats".  If the problem statement says the host has a preference but says the player doesn't know this preference, then we can ask two different questions: what is the actual chance, and what is the apparent chance.


 * I know you want the solution to be simple. The problem is, it's not.  An unconditional version could be stated as an urn problem: three identical balls in an urn, one with a marking that you can trade for $1000, you withdraw one without looking at it, the host withdraws an unmarked one and shows it to you, what is your chance of having the marked one or do you want to trade for the remaining one in the urn?  I think this version has the simple solution you seek (because the balls are identical neither you or the host can tell the two unmarked ones apart).  If you could find a reliable reference that discusses a variant like this, we could include a discussion of the simpler problem perhaps even before discussing the solution to the MH problem.  But lacking a source, it's simply WP:OR to make up variants (note that the variants currently mentioned in the article are all referenced).  My guess is you won't find such a reference because the math for the actual conditional MH problem is not that difficult.   -- Rick Block (talk) 01:31, 20 February 2009 (UTC)


 * So what exactly do you mean by the probability that the player will win if she switches? I do not mean how do you calculate it but what does the word 'probability' mean?  I am aware of two definitions:


 * 'Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence', which is pretty close to your statement "what should the player think her chances are of winning by switching".


 * The other definition that I am aware of is based on frequency and is, 'The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment'. This is the unconditional probability, which we agree for the problem as stated above this is 2/3.


 * Both the above definitions, by the way, come from the WP article on the subject.


 * I do not understand what definition of probability you (or Morgan) are using. Martin Hogbin (talk) 09:31, 20 February 2009 (UTC)


 * I think we're all talking about the "Modern definition" at Probability theory. -- Rick Block (talk) 19:34, 20 February 2009 (UTC)


 * From what I can understand of the article, it would seem that probability is given no intuitive meaning, it is something that can be calculated according to a certain set of rules. That is how mathematicians like to work and that is fine, but the problem for us comes when you try to answer the question posed with such a probability.  For example using the frequency definition, you could say, 'On average you will win 2 times out of every 3 if you take the swap'.  Using the modern definition you could say, 'Your probability of winning, if you take the swap is 2/3' , but what do you reply when asked, 'So does that mean that it is to my advantage to do so and if so why?'  To give any kind of meaningful reply , you need to suddenly revert to one of the 'classical' definitions. Martin Hogbin (talk) 22:04, 20 February 2009 (UTC)


 * I think that I am beginning to understand the "Modern definition" of probability as described at Probability theory and how it allows Morgan to come to the conclusion that they do. I would like to ask some questions in a new section below. 86.133.179.19 (talk) 12:21, 21 February 2009 (UTC)

Simple play
May be a simple example may help. I suggest the following play. I have writtin down the name of a random chosen British person. You may guess wheter it is a man or a woman.


 * Step 1: What is your guess?

After you have answered, I say: the person comes from Yorkshire.


 * Step 2: You may change your choice. Will you change?

After you have answered, I say: the person is a member of the nursing staff in a hospital.


 * Step 3: You may change your choice. Will you change?

After you have answered, I say: the person's christian name is Billy.


 * Step 4: You may change your choice. Will you change?

See how with every step the possibilities are limited. Notice also that in every step a different probability notion is used. Yet the probabilities in step 1 and step 2 are both 1/2. That why people say the probability has not changed. Numerically it has not changed, but by definition it is a different probability (notion). Nijdam (talk) 11:36, 20 February 2009 (UTC)


 * Are you replying to me Nijdam? Martin Hogbin (talk) 18:17, 20 February 2009 (UTC)

@Martin Hogbin: Yes this example is especially meant for you, but others may profit of it as well of course. Nijdam (talk) 18:42, 20 February 2009 (UTC)


 * Yes, I full understand what conditional means. My question is, how does one decide when formulating a question, what events are conditions?  Intuitively we generally have a some idea.  For example, if in your example above I added that between steps 3 an 4 it started to rain, we would not naturally add that into our formulation as a condition.  However, for almost any event, it is probably possible to contrive a circumstance in which a seemingly unimportant event is, in fact, an important condition of the problem.  My example of the condition that Monty used the word 'pick' was exactly that, the exact wording Monty used when he asks the players if they would like to change doors it is seemingly unimportant, unless he were to use say 'pick another door' when the player had originally chosen a car and 'choose another door' when they had chosen a goat.  In that case it becomes a vitally important condition. Martin Hogbin (talk) 19:11, 20 February 2009 (UTC)


 * My understanding is that anything that gives new information that could be used to revise the probability of a particular outcome should be treated as a condition. How we know that is another matter.  Martin Hogbin (talk) 19:19, 20 February 2009 (UTC)


 * One of my complaints about Morgan is not that they sum the two conditional probabilities of the host opening either unopened door or even that they suggest that this is a better way to do things but that they state that this is the only way to do things, even for the random-goat-door version where it is known not to be necessary. If you add every condition that could conceivably be necessary you would get a very complicated problem. Martin Hogbin (talk) 19:39, 20 February 2009 (UTC)

If we don't get too philosofical, any event that limits the outcomes, forms a condition. That's what the simple play tries to show. I didn't introduce events which are no condition, but it's very easy to think of some. As you said, suppose after you have been informed about Yorkshire and made your choice it starts to rain, and the host tells you so, and asks whether you want to reconsider your choice. Does it matter? No, because the people involved, i.e. the Yorkshire people are still the ones to be considered. No limitations. If the host would inform you that a terrible catastrophy has struck Yorkshire anda lot of people has died, this changes the population, so forms a condition. Is this in any way helpfull? Nijdam (talk) 19:55, 20 February 2009 (UTC)


 * I am not looking for help, I am hoping to convince you that that by insisting we must formulate the problem in a certain way, the Morgan paper only serves to obfuscate the central and notable issue, which is that, even for the vos Savant formulation treated unconditionally, most people get it wrong most of the time. This is the only reason there is a WP article on the Monty hall problem; we do not have articles on the odds of winning other game shows. Martin Hogbin (talk) 20:08, 20 February 2009 (UTC)

More Information => Better Odds
You know how the 'left-most door' variant is used as one method to disqualify any unconditional solution as valid for the so-called conditional problem? I think the reasoning goes that since it has an overall probability of 2/3, but individual plays of the game have varying probabilities, then this conditional problem cannot properly be solved using the unconditional proof. Rick Block, it would be much appreciated if you could leave a brief acknowledgment that this is either a correct interpretation or not.


 * Sort of yes. The variant is used to show the conditional and unconditional problems are different.  In any variant, including variants where it happens to produce the correct numerical result, an unconditional solution is addressing the unconditional problem, not the conditional problem. -- Rick Block (talk) 15:44, 20 February 2009 (UTC)

The following assumes that the contestant knows the method Monty uses. Rick Block, it would again be appreciated if you could leave a brief acknowledgment that this is either a correct interpretation or not.


 * It depends on how the problem is phrased, but in this case I'd say no. The problem statement doesn't say anything that would lead us to believe there's any difference between what the contestant knows and what we know. -- Rick Block (talk) 15:44, 20 February 2009 (UTC)

By my reading, with the 'left-most door' constraint added, there are times when the contestant knows exactly where the car is. Rick Block, one more time, please.


 * If the contestant knows what we know, yes. -- Rick Block (talk) 15:44, 20 February 2009 (UTC)

Here's my question. From nothing more than a logical standpoint, does it seem reasonable that since Monty sometimes gives us additional information about the whereabouts of the goat, that the contestant can improve his overall probability of winning the car from the previously unconditionally proven 2/3? Glkanter (talk) 12:23, 20 February 2009 (UTC)


 * Depending on the door the host opens the odds may go up, but they may go down as well. The conditional answer is 1 / (1 + p) where p is the host's preference (between 0 and 1) for the door he's opened.  If p is 0 (the host really hates the door he's opened and opens it only when forced because the player picked a goat and the car is behind the other door, this is the "leftmost" variant) the player's chance of winning is 1 / (1 + 0) meaning if the player switches she wins.  But if p is 1 (the host opens this door whenever given the chance, i.e. both if the player picks the car and if the player picks a goat and the car is behind the other door), the player's chance of winning is 1 / (1 + 1) which is only 1/2.  If the unconditional odds of winning are 2/3 and there are cases where the conditional odds improve, there must be cases where the conditional odds go down as well.


 * Imagine the stage is set up so the player always stands on the left (on the Door 1 side) and Monty has to stand next to her. If she picks door 1 Monty has to walk across the stage to open one of the doors.  Sometimes he's feeling lazy and opens door 2 (because he gets there first) unless the car is behind it.  This makes his preference for door 3 (p) equal to 0 and his preference for door 2 equal to 1.  If he opens door 2 the player has a 50% chance of winning by switching.  If he opens door 3 the player has a 100% chance of winning.  Other times, he wants to walk as much as possible so he always skips right by door 2 (unless the car is behind door 3 and he has to open door 2).  It doesn't matter whether the contestant knows this or not - in a logical sense the probability of winning by switching is always 1 / (1 + p).


 * The K&R version takes this out of Monty's hands, and specifies his preference is 1/2. This means on the show Monty would have to flip a coin (or something) to decide which door to open if the player initially picks the car (and, to avoid this being a dead giveaway that the player has picked the car he'd have to do this in secret).


 * I assume you haven't done the card experiment yet. Please try it (always keep the ace, but discard the two of hearts if you can - keep track of wins/losses by discard).  It shouldn't take more than 10 or 15 minutes to get enough samples to get a meaningful result.  Across all samples (unconditionally) switching will win about 2/3 of the time.  But conditionally, switching will win about 1/2 (if you've discarded the two of hearts, and this will happen about 2/3 of the time). or 100% (if you've discarded the two of diamonds, which will happen about 1/3 of the time).


 * In a probabilistic sense, more information simply makes things different. Might be better, might be worse, might be the same. -- Rick Block (talk) 15:44, 20 February 2009 (UTC)


 * Thanks, Rick, I appreciate your input.


 * I'm unclear on the 2nd response, and therefore the 3rd response. Can you help me?


 * I'd still like your opinion, and anybody else's as well. Given that, on occaison, Monty tells us (and presumably the contestant) where the car is, does it seem reasonable that the liklihood of picking the car would increase? I'd like to defer the actual proofs for just a little bit, please. Thanks. Glkanter (talk) 18:53, 20 February 2009 (UTC)


 * Regarding the 2nd response, does this relate to the same point Martin is raising above at ? What I'm saying is the analysis uses what is given in the problem statement.  We're not limited to what individuals described in the problem statement may or may not know.  The problem may ask "from the perspective of the host ..." - but it doesn't.  Is this what you're asking about?


 * I presume that the MHP is solved from the contestant's point of view. So, what he is told becomes constraints/premises. So, does he know what Monty's up to? Glkanter (talk) 21:01, 20 February 2009 (UTC)


 * We know the unconditional probability is 2/3. If Monty tells us in some case where the car is, the likelihood of picking the car goes up in this case - but because the unconditional probability is 2/3 there must be a matching case where the likelihood of picking the car goes down.  This is like knowing we have some numbers whose average is 2/3.  If one of those numbers is higher, another one better be lower. -- Rick Block (talk) 20:10, 20 February 2009 (UTC)


 * If I read your preceding paragraph correctly, then you've just said that 'overall' the uncondition solution indeed gives you the conditional solution. But I know you don't mean that.


 * I don't think your 'equilibrium' idea is right. I think with new, useful information, like 'the car is behind that door', the overall probability goes up. Glkanter (talk) 21:01, 20 February 2009 (UTC)


 * Surely we agree any problem is solved from the point of view of someone reading the problem. I think we're probably saying the same thing, which is that what is important is what is given in the problem statement.


 * What I said, and what I've been saying all along, is that the unconditional approach gives you the unconditional answer. The "overall conditional solution" is a very weird phrase, but looking past the specific words probably means exactly the same thing as the unconditional solution.  Just like a set of 5 different numbers has 5 things each with their own individual value and an overall average, a set of conditions each have their own conditional probability and there's an "overall" unconditional probability.  In the MHP there are two conditions (like two numbers contributing to an average), a "host opened door 2" condition and a "host opened door 3" condition.  The unconditional probability combines these (sort of like an average).  The probabilities are 1 / (1 + p) and 1 / (1 + (1-p)).  To combine them you don't just add and divide by 2 (like you would to get the average of two numbers), but you can combine them to get the unconditional probability.  If you do combine them, you get 2/3.


 * I think we're at the point where we're not talking about MHP, but the basics of probability theory. -- Rick Block (talk) 00:48, 21 February 2009 (UTC)

I'll try harder to stay on the MHP. You agree that when Monty shows us where the car is NOT, that the chance of winning increases from 1/2 to 2/3, so why doesn't it follow that when Monty additionally shows us where the car IS ('left-most door' constraint) 1/3 of the time, the chance of winning will increase from 2/3 to roughly 78% [(1/3*1)+(2/3*2/3)]? Since it's conditional, there's got to be a plus sign in the equation, right? Glkanter (talk) 11:28, 21 February 2009 (UTC)


 * Let's back all the way up. It sounds like you are very confused about conditional probability.  Is this a term you're familiar with and do you think you know what it means?  If so, please explain it briefly here. -- Rick Block (talk) 16:01, 21 February 2009 (UTC)


 * Take out the first half of my last sentence, then. I'm just trying to have you explain to me, in English, why the overall odds of selecting a car do not go up when the contestant is given additional useful information. Glkanter (talk) 16:50, 21 February 2009 (UTC)


 * I assume by "overall odds" you mean what we'd expect to be the proportion of players who win across many instances of the game (right?). So, if we do this 6000 times and all players switch, by saying the overall odds of winning by switching are 2/3 we're saying we'd expect (about) 4000 to win.  [agree or disagree?]


 * I think I may have said this before, but the leftmost variant DOES NOT CHANGE these overall odds. If we use the rules in this variant, and do the game 6000 times and all players switch, we'd still expect (about) 4000 to win.  This should sound familiar, but the host always shows a goat, players still have a 2/3 chance of initially picking a goat, and if you switch you get the other thing, so 2/3 of the players who switch will win.


 * What this variant does (actually, any variant does this) is split the 6000 players into two groups depending on what door the host opens. In this variant, the door 3 group is given the additional information that a goat is behind door 3 and that a goat is NOT behind door 2 (because the host always opens door 2 if there's a goat there).  But this information is specific to this group (we'll talk about the door 2 group in a moment).  This group, the door 3 group, gets additional useful information - i.e. that the car must be behind door 2 so switching wins (100%, not 78%).  Per the paragraph above the overall odds haven't changed, just the odds for this group.  Does this make sense?


 * Before talking about the door 2 group I have a couple of questions for you. The chance there's a goat behind door 2 is 2/3.  [agree or disagree]


 * Thinking only about the cases where there's a goat behind door 2, what are the chances that the car is behind door 1? Is this 50/50?   [agree or disagree]


 * We need to agree about the above before we can talk about the door 2 group. -- Rick Block (talk) 18:10, 21 February 2009 (UTC)

Yes, the door 3 group is 100%. And the door 2 group is 2/3. This is based on the unconditional MHP. Monty's actions can't REDUCE the contestant's 2/3 probability of winning by switching. You already know my feelings about time travel.

And they occur 1/3 vs 2/3 of the time. So, 1*1/3 + 2/3*2/3 = 78%. No, I do not expect 4,000 to win. I 'expect' 4,667. (Is 4,000 from Morgan, or OR?). That's my point. More info = greater chances. When Monty opens a random door with a goat, the odds go from 1/2 to 2/3. The overall wins didn't stay at 3,000. This is the same thing. More info = more wins.

Here's my original question: From nothing more than a logical standpoint, does it seem reasonable that since Monty sometimes gives us additional information about the whereabouts of the goat, that the contestant can improve his overall probability of winning the car from the previously unconditionally proven 2/3? Glkanter (talk) 12:08, 22 February 2009 (UTC)


 * From a logical standpoint, lets talk for a moment about the overall odds of getting bit by a shark in a year. This has some number, say .001% (I'm making this number up).  If we divide people into those who NEVER go into the ocean and those who do, what happens?  The ones who go into the ocean have a slightly higher chance and the ones who don't have a 0% chance (right?).  This is the essence of conditional probability.  We have two groups making one larger group.  The odds for the larger group (the unconditional probability) don't have to be the same as the odds for each of the smaller groups (which each have their own conditional probability).  If the odds for one of the smaller groups is higher, the odds for one of the other smaller groups must be lower.


 * In the "leftmost variant" I think you agree the chance a goat is behind door 2 is 2/3, and hence the chance of a player being in the door 2 group is 2/3, and the door 3 group 1/3.


 * We agree the players in the door 3 group have a 100% chance of winning. This is the conditional probability of winning for players who see the host open door 3 - i.e. the probability of winning considering only players in this group.  It is not the same as 2/3.


 * Similarly the door 2 group has a conditional chance of winning. You agree if we only think about the players in the door 3 group they have a conditional chance of winning that is not 2/3.  Why are you so reluctant to believe the door 2 group might have a different conditional chance of winning as well?  We're not going back in time and changing the probability for this group, we're just splitting the whole group (by the one act of opening a door) into two subgroups in a way where the chances for one group are higher - but this means the chances for the other group have to be lower.


 * You didn't answer my last question. Thinking only about the 2/3 times there's a goat behind door 2 (before the player picks, before the host opens a door), what is the chance the car is behind door 1 vs door 3?  -- Rick Block (talk) 18:11, 22 February 2009 (UTC)


 * It's my thread. You employ a very effective technique for forestalling improvements to the article. Rather than engage in discussion, which I'm trying to do in a civil manner, you go off on long-winded tangents. You're like Jello. It's impossible to go point by point on any issues, large or small.


 * Here's my original question: From nothing more than a logical standpoint, does it seem reasonable that since Monty sometimes gives us additional information about the whereabouts of the goat, that the contestant can improve his overall probability of winning the car from the previously unconditionally proven 2/3? Glkanter (talk)


 * I'm not forestalling anything, I'm responding. The direct answer to your question is yes.  I've said that already and we already agree about that.  Any other direct questions you'd like to ask? -- Rick Block (talk) 18:46, 22 February 2009 (UTC)


 * Thanks. I honestly was not clear that we agreed on that point, already. Sorry. I expect 4,667 wins. You expect more than 4,000. How can the probability still be stated as 2/3? Glkanter (talk) 19:24, 22 February 2009 (UTC)


 * I'm sorry - I misinterpreted your statement. I agree that if Monty gives us additional information it is possible to improve the overall odds of winning by switching.   I don't agree that this is all that's happening in this variant.  In this variant I expect 4,000 wins by switching.  In the terms you're using I'm saying in this variant Monty gives some players more information and some less, changing the probability of each group.  Rather than explain I'll let you ask direct questions. -- Rick Block (talk) 19:43, 22 February 2009 (UTC)

In the unconditional problem, the constant will always switch. In the 'left-most goat door' variant when Monty reveals door 3, the contestant is best served by switching (with 100% certainty of winning the car), right? So whether Monty has let the cat out of the bag or not, the contestant doesn't change his action, he will switch? Glkanter (talk) 20:05, 22 February 2009 (UTC)


 * Like any other variant, the "leftmost door variant" has unconditional and conditional answers, so I'm not exactly sure what you're asking. Unconditionally, in this and any other variant (where the host always opens a door showing a goat and offers the switch) 2/3 of contestants who switch should win.  Conditionally, the contestant is no worse off switching, but the conditional odds are between .5 and 1 (so it is not necessarily an advantage to switch - but it's never a disadvantage). -- Rick Block (talk) 20:21, 22 February 2009 (UTC)


 * So, Morgan demonstrates that the unconditional solution is not sufficiently rigorous by:
 * Adding a constraint to the original problem - Monty always reveals the left-most door with a goat
 * Which doesn't change the contestant's best strategy - he still always switches
 * Which relies on the unconditional MHP solution - otherwise, you wouldn't start with 4,000 wins and subtract 2,000
 * Then calls this a co-incidence - even though the contestant still switches every time => 4,000 wins
 * And concludes that the unconditional proof lacks sufficient rigor


 * I don't agree with their techniques, and I don't agree with their conclusion. I could do this with any problem, as long as the contestant's behaviour stays the same. Now, if Monty's action caused the contestant to NOT switch sometimes, maybe you'd have something. But still, that would be a new constraint, making it a different problem.
 * What was their point, exactly?
 * What came of Morgan's paper? Did they invent a new theory of probability? Glkanter (talk) 08:30, 23 February 2009 (UTC)


 * Morgan et al., and Gillman as well (not one of the et al.s), analyzed the Parade version of the problem. In this version the host choice when the player picks the car is not specified.  They BOTH say this is a conditional, not unconditional problem.  They BOTH use the "leftmost goat" variant not as an additional constraint, but as one example of a host strategy that influences the conditional probability of winning by switching.  They BOTH conclude (and I assume they arrived at this independently) the chance of winning by switching is 1 / (1 + p).  They BOTH say this means the chance of winning by switching is between 0.5 and 1.  They BOTH say this is 2/3 if p is 1/2.  They BOTH support the overall conclusion that you're no worse off switching.


 * Their point is that as usually stated the MHP is a conditional problem, and that the unconditional solution is insufficient. They BOTH are sources that are cited by many (dozens) of subsequent academic papers.  They BOTH use standard probability theory, which distinguishes unconditional and conditional probabilities.


 * You say you don't agree with their techniques or their conclusion. You said once you're "not arrogant enough to argue math with a math PHD".  You're arguing here with not just someone on Wikipedia who claims to have a math PhD but two independently published math papers written by math professors at major universities.  Arrogant doesn't even come close to describing it. -- Rick Block (talk) 14:42, 23 February 2009 (UTC)

Decision Tree
If you draw a decision tree for the problem, you will see that the final probability is actually 0.5. This is caused by the fact that the rules state that there MUST BE a car behind one of the final two doors.

We can assume that there are two main branches to the tree, one in which the correct door is initially chosen, and one in which an incorrect door is chosen (it doesnt matter which, just that its not correct).

1. In the case that the player has chosen the winning door, only one (1) choice allows the selection of this door. Then the host can choose any of the two remaining doors, so he has 2 choices (or N-1, for N initial doors). Finally the player again has two (2) choices, ie. to switch or stay. This gives a final outcome count of 1X2X2 = 4. Of these four final outcomes, two are switches and two are stays. Two are winners and two are losers. In this case, because the initial choice was correct, staying wins. So, assuming he picked a winner right from the start, he has a 50/50 chance.

2. In the second scenario, where the player has picked a loser, there are N-1 possible choices he could have made. For the current example that means he had a choice of two (2) doors. Here is the clincher: Since the player picked a loser, the Host MUST pick the winning door (according to the rules). This means that for each of the players two possible choices above, the host can only make one (1) choice. This cuts down this side of the decision tree to the same size as the side where the initial choice was a winner. Finally the player once again has two (2) choices to make: switch or stay. So here we get 2X1X2=4 once again. Also, once again, there are two out of four results where the player switched and two where the played stayed. Since the initial decision here was wrong, it means that switching wins. So on this side of the tree, the player also has a 50/50 chance.

So on either side of the tree, there are four results of which two won. On the left staying won and on the right switching won.

This gives a final result of staying winning 2/4 times and switching winning 2/4 times. Therefor, regardless of the players choices, there is a 50/50 chance of winning.

Draw out the tree for yourself and see... —Preceding unsigned comment added by 66.8.57.2 (talk) 10:19, 20 February 2009 (UTC)


 * Two problems with this: Firstly, you seem to have assumed that whenever there are two branches of the tree, they must have equal probability, which is incorrect in the case of the player's initial door selection - that has a 2/3 or (N-1)/N chance of being wrong. Secondly, you haven't mentioned a reliable source that publishes this line of reasoning, which means that we can't really include it in the article, because of our verifiability and nor original research policies.  S HEFFIELD S TEEL TALK 14:01, 20 February 2009 (UTC)

I apologise, I left out one very important fact: The choice that the host is making is not which door to open, its which door to leave closed. This is made clear in the 1000000 door version in the main article. This is why he only has one (1) choice when the player has picked a losing door (regardless of the initial number of doors). The rules state that one of the final two closed doors, MUST have a car behind it. —Preceding unsigned comment added by 66.8.57.2 (talk) 10:31, 20 February 2009 (UTC)


 * Please look at the decision tree (referenced to a reliable source) in the Solution section of the article. -- Rick Block (talk) 14:24, 20 February 2009 (UTC)

Modern definition of probability
I understand the Bayesian and frequency definitions of probability, which I claim would not result in the same conclusions for the MHP as the modern definition, see above. I am new to the modern definition so perhaps someone who knows could answer my questions below:

In the modern definition of probability we refer to some, possible notional, point in time. Everything that happens before that time has a probability of 1 (or 0 if it has not happened); these are the givens. Rather than use the MHP to help me understand how things work, perhaps I could ask some much simpler questions.

I take a fair coin from my pocket and place it face up on the table in front of me, keeping it covered with my hand. What is the probability that it is a head? Are these answers correct?

Bayesian, from my perspective - 1/2 Frequency definition - 1/2 Modern definition - either 0 or 1, we cannot say which.

Martin Hogbin (talk) 12:33, 21 February 2009 (UTC)

As exact wording is, no doubt, important, what about these questions:

I take a fair coin from my pocket and place it face up on the table in front of me, keeping it covered with my hand. What is the probability that it will prove to be head?

I take a fair coin from my pocket and place it face up on the table in front of me, keeping it covered with my hand. What is the probability that I will see a head when I uncover it? Martin Hogbin (talk) 12:38, 21 February 2009 (UTC)
 * Did you have a fair coin in your pocket, and in what way do you place it face up on the table?Nijdam (talk) 13:35, 21 February 2009 (UTC)

Yes, I did have a fair coin, and I simply took it out of my pocket without looking at it, in the normal way that one does. You may take the method as random. Martin Hogbin (talk) 14:05, 21 February 2009 (UTC)


 * If you didn't look at it, how do you know it is a fair coin? Nijdam (talk) 15:19, 21 February 2009 (UTC)


 * It doesn't have to be fair: it has to have two visually identifiable sides that are tactiley indistinguishable. The indistinguishability could be established in separate tests. Brews ohare (talk) 16:10, 21 February 2009 (UTC)


 * May be it doesn't even has to be a coin, as long as it has two tactiley sides. And why should they be indistinguishable? Nijdam (talk) 20:04, 21 February 2009 (UTC)

Martin - you're seriously barking up the wrong tree here (you may even be in the wrong forest). The technical differences between the definitions of probability have no impact on this problem. We're all talking about probability in the sense of a number between 0 and 1 reflecting the chance of some expected outcome which should adhere to the law of large numbers. Asking "what do you mean by probability" here is sort of like asking "what do you mean by number" in the context of a counting problem (Johnny has 3 apples, ...). There is absolutely no need for any confusion here. -- Rick Block (talk) 18:42, 21 February 2009 (UTC)
 * @Rick: don't bother anymore.Nijdam (talk) 20:07, 21 February 2009 (UTC)
 * I'm very, very, very reluctant to suspend WP:AGF. -- Rick Block (talk) 20:30, 21 February 2009 (UTC)


 * Thank you Rick. I can assure you that the questions are asked in good faith.  Could you possible give me the answers please.  I am not sure why some people are getting so hot under the collar here, these are very important points.  There is no point in saying the probability is 2/3 if we cannot ascribe any meaning to that statement.  According to the frequency and Bayesian notions of probability we can give an understandable meaning to that statement but not, it would seem, for the modern definition.  If you do not want to answer, for whatever reason, perhaps you could tell me where I can get answers to my questions. Martin Hogbin (talk) 22:32, 21 February 2009 (UTC)


 * Martin - I think the point Nijdam is making (with his arguably flippant responses) is that the line of questioning you're pursuing is tiresomely irrelevant with respect to the MHP. Yes, the nature of probability can be debated (see Probability interpretations), however Bayesian, and modern, and frequentist interpretations are all essentially specializations of classical probability theory.  All would say the probability of winning by switching in the MHP is 1 / (1 + p) where p is the host's preference for the door he's opened (and in the K&R version, this is 1/2, so in this version the probability is 2/3).  We don't need to ask which technical meaning of probability we're talking about here, because for this problem, the distinctions between these formalisms don't matter.  -- Rick Block (talk) 23:43, 21 February 2009 (UTC)


 * Rick thanks for you patience. I also do appreciate your position, I am a physicist and have spent much time on the physics newsgroups and on WP arguing with people who do not appreciate the many years of thinking by the worlds best physicists on subjects like relativity and believe that they can prove it wrong with a simple argument that no one else has ever thought of before.


 * So, I would like to continue the discussion, which might get a bit philosophical and where I may try and box you into a corner to try and clarify a particular point. As I say to Nijdam below, I get the feeling, which I a groping to put into words properly, that something is wrong, or at least arbitrary, in the Morgan analysis of the problem.  I would not mind this if they were not so adamant that theirs is the only way to tackle the problem. Martin Hogbin (talk) 10:41, 22 February 2009 (UTC)

@Martin: I think Rick is hitting the nail wherever it should be hit, and no, I'm mainly getting cold, not only under my collar, and I would be willing to talk about notions of probability, BUT not mixed with this discussion.Nijdam (talk) 09:42, 22 February 2009 (UTC)

@Nijdam, I am not demanding that any particular person should reply. Rick has been kind enough to reply to my genuine, but possibly misguided, questions. You are welcome to reply if you wish, or not if you do not. It would seem that both you and Rick are more knowledgeable about the theory of probability that I am but I still think that there are problems with Morgan's approach to the MHP, however, I can be persuaded by logical argument. You are welcome to contribute if you wish. I think that this page is an appropriate place for this discussion, it is certainly better that using the article itself as a mouthpiece for opinion. Martin Hogbin (talk) 10:41, 22 February 2009 (UTC)

@Martin, sorry, didn't mean to offend you. But I think the problem originally was not meant to take into account all kind of more or less philosophical aspects. And it isn't much help for the planned revision. Yet it is of course of interest what kind of alternatives one can think of. I've seen a paper in which at least some of your questions are treated; if I refind it I'll let you know. Nijdam (talk) 11:41, 22 February 2009 (UTC)

The paper is: Marc C. Steinbach, Autos, Ziegen und Streithahne, I found it on the internet.Nijdam (talk) 15:52, 22 February 2009 (UTC)

Martin - can we skip to the chase here? My impression is you're looking for a probabilistic formalism supporting an unconditional interpretation of the MHP. It is a well known, very mainstream conditional problem in simple probability theory, so even if you can convince yourself it's reasonable to interpret it as an unconditional problem in some formalism (and, to keep out of trouble with WP:OR, find a suitable reference that talks about MHP in this form) you'll be in WP:UNDUE territory meaning this interpretation shouldn't get much more than a footnote in the article. You're certainly free to pursue this, but the more esoteric and less mainstream you go the less it should be discussed in the article. -- Rick Block (talk) 20:23, 23 February 2009 (UTC)

Excel simulation of difference between "random goat" and "leftmost goat" variants
I've written a little simulation in Excel that randomly picks a car location (between 1,2, and 3), and for each pick simulates both a "random goat" host strategy and a "leftmost goat" host strategy (assuming the player initially picks door 1). I can run this over and over. The results don't stray too far from the following:

The top section of this table says how many times the simulation was run and the total number of times the car was behind each door. If the car is behind door 1 switching loses, and if the car is behind either of the other doors switching wins.

The next section shows the simluated results of a host following the behavior specified in the Krauss and Wang statement of the problem, i.e. always shows a goat, but picks between two goats randomly if given the chance. The row labeled Random host opens door 2 says this host opened door 2 296 times (out of the 600) and of these 296 times the car was behind door 3 195 times. For this simulation, the conditional probability of winning by switching (given this host and given the host opened door 2) is 65.88%.

The next section shows the results using the same initial car location of a host following the "leftmost variant" behavior, i.e. always shows a goat, but always opens the door 2 if possible. The row labeled Leftmost host opens door 2 says this host opened door 2 401 times and of these 401 times the car was behind door 3 195 times. For this simulation, the conditional probability of winning by switching (given this host and given the host opened door 2) is 48.63%.

Some things to note about this simulation:
 * There is only one set of 600 cases. Both host strategies are played against the same cases.
 * The "overall odds" of winning by switching are exactly the same for both hosts. In this particular simulation, this is 65.67%.
 * If the car is behind door 2 (which it was 199 times in this simulation), both hosts open door 3 and switching wins.
 * If the car is behind door 3 (195 times), both hosts open door 2 and switching wins.

The only difference between these hosts is what happens if the car is behind door 1. The "random goat" host opens door 2 about half the time in this case (101 out of 206 times). The "leftmost goat" host opens door 2 every time the car is behind door 1.

If the player does not know which host she's encountered, after the host has opened a door she doesn't know her (conditional) probability of winning by switching. It could be as low as 50% or as high as 100%. If she knows she's encountered a K&R host, the probability is 2/3.

I can make the Excel file available to anyone who might be interested in verifying the formulas are correct. -- Rick Block (talk) 22:58, 22 February 2009 (UTC)

"Choice" vs. "Guess"
Shortly after Marilyn vos Savant wrote in Parade magazine on this topic, many others republished it themselves and it was via one of the other articles I first read this problem. The article I read had the problem framed as the odds of a "guess" rather than the odds of a "choice". I submit to this talk page that when the problem is framed as "guess", the odds are 50/50 based on the fact that a "guess" and an "educated guess" are not the same thing. If one is told they must honestly guess, if they are honest, they will not take any information derived from the host's action into account. And without the information derived from the removal of one door from the pool of choices, one can't calculate that the odds of switching are better than the odds of staying. Suffice it to say, I am confident that an accurate review of the history of the propagation of this conundrum throughout the common discourse will find that many times this problem has been sneakily misframed from a math problem to a word problem via the word "guess". If one is actually "guessing", that is blindly guessing, the odds that the information one is using will help you find the car is 50/50 - because your information is not enhanced by that gleaned from the host's action. But, when you enhance your information by taking the host's actions into account, your odds of finding the car do increase by switching. But it's the accuracy of your information which increases, not your car/door ratio. When you have two doors and one car left, you are left with 1 car / 2 doors or a 1/2 chance that the car is behind a door. However, your information about which door the car is behind improves when you take the hosts actions into account. Therefore, this problem fails if you are not allowed to take that information into account. Hence, if you are asked to "guess", the question rises and falls on how you define guess, "blind" or "educated". Interestingly enough, over at the Gambler's fallacy article, a similar question about odds is being raised - that of "absolute odds" vs. "localized odds derived from current information". Anyone interested in absolute vs. localized might want to see the talk page over there as well. It's my contention that localized odds change when information is added and I feel that this premise, that of localized odds changing with information, contradicts the premise of the Gambler's fallacy article. I feel that both this Monty Hall problem problem and the Gambler's fallacy problem are in essence "overlay" problems, in that then the odds of something occurring and the odds of you detecting it are not the same thing. In essence, the arguments occur because the parties who disagree are arguing about the base occurrence vs the overlay of improved information, i.e.; the absolute vs the localized. As the Monty Hall article proves, localized odds of detection improve based on localized information, but the exact opposite is being argued by some over at the Gambler's fallacy article. 216.153.214.89 (talk) 14:11, 24 February 2009 (UTC)
 * Well, of blindly guessing is the same as 50/50 you're by definition right. So, why bother?Nijdam (talk) 16:33, 24 February 2009 (UTC)


 * There's a misunderstanding in the above. If the scenario follows the usual rules (including that the host picks between two goats evenly), then at the point the player is asked whether to switch the odds are split 1/3 player door, and 2/3 other door whether the player knows it or not.   It's not the player's knowledge that makes these the odds, it's that the host opened the door following the prescribed rules.  There's two doors, and a car and a goat behind them, but the goat is twice as likely to be behind the player's door than the other one.  If the player doesn't know this and simply guesses, the player has a 50-50 chance of ending up with the car but this doesn't mean the probability of the car being behind the player's door changes from 1/3 to 1/2. -- Rick Block (talk) 00:24, 25 February 2009 (UTC)

The Monty Hall problem is a trick question. The odds of finding the car can only be re-calculated to recognize the greater chances gleaned by switching if the contestant takes into account the fact that the removed door does not have the car behind it. Without that certainty, there can be no improvement in the calculated odds of a winning selection. That's why when the requirement is to "guess", the honest and correct answer is 50/50. The calculated odds of detection only improve if one is allowed to switch away from a certain fail, that is the removed door - which is a known loser. It's this knowledge, that the removed door is a known loser, which allows the switcher to be confident that the switch improves his odds, which it does. However, if a 2nd person, without any knowledge of a door having been removed, were to choose between the 2 remaining doors, his odds of detection are the same percentage as the incidence of occurrence, 1 car / 2 doors or 50%. You can only improve your odds by gleaning information from the host's action. And this information, if used, violates a true "guess" transforming it to an "educated guess" which is not the same thing. There are two data sets at work here: a) the statistical incidence of cars per available doors and b) the likelyhood of choosing the right door. The car per door ratio is only divorced from the likelhood of finding it when additional information is gleaned from the host and the guess is no longer blind. In this game, as played, the 1st guess is always blind, the 2nd guess is educated. Those players who insist that the odds of a winning selction remain at 50/50 are under the mistaken impression that the 2nd choice must also be a blind guess. And, as I stated in my comments, one of the reasons for this is that some who published the puzzle specifically used the term "guess". I first read this problem in Bostonia Magazine (now defunct) back in 1991 and the author of that article (a MIT professor) specifically used the word "guess". That word, as I say, is the linchpin in why people misunderstand this. At the time I read the Bostonia article, it became clear to me that the author (Massimo Piattelli-Palmarini) was prevaricating in his presentation of the puzzles so as to prompt feedback for a book. I was one of about 20 letter writers that the author replied to via the magazine and was solicited for further rebuttal. My responses took me a 16-hour jam session to compose and I couried them over with less than 1 hour to spare before the cut-off time. It was after the author never replied to the rebuttals (not to my knowledge at least) that I realized he was gathering material for a book. If you could get your hands on a copy of the magazine article (I've long ago thrown mine out), you'd see how sneaky some people are with these types of questions. Here's a few of Massimo's brain teasers which I remember, did debunk, but never heard back from (not in their original order, paraphrased from memory):
 * 1) Using an ELISA test to make a particular detection, but the stated premise of the problem was impossible. He gave us a testing accuracy of 80%, but an incidence of occurence of 5%. The simple fact is that a test that inaccurate can't be detect for certain the actual incidence of occurence without more iterations than an ELISA test performs. Using his starting data, the question regarding false positives could not be answered as posed.
 * 2) Massimo used Bayes' theorem the "prove" that a witness who stated the color of a car must be most likely mistaken because he said "green" when 95% of the cars are blue, so hocus-pocus, the witness is wrong. This of course is pure BS because the correct thing to measure is the visual acuity and color recognition abilities of the witness. Using same distance and lighting, you run 19 blue and 1 green car past him in varying order 25 times. If he says "green" when it is in fact green 23 or more times, he's 90% (or better) accurate and a good witness.
 * 3) This was the Monty Hall problem. At the time, an acquaintence of mine and I discussed it and he happened to be a math expert who flow-charted it out on a spreadsheet (basically the same as the block diagram in the article). It was reviewing his chart that helped me see the trick nature of the problem as I found it, that being "guess". However, if you read the original article by Marilyn, you'll see that she shrewdly omits the word "guess". Others, by accident or design, including Massimo, did not.

Here's a google search on the title of the original Bostonia article. I know of no copy at any library near me. If you can get a .pdf or other copy of it, I'll give you my gmail address. I'd like to read it again. 216.153.214.89 (talk) 04:49, 25 February 2009 (UTC)


 * I'm not sure whether we're agreeing or not. What I'm saying is that if we follow the usual rules, then when the host opens a door the probabilities are that the car is behind the player's door with a 1/3 chance and behind the other door with a 2/3 chance.  These chances reflect the statistical incidence of cars per door (as you put it).  Players who ignore this and then make a 50-50 guess whether to switch have a 50-50 chance of ending up with car, but of the ones who choose to switch roughly 2/3 will win the car vs. roughly 1/3 who those who choose not to switch.  Because you're randomly picking which group to be in, your overall odds of winning are (1/2)(2/3) + (1/2)(1/6) which is 1/2.  Anytime you make a random choice between two things, you have a 50/50 chance of picking either one but your random choice is making this happen, not the relative probability of what you're choosing between.  (1/2)p + (1/2)(1-p) is 1/2 for all p. -- Rick Block (talk) 09:00, 25 February 2009 (UTC)

Rick - do you agree that there are two data sets involved, a) the set of doors per car and b) the set of information regarding how to best find the car? If you do, then you will see my point that the Monty Hall problem is a trick question. Brain-teaser fans recognize that "b" is the data set which contains the answer they seek, while ordinary people count the cars and divide by doors. The problem is designed to trick people into answering with the door / car ratio which indeed starts at 1/3 and does change to 1/2. But the door / car ratio is not the same as the odds of locating the car, hence two data sets. That's why if the question is framed with the word "guess" it's a misleading question due to the variance between "blind guess" and "educated guess". Math has strict standards with no deviations. A math formula is always what it is. There are no analogs in formula composition, only in variables and data. All formulas are binary propositions - at any given time, a formula as stated is what it is. The problem with this type of puzzle is that it purports to "prove" that people misread odds or misunderstand math when in fact the "gotcha" is verbal - the puzzle framer either casually obfuscates the formula by using fuzzy English or flat-out intentionally misleads the reader by using words like "guess" in a trick manner. In short, when there are two doors left, the odds of the car being behind any given door is 1/2 or 50%, but the odds of finding it rise when you count the 1/3 removal and reallocate your search pattern in light of that information. It's the reallocation of your search pattern that makes switching work better than staying. It's your odds of finding the car which rise, not the odds of it being there. 216.153.214.89 (talk) 17:51, 25 February 2009 (UTC)


 * I agree we can talk about the two data sets you're distinguishing but I think you've got the probabilities involved exactly backwards. What I'm saying is that in the usual interpretation, after the host has opened a door, the actual  probabilities of where the car is located (not the probability of winning the car according to some method of determining whether to switch) are 1/3 player door and 2/3 other unopened door.  This is not just some trick question depending on the player's perception.  When the host opens a door the original situation (where the car is behind each of the unopened doors with equal probability) splits into two subcases where (in each) the conditional probability of the car being behind the remaining two closed doors is no longer equal.  Look at the figure at Monty Hall problem.  The third row from the bottom lists the actual (no trickery) total probabilities.  If the player chooses door 1 and the host has opens door 3, the total probabilities in this case are 1/6 door 1 and 1/3 door 2 - meaning the conditional probabilities are 1/3 door 1 and 2/3 door 2.  If you're standing in front of these two doors and blindly guess whether to switch you'll end up with the car only half the time, but the odds really are 1/3 door 1 and 2/3 door 2. -- Rick Block (talk) 19:09, 25 February 2009 (UTC)

Rick - this is 216 answering you from my laptop while away from my desktop pc. The word "probability" is clouding the discussion here. Think along the lines of insurance risk and use the term "incidence of occurence". It's verbally redundant, but incidence has multiple uses and I am referring to "occurence". Think about it for a minute: If no doors are removed the incidence of cars per doors is fixed at 1/3. Incidence is always a ratio between the occurence and the context. Five cows per acre, two cars per garage, 2 chickens in every pot. Incedence is an actuarial fact which is measured and is always uniformly spread out over all the context (possibilities). After a door is removed, how many possible locations remain? Two. How many cars? One. One car, two doors, 1/2, 50% occurence aka incidence. The word probability is pregnant with the tinge of "likely" which is not the same as possible. The possibilities of physical location never change except uniformly over the entire available context, but the probability of detection does vary based on the formula used. When a door is removed as per the rules, one ceases to use the simple car/doors formula and instead divides the total doors into 2 groups. But you can only conceive that the doors are in 2 groups AFTER a door is removed. Then you have two groups of doors - one group, the removed group definately is impossible to have the car. This impossibility allows you to change the formula to exclude that door from the search set (likely). There are two sets of data: One for possible, one for likely. The "likely" set is subdivided into search groups after a door is removed - the "searchable" and the "not searchable". When you exclude the "not searchable" group, you can extrapolate a divergence between the values of the original choice and the switch. But what you are measuring is the value of the choice, not the incidence of cars (ocurrence ratio) 66.236.180.1 (talk) 01:23, 26 February 2009 (UTC) (aka 216.153.214.89 (talk) - via my laptop)
 * I'm thinking of probability in the sense of what we'd expect the proportions to be over many trials. Lets say we do this 6000 times.  We'd expect the car to be behind door 1 and door 2 and door 3 about the same number of times, i.e. 2000 times each.  The way the rules work, if all 6000 players initially pick door 1 the host opens door 2 about 3000 times and door 3 about 3000 times.  When the player is asked to switch she's in one of two groups - the door 2 group or the door 3 group and roughly half of the players end up in each group.  In the door 3 group (when the host opens door 3) we have ALL of the cases where the car is behind door 2 (roughly 2000) plus roughly half of the cases where the car is behind door 1 (roughly 1000) - the other roughly 1000 cases where the car is behind door 1 are in the other group (the door 2 group).  In the 3000 cases where the host has opened door 3 the 3000 cars are not evenly distributed behind the 2 doors.  The "occurrence ratio" is 1/3 behind door 1 and 2/3 behind door 2.  Incidence is NOT always uniformly distributed.  In this case it's not 50/50.  If you do this 6000 times, stop the game at the point player is asked to switch, and bring in a person off the street and have them switch, roughly 4000 will win (not 3000).-- Rick Block (talk) 03:43, 26 February 2009 (UTC)

I agree with what you say except that you havn't made the connection that it's your knowledge which changes, not the distribution formula. The cars are always distributed as car / doors. That never changes. Only the accuracy of your knowledge changes. Your knowledge becomes more accurate after you determine that one of the doors becomes excluded from consideration. As an alternative, let's look at the opposite: Let's say there's three doors, one car and you add another door. Provided that you are alllowed to track the original 3 doors, so you know which ones they are, does the addition of another door reduce your "odds" when you switch? The answer is no, because since you know which are the original doors, you exclude the new door from consideration. So the "odds" of you finding the car stays at 1/3. However, the distribution ratio changes from 1/3 to 1/4. And if adding a door changes the distribution ratio from 1/3 to 1/4, then subtracting a door changes the ratio from 1/3 to 1/2. The distribution ratio (the "incidence") is always the same formula: car / available locations. It's what you know about the distribution that aids your selection, not the actual ratio of the distribution itself. It's the exact principle behind Minesweeper (computer game). In that game, numbers appear on blocks and those numbers supply information about the surrounding blocks. The distribution ratio of the bombs is always bombs/unchosen blocks. But, armed with the information supplied by the numbers you can deduce where not to go and by process of elimination, arrive at the only possible locations where you can go. The only reason why the Monty Hall game is not as precise is because you are not able to gather any additional information beyond the exclusion of 1 door. It's the same principal in operation - as information increases, the accuracy of your selction can be refined. In the Monty Hall game, the improvement in the "odds" is measured across the information, not the distribution ratio formula. Two data sets: a) distribution ratio b) your knowledge. In any discovery situation, the acquired knowledge is an overlay which rests on the underlying facts. As the knowledge becomes more accurate, the ability to navigate around underlying facts to reach a desired conclusion improves. Sort of like mapping unknown territory. The territory is what it is. There either is a pond over the next hill or there is not. What you know about that doesn't change it. But if there's 3 hills and 1 pond and just as you begin your search (you have chosen hill #1 to search 1st) a reliable local informs you it's not behind hill #2, then of course, based on the MH math, switching to #3 is better. But think about this, what if you have chosen #1 and you are informed reliably that it's not #1, what are your "odds" then? In this instance, your odds of coreect selection exactly correlate to the distribution ratio 1/2. If your choice can possibly affect the distribution ratio, why doesn't your choice here do that? Because that ratio is independant of your knowledge - two data sets, each independant, sometimes they correlate, sometimes they don't. Why? Because they are distinct sets. 216.153.214.89 (talk) 05:38, 26 February 2009 (UTC)


 * What you're missing is that there's also a "best analysis considering all given information" probability, which may or may not be the same as the n/m distribution ratio you're talking about. There are 3 different things - the "best analysis" probability, your uniform distribution ratio, and the player's knowledge.  I'm saying the Monty Hall problem is more about the first two than anything involving the third.  If all we know is that n things are distributed randomly over m locations, then the ratio n/m is the same as the best analysis probability.  If, for example, it's given that a car has been randomly placed behind 3 doors the distribution ratio is 1/3 and the best analysis probability of the car being behind each door is also 1/3.  If we now add another door the distribution ratio is 1/4 but we know based on what's given that the probability of the car being behind the original 3 doors is still 1/3 and the probability of it being behind the 4th door is 0.  This is not a "player knowledge" thing, but more like the "actual" probability.  Another way to add a 4th door is to let the user pick an initial door (say door 1), add another door, and now if the car was behind doors 2 or 3, pick a new random spot for the car behind any of doors 2,3 or 4.  By doing this, we now have a more complicated scenario.  The distribution ratio is still 1/4, but our "best analysis" probability would say door 1 still has a 1/3 probability and the remaining 2/3 is evenly split between the other 3 doors, so these doors each have a probability of 2/9 (1/3 + 2/9 + 2/9 + 2/9 = 1).  These are the "actual" probabilities given the description of what we've done.  What this means is if we do this 9000 times we'd expect the car to be under door 1 roughly 3000 times and behind each of the other three doors roughly 2000 times.


 * Similarly, in the MHP, the "best analysis" probability says before the host opens a door the probabilities are 1/3, 1/3, 1/3 but after the host opens a door (given the usual rules and that the host opens door 3) we're only looking at half the total cases and the probabilities are 1/3, 2/3, 0 NOT 1/2, 1/2, 0 (and, if we look at the other half of the total cases the probabilities are 1/3, 0, 2/3 NOT 1/2, 0, 1/2). Doors 1 & 2 (or 1 & 3), which had equal probabilities before the host opens a door, have different probabilities after the host opens a door.  The combination of your choice and the host opening a door changes the probability distribution between these doors (which I think is what you really mean by distribution ratio).   This is the veridical paradox of the MHP.  -- Rick Block (talk) 01:22, 27 February 2009 (UTC)

Rick you are sooooo close. When you say "probability distribution", ask yourself "probability of what?" Finding the car or the car being "there"? As doors are removed (or added), the "theres" available for the car to "be" behind changes but the formula of distribution is always car / doors. The only reason why it appears that this changes is because you are conflating two data sets. What you know about where the car is and where it can possibly be are simply not in the same data set. Don't you see? Even in your last reply you are talking about "player knowledge" and "best analysis". Those are information which you develop. Information you develop is always in a distinct data set from the underlying facts you are studying. The distribution ratio is a distinct fact from the discovery odds. Something is either true or it's not. Whether you discover a better way to put that truth to use (or not) and whether you confirm various truths to be so (or not) does not affect the external nature of any truths which exist beyond you. There is water in the ocean. That is a true fact which exists beyond you and no knowledge you glean about the ocean, its uses or properties will ever change that fact. There is one car and there are multiple doors. This is always represented by car / doors. Now which particular door it's most likely to be found behind can be stated with an increasing degree of assurance as information develops, but all of that: "found", "stated", "assurance" - that's all information which pertains to our knowledge of precisely how the random distribution of car was finalized behind various doors. It's 2 sets a) car / doors and b) what we know about it. If you don't abstractly assign the immutable ratio of item / location(s) to a distribution pattern, you never can make that distinction. Most people won't do this because they are too excited that they can "prove" it's not so, when in fact they are proving no such thing. All they are proving is that they are able to refine their search pattern, nothing more. 216.153.214.89 (talk) 08:00, 27 February 2009 (UTC)


 * I'm talking about the conditional probability of the car being there given what is stated about the situation, not the probability of finding it. If we do the MHP 6000 times the probability of the car being behind each door is actually 1/3 - consistently with a uniform probability distribution.  This means if we look at an actual data set of 6000 iterations, we'll likely observe the car was behind each door 1/3 of the time.  With this actual data set, there will be about 3000 cases where the host opened door 3.  Of these actual 3000 cases the car is never behind door 3, and it is behind door 1 about 1/3 (1000) and behind door 2 about 2/3 (2000).  The conditional probability of the car being behind door 1 given the host opens door 3 is actually 1/3 - the probability is not uniformly distributed between door 1 and door 2.  What we've done is carefully select two subgroups out of a uniformly distributed group ending up with two equally (and compensatingly) non-uniform groups.  The 6000 cases are still uniformly distributed but we're no longer talking about the whole group - only our carefully selected subgroup of 3000.  This principle is what conditional probability is all about.  If you don't understand this you are extremely confused. -- Rick Block (talk) 14:49, 27 February 2009 (UTC)

216 here, replying from my laptop. Confused isn't the right term. Inarticulate in the vernacular of Probability Theory would be more correct. But thanks for the conditional probability link. I've read it and here's my answer: Please look a the articles Event (probability theory) and Sample space. As you can see from the Venn Diagram image on the event article page, there are indeed 2 data sets. The "sample space" (distribution) and the "event" (probability). If you read the articles, you'll see that only non-infinite sample spaces can be measured as a definate starting point. Look again at the Venn. The distribution ratio of cars per doors is 1 car per 3 doors 1/3 in the large circle and 1 car per 2 doors 1/2 during the event, the small circle. These numbers are NOT the numbers for the probability and yet they do exist. So please, don't call me confused. I understand the math just fine. My contention is that you are simply ignoring the point I've made which is that one can't deduce the improved choice odds without doing a math calculation and the requirement to do that math would violate a requirement to "guess" (see above). Do you deny that the distribution ratio of cars to doors changes from 1/3 to 1/2? If not, we have no disagreement. 216.204.235.66 (talk) 16:23, 27 February 2009 (UTC) (aka 216.153.214.89 (talk))


 * I'm sorry, but it certainly sounds like you're saying the probability at the point the player is asked whether to switch (i.e. after the host has opened a door) is 1/2 because there's one car and two doors. If this is what you're saying I definitely do not agree.  I do agree that a player who guesses where the car is has a 1/2 chance.  In this case we're assessing the probability of a 1 out of 2 guess being correct, which reflects the ratio you're talking about (which is the probability of a random selection between any two alternatives, no matter how unequal their probabilities might be).  However, this has next to nothing to do with the MHP.  Someone who phrases the MHP in a way where the player "guesses" at the end has gone to a lot of trouble to create an intensely counterintuitive non-uniform probability only to require that it be completely ignored.  -- Rick Block (talk) 03:36, 28 February 2009 (UTC)

No, I am not saying the probability is 1/2, I am saying the distribution is 1/2. The probability is an information contruct which cannot be arrived at as being distinct from the sample space (base distribution) without additional information. At its essence, the MHP is trick question designed to elicit an answer based on the statistics of the distribution (1/2), when in fact additional math which takes into account the removed door can be performed. This is a probability calculation which uses the information gleaned about the removal of the door (a known negative). In all detection scenarios, whether via mathematical probability extrapolations or a literal turning over of stones, it's the information acquired which increases the odds of success. Ordinary people are tricked by this question - some overtly by the word "quess", others by the fact that no effort is made to teach with this problem. It's a cheap trick based on a "gotcha" which blurs the distinction between statistics (distribution) and probability (detection). When I said "overlay", I meant it literally as an illustration. Think about a transparant sheet over another. There's facts on the base sheet. That sheet holds the distribution pattern. As we acquire information about those facts, we add information to our overlay sheet - the top level clear one. The probability MUST reside on the top sheet, because without the information on the top sheet, we can't do any probability calculation. That's why a new player without the information about the removed door or original choice faces fully correlating odds and distribution. He's got no information which would allow him to refine his odds. That's why I say probability is information based, not fact based. It's on the top sheet, not the bottom one - if you follow my illustration. Anyway, I think we both understand each other now. I just don't like "gotcha" problems which ultimately teach people nothing. Perhaps Massimo's crass efforts to get material for a book and his obviously intentional use of the word "guess" sours me on this problem. However, I do think this article could stand improvement by showing some information (venn diagram?) about the event of the 2nd choice and the change in the composition of the sample space after the 1st door is removed. And the sample space does indeed change because one door is removed from consideration by the rules of the game - so by definition, it's no longer in the sample space. A typical person, not knowing the nuances here, will see the sample space shrink from 3 to 2 while the car stays at 1 and think that the "odds" went for 1/3 to 1/2, when in fact only the distibution did. People who say 1/2 aren't stupid, they are just being tricked by the question. Remember, the facts on the base sheet change when the host removes one door from consideration - that changes the sample set from 1/3 to 1/2. Additionally, we can go beyond 1/2 accuracy in our selection because we know about the removal, and because we know we can extraoplate diverging odds because the removal of the door tells us for sure one location where the door is not. And, we also have information because we know which door we originally chose. Without those two pieces of information, the new probability can't be known and if can't be known, it's never more precise than the underlying set, 1/2. With information, we can arrive at 2/3 probability. Without information, we can't rise beyond the post-removal base rate of 1/2. Probability is an information construct which is distinct from the underlying facts. Two data sets. 216.153.214.89 (talk) 05:17, 28 February 2009 (UTC)


 * Perhaps I'm not understanding what you're saying, but it still sounds like you're saying what I'm calling the probability is 1/2. If the host rolls a die and initially places the car behind door 1 if the die is 1, behind door 2 if the die is 2, and behind door 3 if the die is any of 3, 4, 5, or 6, are you saying (one car, three doors) the "distribution" is 1/3 per door?  I'm saying the probability in this case is 1/6, 1/6, 2/3.  The car is not randomly located behind these three doors.  The 1/3 distribution (which would be the probability if the car were randomly placed) has no particular relevance.  If someone were to randomly guess which door the car is behind they would have a 1/3 chance of guessing correctly (for the reasons you state - and this can also be shown using a conditional probability analysis).  Similarly, after the host has opened a door in the MHP the probability is 1/3 vs. 2/3.  Which door the car is actually behind is not equivalent to a random choice of 1 object from 2 possibilities.  Probability and statistics are related.  Probability theory analyzes what's happening on the bottom sheet in your analogy, not the top sheet.   You're almost right about the Monty Hall problem, but it isn't actually a trick question.  It's a question designed to powerfully resonate to the "it must be 1/2" intuition that it sounds like you are stating is an immutable fact.  -- Rick Block (talk) 05:57, 28 February 2009 (UTC)

Prior to any additional information being obtained, there is no top sheet. We have only the bottom sheet which tells us there is 1 car / 3 doors, a 1/3 distribution (sample space). Subsequently, that distribution (sample space) is changed to 1/2 when door is removed. Also, if the door were removed prior to 1st our choice, our choice would give us no information. But because we are allowed to interact with the sample space prior to it being modified, we are able to fix the odds of one door (by choosing it) at it's original sample space distribution rate of 1/3. This is information element #1 which we develop. This datum goes on the top sheet. Next, a door is removed and we now know that 100% of the car location possibilities are confined to remaining two doors. This is datum #2 and it also goes on the top sheet. Now, because 100% of the location possibilities remain and we have fixed the odds of the 1st choice door at 1/3, we know that the 2nd remaining door has what's left of the original odds which is 1 minus 1/3 or 2/3. That is all very simple, but the way the problem is formed and this article is written both mislead people. This partial sentence "It also makes clear that the host's showing of the goat behind Door 3 has the effect of transferring the 1/3 of winning probability a-priori associated with that door to the remaining unselected and unopened one..." is simply not true as posed. There is no probability transferrence of any kind. Instead, the reason why the switch-to door has higher odds is because the removed door has zero odds and the 1st door has 1/3 odd by virtue of the actions we took against them. Our actions made it impossible for those to be the most likely locations, hence the last door has to be more likely. There's not "odds transferrence", there's only information acquisition and state setting. We set the removed door to a 0 state and the 1st door in a 1/3 state. The odds didn't transfer to the last door. Rather, the last door became the last unturned stone. Probability of finding it increased by our actions, not by transference. We assign the greater value correctly to the last door because we correctly assigned the values to the other doors 1st due to certain knowledge. An assignment is not a transfer. Based on what we 1st know, we originally assign 1/3 odds, based on what we learn, we are able to later assign 2/3rds odds. Those odds (the probability) definately rests on the top sheet because we develop our accuracy to assign it correctly only after we acquire information. The acquired information is not the sample space. And, if you don't allow for the bottom sheet to be the sample space, then the Venn Diagram I directed you do is invalid. But if you do allow the sample space to be the bottom sheet, then each datum we develop must be on the top sheet. There are two data sets, the underlying facts on the bottom sheet and the data we develop on the top sheet. The developed data allows the event, but the event is and must be distinct from the sample space. If you refuse to accept this, we are at an impasse. The illict act of fixing the odds of our 1st choice prior to modifying the sample space is what makes this a trick questiton. The act of modifying the sample set while holding onto prior definate knowledge is what modifies the odds - there's no transferrence and any such notions suggesting there is a transferrence is metaphysical bunk. BTW, you do know that the Probability article states "Probability, or chance, is a way of expressing knowledge or belief that an event will occur or has occurred.", which is EXACTLY what I've been saying. It's information based - based on the information we develop (and in the MHP, the states we set) it doesn't modify the underlying sample space. Rather, it only reflect what we know and can deduce about the sample space. Our combined information and calculations are called an event. Events and samples spaces are not the same thing. Now that I am using the correct vernacular, you are compelled to concede this point. You have no rational argument against this overlay model. There are two data sets a) the base distribtion (sample space) originally 1/3, then 1/2 and b) the information/calculations overlay, aka odds, aka probability (event), originally 1/3, then 2/3. The 1/2 and the 2/3 both exists simultaneously. There is no conflict here. The 1/2 is the distribution rate after the sample set is modified by door removal and the 2/3 is the probability of the car being found behind the swtich-to door after we use acquired knowledge to set lower values for the other doors. There are two concurrent data sets and the best way to think about them is as an overlay - as if the small cicle on the Venn Diagram rests on top of the other, except I like the mental image of an actual transparent sheet with wax crayon markings on it - just like an overhead projector transparency would seem if laid on top of a typewritten letter. So, are we done yet? Do you concede that the 1/2 and 2/3 exist at the same time and this is rational explanation as to where the fooled people go wrong? They look to the 1/2 which does exist and don't looks any farther to find the 2/3 which can be determined. And frankly, I feel that trick words like "guess" and "transferring" along with illicit modifications of the sample state after the event has begun make the MHP an intentional fakery. 216.153.214.89 (talk) 07:15, 28 February 2009 (UTC)

2/28 - ease of editing indent

 * An event is a partitioning of a sample space, not an overlay view on top, resulting in a new possibly smaller sample space. The probabilities in the original (possibly larger) sample space are called unconditional probabilities.  Probabilities in the new possibly smaller sample space are called conditional probabilities.  The probabilities do not have to be the same in both cases.  The probabilities in the sample space corresponding to an event depend on how the original sample space is partitioned (i.e. depending on the event).  We can model any particular state of knowledge about the original sample space or the event as an overlay, but the event itself occurs at the same level as the sample space.


 * For example, imagine the sample space is a 60x100 grid of 6000 iterations of the MHP where the player has initially picked door 1. If we've followed the rules as given, the distribution of times the car is behind each door in this grid is 1/3, 1/3, 1/3. If we pick a random selection of 600 of these iterations we'd get a smaller sample space (which means this selection can be considered an event) in which we'd expect the distribution to still be 1/3, 1/3, 1/3.  Any event results in a potentially smaller sample space.  Conditional probability provides techniques for analyzing probabilities in these smaller sample spaces.


 * The analysis of the MHP resulting in the 1/3 vs. 2/3 split is at the sample space level, not with respect to some overlay state of knowledge. In the whole sample space the player has picked door 1 and has a 1/3 chance of having picked the car, i.e. in roughly 2000 of our iterations.  The unconditional probability of the player's initial pick being the car is 1/3.


 * The host now opens a goat door. There are two events (two sub-sample spaces) created by this action.  In one, the host has opened door 2.  In the other the host has opened door 3.  This action doesn't affect any of the unconditonal probabilities.  Looking at the entire sample space there's still a 1/3, 1/3, 1/3 ratio.  If we pick a random set of 3000 of our 6000 iterations we'd still have a 1/3, 1/3, 1/3 ratio - but we didn't do this because we know in one of our samples there's always a goat behind door 3 (and in the other there's always a goat behind door 2).  So, how exactly did we divide up the original sample space?


 * Please think about this and let me know what you come up with. -- Rick Block (talk) 16:02, 28 February 2009 (UTC)


 * Note that one or the other of the "host opens door 2" or "host opens door 3" events happens in all cases, i.e. the union of these two events (smaller sample spaces) is the same as the original sample space. Also note these two events are disjoint as well.  -- Rick Block (talk) 16:11, 28 February 2009 (UTC)

Rick, I am enjoying our dialog - thank you for participating. According to Sample space here is a starting premise for our reasonsing: "In probability theory, the sample space or universal sample space, often denoted S, Ω, or U (for "universe"), of an experiment or random trial is the set of all possible outcomes." Note that it says all "possible outcomes". If the host honestly removes the door from the game (in that the removed door doesn't have the car), not only is he is removing it from the sample, but since you are not longer allowed to choose it and we honestly know it doesn't have the car, it's "not possible" for the removed door to have the car. And since, at the point of the 2nd choice (the event at which we calculate 2/3 odds) the removed door is not a possible location, it's "not" part of the sample set at the time of the 2nd chice event. Please not that "sample set" refers to "possible" locations and by the rules of the game, the removed door is impossible as a location. Once that door is removed, it's simply "not" possible that the outcome could include finding the car behind it. Rememebr, the odds we are calculating is the odds the car being found behind any particular door which we can choose from and we "can't" chose the removed door. At the pount of our 2nd choice, the possible outcomes which remain are (car at door 2) or (car at door 3) as door 1 is already removed. Count the possible outcomes. One, Two. Count the car, One. One car, two doors, the sample set distibution ratio at the time of the 2nd chjoice is one car / two doors or 1/2. This ratio exists independantly of the calculation of the odds. There are two data sets. One on the Sample space level and one on the Probability level. They are not the same thing. Possible outcomes and probable outcomes are not the same thing. Probability refers to the most likely ourcome of the possible ones. Possible is the base set, probable is the overlayed sub-set which zooms in on a partiular focus point as a consequence of information. We ask ourselves, "based on what we know - which is the best chice?". We don't ask ourselves "based on what is, which is the best choice?". They key when deciding what we know, is to use all the available information. But it's the information we use that chnges the odds, not the underlying facts. Because if we had no information, we'd be doing nothing but making a blind guess and we've already seen that blind guesses don't increase the odds. The information goes on the top layer and so does the probability. The sample space is always it's own data set, below. Two distinct sets. 216.153.214.89 (talk) 16:55, 28 February 2009 (UTC)


 * I'm fine with discussing this as well and hope I'm not coming across as arguing. I think we're getting closer.  I'm saying we should consider the universal sample space to be a set of cases just before the host opens a door.  This is a sample space where the odds are clearly 1/3, 1/3, 1/3 (because we said they were) and we can assume the player has picked door 1 (because we can simply rename the doors if the player hasn't picked door 1 - or we can run the game enough times that we have a significant sample where the player indeed did choose door 1).  I think the whole question of the MHP is what happens next.  When the host opens a door, the host splits this sample space into two smaller ones.  It is in these smaller sample spaces, not the original one, that a given door has been removed.  If we're in the "host opened door 3" sample space there is indeed no outcome where the car is behind door 3, but we only got here from the original (larger) sample space where there was an obvious 1/3, 1/3, 1/3 distribution.  According to the rules we've established, in our larger universe the host always opens door 2 or door 3 (so, in the 6000 cases above one or the other of these has happened), but the question is if we're in (say) the smaller sample where the host has opened door 3 what are the odds?  I'd take this a little farther and say that to reasonably talk about this sample we HAVE to start with a universe where the odds are 1/3, 1/3, 1/3.  Either it matters or it doesn't, but could you indulge me and start with this as the universe and let me know what you think happens? -- Rick Block (talk) 17:25, 28 February 2009 (UTC)

I agree we are getting closer. And though I won't "indulge" you (I don't like that word), I will address the points you raise. Here's where I rely on the premise of the definition of probability. For something to be probable, it has to be possible. If it's not possible, the probability is always 0. With that in mind, the removed door, by virtue of 100% certainty of not having the car behind it, not only is it's probability 0, but it's possibility is also 0. That being the case, it's not part of the sample space. Any why not? Because nobody would search for something where the've already confirmed it can't be and isn't. There's no bifurcating of the sample space into two parts. Rather, the removed door is no longer under consideration and is not part of the sample space. Read the definition of sample space "In probability theory, the sample space or universal sample space, often denoted S, Ω, or U (for "universe"), of an experiment or random trial is the set of all possible outcomes". As you can see it's the set of all possible outcomes. That door is not a possible outcome (can't have the car), so it's not in the sample space. This is why the distibution ratio of the sample space changes from 1/3 to 1/2 - the sample space shrinks. And it's also why the MHP is illigitimate - when we are alllowed to carry-over the knowledge of a) which door we 1st chose and b) which door doesn't have the car, we have more information that we would otherwise have with a 1/3 or 1/2 sample space. It's that additional information which allows us to pick the better door. The MHP is a trick question with a deceitful and dishonest premise. It's a conflation of a before and after state which allows an illigitimate information carry-over. Keep this in mind: "In probability theory, an event is a set of outcomes (a subset of the sample space) to which a probability is assigned." The 1st choice is an event in the originally sized sample space, the 2nd choice is a 2nd event in the now adjusted-size sample space. But the information carried over from the 1st choice has higher informational value, because in was gleaned from a larger set with more information in it. And the definition of "event" says it's a sub-set of the "sample space" singular, not plural - ie, not "spaces" and not "events". Well, if the removed door isn't possible to be the location, it can't be probable. And by definition, inclusion in the sample space requires possibility. Impossibilities can't be in a sample space. At the point of the 2nd choice is where the "gotcha" is. At that point in the proccess, the removed door is just not part of the 2nd event's sample space - but the information acquired earlier is being carried over. That's why so many people object to this question. It's not that it fools them it's that they intuitively understand there's a cheat going on, but they can't explain it. The carried over information is what allows the 2/3 odds upgrade. There is no "transferrence" of probabilities. It's just a trick question and there are indeed two data sets, a) the distribution ratio of 1/3, then 1/2 and b) the odds of 1/3, then 2/3. The distribution ratio and odds exist concurrently and on seperate layers of base information/acquired knowledge. The odds are an overlay. It's the only way to see the truth here accurately. MHP is an illegitimate trick question. 216.153.214.89 (talk) 04:40, 1 March 2009 (UTC)


 * I think we're agreeing there are two different sample spaces. The original (larger) one encompassing cases where the host opens door 2 or door 3, in which (I think we agree) the odds of the car being behind any door are clearly 1/3.  I also think we agree when the host opens (say) door 3, we're no longer talking about this sample space but a subset of the original sample space and viewing this subset as a "new' sample space the probability of the car being behind door 3 is 0 (completely impossible).


 * I also agree most people don't realize their first choice (picking door 1, which clearly has a 1/3 chance of resulting in picking the car) and their second choice (whether to switch) are occurring in these two different sample spaces. However, I don't think there's any particular trickery involved.  It's just that people are generally very bad with conditional probability.


 * If we can return to the original sample space for a moment there are two events that might happen: host opens door 2 or host opens door 3. Both of these are subsets of the original sample space (right?).  If the question is what are the odds of switching if the host opens door 3 we're talking about one of them, and if the host opens door 2 we're talking about the other.  These subsets didn't spring into existence from nowhere - they only exist (we can only analyze them) as events occurring in the original, clearly 1/3-1/3-1/3, sample space.  We can think about them one at a time (and the "switch or not" question must be evaluated from the context of one of them), but they are both possible outcomes   from our original sample space.


 * So, I think the question boils down to how the action of opening a door affects the original (1/3 each) sample space, specifically what is the composition of the subset sample space. Do you agree with this so far?


 * Assuming you do, I think we're now at the precise point where we disagree. I think the following question will likely highlight this.  If the original sample space includes 6000 iterations of the MHP, about how many of these would you expect to be in the smaller sample space corresponding to the "host opens door 3" event?


 * Based on what you seem to be saying, I think your answer is 4000 (2000 where the car is behind door 1 plus 2000 where the car is behind door 2). My answer is 3000 (1000 plus 2000).


 * Is this where we disagree? -- Rick Block (talk) 16:36, 1 March 2009 (UTC)

I don't think we disagree on the 2/3 odds of the final choice. Rather, what I am saying is that most people who answer this question expect that there should be no cheating involved and yet there is. At the time of the 1st choice from the larger sample space, information is developed. That information is allowed to carry-over to the 2nd choice in the 2nd sample space, but this carry-over is illicit. With probability, it's the odds of a possible choice being correct. But MHP carries-over information about which door is not correct (not possible and thereofore 0 probable) and armed with this information and the fixed 1/3 odds of our 1st choice, we can deduce that the reciprocal of the non-0 remaining doors (in aggregate) must be 1 and since our door is fixed at 1/3, well there you have it. But it's the illicit use of the information acquired from the 1st sample space that allowes us to accurize our odds for the 2nd choice. We are allowed to used information about an impossible choice to assign odds to the two remaining possible choices. This contradicts the premise of sample space which is the set of possible outcomes - which by definition does not consider impossible outcomes. It's a trick question. That's why people confuse the distribution ratio of the sample space, which clearly is 1/2 for the odds of switching which is 2/3 - the illicit comingling of information which by definition should be excluded deceives them because the rules of probablility are not being followed. It's not an honest probablility question - it's a trick question which can only be solved to 2/3 with the help of the "overlay" of the additional information which is acquired outside the scope of the 2nd sample space. Non-cheaters rightly confine their "swtich" calculations to the 2nd sample space only and that's why they say 1/2. to be intellectually honest, when the sample space was downsized, the information store to-date should be cleaned of any information which could not be acquired from the smaller sample space. If not, we are answering a probability question which is not pure as it includes information about impossible outcomes which by definition do not fall into probablity questions as they are excluded from the sample space. It's a trick question. It's bastardized probability and proves nothing other than cheaters are able to deceive honest people. 216.153.214.89 (talk) 19:02, 1 March 2009 (UTC)

Discussion that was here moved to new section

So basically, the math whizzes on this page contend that a sample space is a set and something that's not in the current set, can still be used to calculate the odds on the current set. That's not how I read the rules, but I won't dispute if you fellows are right on this point. However, it's still a trick question as the average person knows nothing about what the rules are and for that reason, doesn't think they are allowed to glean information from the removed door. Suffice it to say removing 1/3 does indeed leave 2/3 and with one door fixed at 1/3, well the other has to be 2/3. The math isn't hard at all. Rather, it's the lack of knowledge of the rules which makes adhering to them hard. Sort of like the foolish man who represented himself in court and annoyed the judge in doing so. When the man stopped to shuffle papers and think for a moment, the judge shrewdly asked "would you like to rest?" Of course, since the man was unaware of the significance of the word "rest" in that context, he was surprised when after he said "yes", the judge promptly gaveled the case closed and ruled against him. Farther on down this page, I extracted agreement that the distribution ratio was 1/3 and becomes 1/2. This is what most people have on their mind and lesser informed ones like myself think (wrongly it seems) that "impossible" (and associated info) should be excluded from the calculations. 216.153.214.89 (talk) 07:55, 2 March 2009 (UTC)

Getting down to Brass tacks

 * I think we're not going to get to a common understanding unless you answer my question. Is the size of smaller sample space 4000 or 3000? -- Rick Block (talk) 19:10, 1 March 2009 (UTC)

Rick, everything we need to see the illicit nature of the MHP is available to us using the actual sizes of 3 and 2. There's no need to go down your rabbit trail - other than to allow you to ignore the points I'm raising. Yes or no, do you think it's legitimate to include information about an impossible outcome when calculating the odds of an Event? I say no, because an event is a subset of a Sample space and an sample space is limited to possible outcomes. For this reason, an event can't include impossible outcomes - as the impossible never exists in any sample space and events are a subset of that - a subset of the non-impossible. The 2nd choice, if it uses information acquired outside the scope of the 2nd sample space, violates the rules for an event. It's a trick question which relies on an information overlay - exactly as I've been saying for some time now. It's your refusal to acknowledge that a probability theory rules violation occurs which is the stumbling block here. Simply put, one can arrive at the 2/3 odds calculation, but can't do so without cheating. 216.153.214.89 (talk) 19:22, 1 March 2009 (UTC)


 * Do I think it's legitimate to include information about an impossible outcome when calculating the odds of an Event? This is not a simple question to answer, mostly because I'm not exactly sure what you mean by this.  I wholeheartedly agree that an event is a subset of a sample space and a sample space is limited to possible outcomes.  I think we are on exactly the same page here.  However, a (subset) sample space corresponding to an event can certainly be (generally is) within a larger sample space where other outcomes are possible.  In particular, in the larger sample space of the MHP the car can be behind any door.  The event "host opens door 3" creates a smaller sample space (the one we're actually interested in) where the host has opened door 3 so in this smaller sample space the car must be behind one of the other doors.  I don't think we disagree yet (right?).  Any given event always divides a sample space into two other ones - one where the event occurs and one where it doesn't.  The union of these two subset sample spaces is always the original (larger) sample space.  If the event happens with probability p, it doesn't happen with probability 1-p, size(original sample size)p is the size of the sample space where the event occurs and size(original sample space)(1-p) is the size of the other sample space.  So, is it legitimate to include information about impossible outcomes when calculating the odds of an event?  If we know the sample space in which the event occurs, and we know the odds of NOT(the event) we know the odds of the event as well (1 minus the other) so in a strict sense yes (but again, I'm not sure exactly what you mean).


 * I'm asking about a sample space of the MHP with 6000 iterations not to try to trick you but because it's large enough to actually see what's going on. When we scale it down to 3 and 2 the probabilities hold but it becomes very difficult to understand.  I'm trying not to ignore your point about information overlay.  Very directly, I completely disagree that this is what is going on.  It's not a trick question.  There is no cheating.  What you seem to be saying is the MHP is some black hole outside the bounds of probability theory.  I'm saying probability theory exactly explains it.  The probability of winning the car by switching (in the sense we've been talking about), and the probability distribution between door 1 and door 2 are the same thing (pretty much by definition).  One of these cannot be 2/3 if the other is 1/2.  -- Rick Block (talk) 21:38, 1 March 2009 (UTC)

Without taking a position on the legitimacy of using the initially acquired information, you won't be able to accept my contention as valid or not. You have to understand this problem from the context it presents. A player is asked to choose or stay and is also asked if the odds are 50/50 or not. If the player confines his decision tree to only the information available from the possible locations of the car, he can't arrive at the conclusion that the odds are 2/3 by switching. That conclusion can only be reached if the additional information from the larger set is included. It's my contention that the very definition of probability, in that probability only deals with what's possible, precludes the extra initially acquired information from being used to determine the 2nd choice. The most accurate answer to the question of is it better to swtich or stay is: What information am I allowed to take into account when making that decision? If I am right and the definitions of Sample space and probability deal with only what's possible, you can't take into account the impossible on the 2nd choice. And without that information, one can't do the math which enables the conclusion that switching is better. At the point of the 2nd choice, there is clearly 1 car per 2 doors or a 1/2 distribution. Now accepting for a moment that this is true, that the distribution is 1/2, how is it that the odds calculation deviates from the distribution? It's because the additional information is factored in. Without that information, you are stuck at the blind guess level. That's why if you remove the original player and his information and insert a replacement player in possesion of no information, that player will only find the car 50% of the time - because without the knowledge of a pre-existing 1/3 starting point (the 1st choice of the orignal player) and without the 0 value of the removed door, there's no way to assign the 2/3 to any particular door. Probability is all about the assignments of the odds and the assignments are made based on information. No information = no math = blind guess = no deviation from the distribution ratio. The deviation from 1/2 to 2/3 is accomplished by the illicit inclusion of the information acquired from the larger sample space. Beyond this I don't know what to tell you as I am beginning to repeat myself. BTW: The two "layers" which I refer to are that which is and that which you know (about what is). It seems that you are not able to make this distinction. 216.153.214.89 (talk) 22:39, 1 March 2009 (UTC)

On the importance of rules
Let's say for a minute, you are sitting in a room, blindfolded, ears covered with nose and mouth covered by an oxygen mask. The rules of the game are you must not open your eyes, uncover your ears, unmask your face, rise from your chair or move your chair. Now with those rules in place, I bring a nice oven fresh pizza into the room. You can't smell it, you can't see it, you can't taste it, you can't hear me exclaim how delicious it is and you can't touch it. What are the odds that you will eat a slice of pizza before it's gone? The answer is, without more information, we can't know. Am I allowed to unmask your face? Are you allowed to cheat? Without more information, we simply can't know.

Now, with that thought in mind, let's remove our original MHP player just before he makes choice #2 (stay or switch) and insert a new player with no knowledge of what's transpired before. These two players face different odds and the only difference between them is the information they possess.

At the point of the 2nd choice, the ratio of extant cars (1) to choosable doors (2) is 1/2. But the 1st player, via the use of clever math which takes into account information unavailable to the 2nd player, can more readily find the car than the 2nd player who lacks that information. We represent this detection improvement as "odds". The 1st player has better odds, better probability only because he has better information.

The probability exists on the information level and the "information level" rests on top of the "extant fact" level. What "is" and "what you know" are not the same thing - that's why both the 1/2 and the 2/3 exist and are measurable - you are measuring two different things.

This is why clarity on the rules is paramount. If I am allowed to take into account the information derived from the large sample space, I am able to construct a math formula and better determine the location than if I'm not allowed.

This is the linchpin of the entire question.

I contend that the definitions referred to above (by me) prohibit the use of that information and for that reason, this is a trick question because the solution cheats by making a calculation which is prohibited by the rules. 216.153.214.89 (talk) 23:19, 1 March 2009 (UTC)


 * (written before some of the above responses, but appropriate here as well) Yes, we keep getting to this point and you keep apparently not believing me. What I'm saying is at the point the player is asked whether to switch, the odds (that which is as you put it) really are 1/3 vs. 2/3 (not 50/50).  I understand the distinction you're making between "that which is" and "that which you know (about what is)" and what I'm saying is that your belief that the "that which is" odds are 50/50 is incorrect.  If you insert a replacement player with no information (or switch/stay based on the flip of a coin) this player will indeed win 50% of the time, but even 2/3 of these players who switch and 1/3 who don't will win.  Compare this with actually opening door 2 as well and bringing in a blindfolded player and having her guess door 1 or door 2.  This blindfolded player will also win 50% of the time even though the "that which is" probability is now 0 or 1 (the point is that the fact that such a player wins 50% of the time has NOTHING to do with the "that which is" probability).  If you agree 2/3 of the blind switchers (as opposed to 2/3 of the players who switch blindly) will win (you may not) this means the probability distribution ("that which is") is 1/3 door 1 and 2/3 door 2.  I'm saying at the point the player is asked to switch it's like starting with two doors, rolling a die, and putting the car behind door 1 if the die is 1 or 2, and putting the car behind door 2 otherwise (3,4,5, or 6).  My impression is you have thought about the problem the way you're thinking about it for a long time and are extremely reluctant to consider the possibility you might be mistaken.   You may not have any particular reason to think I know what I'm talking about, but I assure you I do.  I have absolutely no agenda here other than trying to help people understand this problem.  I've been doing it for a fairly long time.  I suppose this ultimately may boil down to you'll believe what you want to believe. -- Rick Block (talk) 00:11, 2 March 2009 (UTC)
 * Marilyn introduced a spaceship landing on the stage and a little green man getting out. For him the odds are 50/50. The odds are not assigned to the doors but to the players. But this will be known to you. Nijdam (talk) 22:29, 2 March 2009 (UTC)
 * To extend this, the green man has to know the rules, but is unaware of the history. It is not sufficient he is just ignorant of all, because then we cannot say anythimg about his odds. The odds being 50/50 means that on the average green men, landing on the stage and finding door 3 open, half of the time find the car behind door 1 and in the other half behind door 2. Nijdam (talk) 22:41, 2 March 2009 (UTC)

Rick, I had to think about this section of yours longer before replying. In the above, you say "but even 2/3 of these players who switch and 1/3 who don't will win". That's only true if we also carry over the knowledge of which door was originally selected or more accurately, that a door was originally selected. In a true substitution, where the new player has 0 knowledge, the "chosen door" state is reset to null and therefore there is nothing to switch away from. Our knowledge tells us that a "switch" is possible, but the new player can't switch because he doesn't have a starting point. He's starting from a blank slate and he's told to pick a door, nothing more. Now other than brain hemisphere bias, the 2nd player is going to pick door #2 50% of the time and #3 50% of the time. We are not player #2. Player #2 can never refine his known odds beyond 50/50 - only we can do that because we have more information than him. It's irrelevant that you say one door has better oods, player #2 will only pick that door 50% of the time. Player #2 can't correlate his selection and align it with the odds you say are there without more information - the information we are denying him, but ourselves possess. Now, if we also deny ourselves the information, the same way we deny player #2 that's analogous to being struck dumb, or getting a stroke. That area of information ceases to be available to us, so we can't know it. And if we can't know it, our choice can't rise in accuracy beyond the blind guess of 50-50. The odds are on the information overlay level. 216.153.214.89 (talk) 18:30, 2 March 2009 (UTC)

End game is approaching
Rick, you are clearly being intentionally obtuse and are ascribing to me statements which I did not make. If you won't even concede that a blind guess made when there is 1 car and 2 doors has 50/50 odds, there is nothing to talk about. It's the gathered information which allows you to assign the better odds to the switch-to door. I've explained it to you 5 ways to Sunday, but you can't understand me. There is nothing further to be gained by this conversation. Beyond this point, I will only dialog with you if you answer yes or no questions, one at a time, and stay with me until I am done asking questions. You can ask me questions only when I am done questioning you. If you agree to that, I'll talk to you, if not, I won't. Here is my 1st question: Yes or no, are you willing to accept the usage of the term "distribution ratio" in referrence of the number of extant cars divided by the numbers of choosable doors, expressed in math as car/doors? 216.153.214.89 (talk) 01:38, 2 March 2009 (UTC)


 * OK, I'll play. First answer is yes. -- Rick Block (talk) 01:44, 2 March 2009 (UTC)

Good, question #2: At the point in time when the player has the option of switching or staying, yes or no, are there two (2) and only two (2) choosable doors left in the game, provided of course that you define staying as a choice (which I do)? 216.153.214.89 (talk) 01:48, 2 March 2009 (UTC)


 * Of course. -- Rick Block (talk) 02:08, 2 March 2009 (UTC)


 * ... and therefore the distribution ratio as you've defined it is 1/2. -- Rick Block (talk) 02:16, 2 March 2009 (UTC)

Not just as I defined it, but as you have agreed to accept it. Now for question #3: Does the fraction 1/2 = the fraction 2/3, yes or no? 216.153.214.89 (talk) 02:18, 2 March 2009 (UTC)


 * No. -- Rick Block (talk) 02:22, 2 March 2009 (UTC)

Great, then I win and you lose. Here's why:


 * a) You've agreed that "distribution ratio" is car/doors
 * b) You've agreed that at the point of the 2nd choice, the disribution ratio is 1/2
 * c) You've agreed that 1/2 does not equal 2/3.

Well, if the distribution ratio is 1/2 and the odds of success by switching is 2/3, then as I've been saying, these must be referring to different things - there are two data sets. Now for question #4: Yes or no, do you agree there are two data sets?

If you agree, I win. If you don't agree, you contradict your previous 3 answers. 216.153.214.89 (talk) 02:40, 2 March 2009 (UTC)


 * I agree. Can I ask questions yet?  I'm still playing but will be offline for a while. -- Rick Block (talk) 02:43, 2 March 2009 (UTC)

Possibly, I have to think through where else I want to pin you down before I turn over the floor. Let's resume Monday. By then, I'll either continue with questions or yield I'll post that I yield the floor. 216.153.214.89 (talk) 04:02, 2 March 2009 (UTC)

Rick, one more question for now (#5): Yes or no, do you agree I have proved that the 1/2 distribution and the 2/3 odds exist at the same time? 216.153.214.89 (talk) 06:15, 2 March 2009 (UTC)


 * I wouldn't exactly say "proved", but yes the 1/2 distribution and 2/3 odds exist at the same time. BTW - even after you've yielded the floor, I'll let you ask more direct questions whenever you like (although I'll ask that you answer the current question first before asking a question in return). -- Rick Block (talk) 06:41, 2 March 2009 (UTC)

Well, if the 1/2 and the 2/3 exist at them same time, can you see why I say "overlay"? The 2/3 isn't extrapolated, derived or calculated from any information the 1/2 provides. Rather, the 2/3 comes about as a result of factoring information acquired from the 1/3 set which is no longer fully choosable. In essence, the 1/2 is a topology map of two mountains and the 2/3 is an overlay of information - a distinct treasure map which helps you best determine which mountain has the gold mine. Two separate maps, two data sets. And remember: if somethings not choosable, such as the door that's not choosable, it's not a possible location. And since it's not possible, it's not part of the sample space. My contention is that no information derived from outside of the smaller sample space should be allowed (the removed door), nor should any information acquired prior to the downsizing of the sample space be allowed (the fixing of 1 door at 1/3) -and I've explained why- it's because doing those things violates the rules of calculating probability (see above and linked-to articles). If I'm wrong about there being a probability rules violation, my point is moot and people who say 1/2 are just ill-informed. But if I am right, some of the smarter ones who should be able to calc to 2/3, but still say 1/2 are doing so because they sense, but perhaps didn't articulate, what I am detailing now. Based on what I've read about the rules of probability, I think those rules prohibit using the treasure map and if I am right, this is a trick question.216.153.214.89 (talk) 07:26, 2 March 2009 (UTC)

I yield the floor. 216.153.214.89 (talk) 07:59, 2 March 2009 (UTC)

Rick's questions
My rules are you need to be willing to answer yes/no and sometimes numeric questions. I promise I'm not trying to trick you into anything. My guess is we'll run into some things you will flat out disbelieve. In these cases I may suggest simple experiments involving a deck of playing cards which you promise to do. First question. Are you OK with these rules (yes/no)? -- Rick Block (talk) 14:52, 2 March 2009 (UTC)

And, to make this more than fair, I'll tell you exactly where I'm going with this. It is my contention that we can talk about inherent probabilities of physical reality, independent of any external state of knowledge, which I think corresponds to what you called in bold above "that which is" and that what you're calling the "distribution ratio" is something different from this. Specifically, the distribution ratio is an overlay (using your terms) given a particular state of knowledge and NOT the underlying, inherent, true, physical (whatever you want to call it) probability. Moreover, in the MHP at the point the player is deciding whether to switch the underlying, inherent, true (whatever) probability is actually 1/3 door 1 and 2/3 door 2. If you stay with me, I am very confident you will eventually agree with this (and I suspect at this point you couldn't disagree more). -- Rick Block (talk) 16:10, 2 March 2009 (UTC)


 * Ok, let's try. But think about this: At any given moment, there are x fish in the ocean. We can represent that as F (fish) / W (cubic miles of water) or F/W. That formula never changes. At at any given moment of time, if you were to be able to count the fish, you'd get that number "x" (F) and if you were able to also at that moment of time measure the cubic miles of water you'd get (W). Those numbers exists independantly of whether we count them or not. You keep thinking that I am suggesting that this fact level has "probability". I am not. My contention is that no probability is assigned to any aspect of the ocean, until we make that assignment. Such as when we search for particular fish - say sardines, or whale sharks. Then, based on what we know about that fish's habits and territories, along with our best estimate of the population count, ocean conditions and other factors - including inspection ability, we can say "there's a 40% chance we'll see a sardine bait ball off the coast of Durban, South Africa today". Our odds represent our best assertion of likelyhood, based on the information we have. Nothing we know about those sardines changes any aspect of their extant state. They are extant before we know and they will continue to be extant after we know. This is why I assert that the extant state, knowable or not, is the base level and what we know about it is the "overlay".


 * I suppose you could say that a fact base is akin to an actuarial table - what has already been measured and confirmed to be and an informational overlay is like recent claims experience... Though precise as they are, actuarial tables are not an exact measurement of the underlying facts, they are only a compilation of prior claims experience and MIB data. The reason why I am so emphatic about the "distribution ratio" in MHP being distinct from the probability is that the doors and car are precisely measurable through direct observation, no calculations are required to observe how many doors there are. The only issue for me is, at any given time, as per the rules of the odds calculation, how many doors are properly considered to still be in the game and for that reason, what information are we allowed to include in our calculations? I still currently believe that the rules of probability require we not use any data acquired or derived from outside our sample space. And, I am still not persuaded that the sample space doesn't drop from DDD to DD when a door is removed. I am going to call some MIT guys or other local math brainiacs about this myself and see what they say. In the meantime though, have at it, I'm all ears. 216.153.214.89 (talk) 17:18, 2 March 2009 (UTC)

Rick, here's what you are saying, in the form of a joke (our roles are reversed in this joke):


 * Me: I'm going to prove you are not here.
 * You: Ok, try.
 * Me: Are you in Bejing?
 * You: No.
 * Me: Are you in Antarctica?
 * You: No.
 * Me: Are you on the Moon?
 * You: No.
 * Me: Well, if you are not in any of those places, you must be someplace else, right?
 * You: Right.
 * Me: Well, if you are someplace else, you can't be here.
 * You: Right.

Whereas, my contention is that the concept of "someplace else" must a) truly mean you are not here and b) be an allowable location per our rules, or else my placing you there doesn't take you from here - even if you agree that you are "someplace else".

It's not what we know or think or agree is true, it's what really is true that forms the base. You are getting confused because in most of your realm of knowledge, direct measurements and direct corroboration are not possible. My contention is that when available, accurate direct measurements and corroboration are best. Of course, that's not always available to us. My next contention is that clarity of rules is paramount. If the rules allow your calculation, MHP is not a trick question, but if not, then it is. So I guess this really is a rules debate, not a math debate. And not only is it a rules debate, but it's a rules application debate - are the true rules for probability being correctly applied if we use the information which I say should be excluded? As I see it, that's the crux of the matter here. 216.153.214.89 (talk) 17:43, 2 March 2009 (UTC)

Dueling questions
So far, you're really not very good at this game. I asked a simple yes/no question and you've entered two long paragraphs. My actual starting question is whether you think it's reasonable to talk about what I'll call "inherent" probabilities that are independent of what you're calling overlays (which we'll get to in a minute). I think this corresponds to your notion of "what it is" - i.e. the true, verifiable probability of something. Some examples: a flip of a fair coin has a 50/50 chance of being heads or tails, randomly selecting one card from a deck of 52 cards has a 1/52 chance of being the ace of spades. Agree (yes/no)? -- Rick Block (talk) 19:34, 2 March 2009 (UTC)


 * Not very good?... Your question presumes evidence not before the court... "My actual starting question is whether you think it's reasonable to talk about what I'll call "inherent" probabilities that are independent of what you're calling overlays (which we'll get to in a minute)." If I understand you right, you're making a proffer as I've not yet accepted your premise. Indeed, I reject your premise, so of course my answer is "no" it's not reasonable. Even so, reasonable or not, I'm interested to hear what you have to say. So, yes I agree to listen amd reply as best as I can. FYI: I do NOT define "what it is" - i.e. (as) the true, verifiable probability of something. I am saying extant -an extancy which can be measured and known. You keep substituting probability. It's my contention that they are not the same thing. Unless you address that distinction and resolve it to our joint agreement, I'm somewhat sure we'll still be on two different topics. There are two doors and the car is extant behind one of them. No odds are involved in that statement of fact. Do you see why I am having trouble giving you an honest hearing? I want to resolve a distinction and you keep blurring the lines. 216.153.214.89 (talk) 20:39, 2 March 2009 (UTC)

Well, perhaps we can call it "extant" (although "probability" is the word usually used for this). What I'm talking about is measurable in the sense of running an experiment many, many times. If we have a coin that we flip 1,000,000 times and count how many times it ends up heads and how many times it ends up tails, what I'm talking about is the ratio of heads to total trials or tails to total trials. It is this sort of ratio that I'm talking about. Note that this sort of ratio both can be measured for things that have happened in the past and serves as a prediction for what we would expect to happen in the future (if we repeat an experiment lots and lots of times - this is the Law of large numbers). Whatever we call it, do you accept that such ratios are factual in much the same manner as 2+2=4, i.e. that such ratios are not "overlays" but are in the realm of "what is" (as opposed to "what is known about what is") - yes or no? -- Rick Block (talk) 02:45, 3 March 2009 (UTC)

And (trying to speed things up), if yes, can we call this Probability (with a capital P - as distinct from "probability" which we could leave undefined for now)? -- Rick Block (talk) 02:45, 3 March 2009 (UTC)


 * Rick, I am familiar enough with P related math princples to concede that they are undergirded by unyielding truth and a form of "that which is" all by themselves. So to that extent I agree to the usage of P. But this comment of yours "measurable in the sense of running an experiment" make me think you want to count the dogs in my back yard via formula rather than looking out the window. If I own three dogs, my vision is good and my view of the entire yard unobscured, it's easier and more accurate to look out the window to be sure they are there instead of what you propose which clearly is a "dog", "escape frequency", "amount of time since last inspection" formula. In the game, when there are two doors (as a base premise of the game) I already have the certainty that there's one car. There is a statistical relationship there of 1 to 2. That's not a probability relationship and yet it "is" all by itself. We are getting caught up in the Clintonian puzzle on the meaning of "is". That said, please continue. 216.153.214.89 (talk) 06:37, 3 March 2009 (UTC)

I'll take the above as "yes" (yes or no, remember?), but since you seem to be quibbling perhaps we should clarify. I think the P of a given coin ending up heads when it is flipped is am intrinsic property of the coin itself, as opposed to a state of knowledge about the coin. By this I mean if I flip the coin 1,000,000 times, and then separately (independently) you without knowing anything about what I did flip it 1,000,000 times, our counts of the number of times it ends up heads will be very similar (likely identical to 2 or 3 significant digits given this many trials). Do you agree (yes/no)? -- Rick Block (talk) 13:32, 3 March 2009 (UTC)


 * Rick - your problem is that the questions you ask are not the ones you seek an answer to. You should be asking me, "will you accept this premise?", rather than "do you agree?" My answer is IFF (if and only if) you are saying that outcomes/possibilities = P, then yes, I agree that "P" is an "intrinsic property" and have basically said as much already. Is that what you are asking me? 216.153.214.89 (talk) 14:51, 3 March 2009 (UTC)
 * Counter-question to Rick (as per your allowance to me - see above). Rick, yes or no, the 2/3 value of switching, do we arrive at that number by extrapolation or not? In other words, the process we used to arrive at that number, was it an extrapolative proccess? In other words, do you concede that we "constructed that data outside a discrete set of known data"? 216.153.214.89 (talk) 15:07, 3 March 2009 (UTC)
 * No, I don't. -- Rick Block (talk) 15:17, 3 March 2009 (UTC)

I'm asking whether you agree the P I'm talking about is a property of the coin that can be discovered by experimentation. Different coins might have different values for this property, but the likelihood of a flip resulting in heads for a specific coin is what it is. The property is attached to the coin as surely as the fact that it has both a head side and a tail side. There are two sides, one of which is the head side but this ratio (1/2) is not what I'm talking about. I'm talking about the propensity of a particular coin to land showing heads when flipped  These are different things (not all coins are "fair"). I'm not talking about premises, but facts (like 2+2=4). Do you accept as a fact that the propensity of a particular coin to end up head side up when flipped is an intrinsic property of the coin itself (yes/no)? -- Rick Block (talk) 15:11, 3 March 2009 (UTC)


 * I'm not so sure if the values are "discovered" or assigned. For example, we assign a 50/50 value to coins, but no one really knows the true numbers. Some coins, when flipped, have been known to land on edge. Perhaps a die (dice) would always have to land 1/6 up, but a coin doesn't have to. But for the sake of this discussion, let's assume we are flipping onto a micro-vibrating surface, the lateral thrust of which is sufficient to tip our perfectly beveled-edged coin past the tipping point, should it land on edge. With that hypothetical coin/table combination, yes, the odds are exactly fixed, same as if the h/t results were indeed a property of that coin. This also presumes that motion is applied to the coin. A coin at rest has no odds. If it stays at rest, it will always display the same surface. So if you are saying that a perfectly designed coin, flipping onto a perfectly designed table (until that coin wears out and becomes lopsided) will always behave exactly the same way on a flip, so much so that P is indistinguishable from the coin itself, then yes, I agree. Also, I am presuming that observation is a form of discovery, yes? 216.153.214.89 (talk) 18:15, 3 March 2009 (UTC)


 * Ok, since I've answered your latest (above), here's my next: Are you contending that odds (probabilities) are discovered by the math and if you are sayng that, can you ever discover any probabilties without the math? (And by "math", I include logic diagrams and all other methods where even so little a calculation as 1+1 is performed.) 216.153.214.89 (talk) 18:20, 3 March 2009 (UTC)
 * No, I'm not contending this. -- 20:58, 3 March 2009 (UTC)

Now I can't tell if you're intentionally being obtuse or simply enjoying being difficult. Reach into your pocket. Take out a coin. This coin (not some idealized "perfect" coin) has real physical properties that have the result of this coin having its own intrinsic propensity of turning up heads when flipped. We could design an experiment to measure this propensity but the propensity itself is a property of the coin. We may or may not be able to accurately measure it, but it exists in as a real a sense as the fact that it has a head side and a tail side. Do you agree with this (yes/no)? -- Rick Block (talk) 20:58, 3 March 2009 (UTC)


 * Provided you aren't trying to fix the P of an ordinary coin at exactly 50/50, I agree subject to the caveat that coins do sometimes land on edge. 216.153.214.89 (talk) 22:36, 3 March 2009 (UTC)

My next question: Are you contending that odds (aka probabilities) can be understood by the human mind without the aid of logic maps or math? 216.153.214.89 (talk) 22:36, 3 March 2009 (UTC)
 * No, but I'm saying they exist whether they're understood or not. Is this question related to this thread? -- Rick Block (talk) 01:17, 4 March 2009 (UTC)

I'm not trying to fix the P of an ordinary coin at 50/50. And if your coin lands on edge one time out of 10,000 (say) that would be due to its intrinsic properties as well. What you've agreed to here is that there are physical Probabilities at the same level as your distribution ratio - i.e. not due to some state of knowledge overlay. Next step (and I hope this isn't as hard or going too far), there are games we can set up that have P values that are similarly real. For example, if my "game" is to take the ace of spades out of a deck of cards and put it face down on a table its P value of being the ace of spades is 100%. Every time we play this game the result is the same. The result is caused by the rules. Similarly if I look through the deck of cards, put the ace of spades face down on the spot 1 and some other non-ace card face down on the spot 2, the card on spot 1 will always be the ace. No overlay can change this (unless we redefine "spot 1" or "ace" or "on" - but I'm not trying to trick you and would appreciate it if we didn't have to discuss pointless nonsense like this). Just like no overlay can change the fact that there are now 2 cards on the table and only one of them is the ace, so the ace/cards ratio is 1/2 (unless we similarly redefine "ace" or "card" or "2" - again, the point is let's not be stupid about this). None of these examples are "overlays" - they're all plain facts. We can verify the ratio by counting. We can verify the P value by designing an experiment that follows the rules lots of times and counting the results. If you and I both do the experiment we will get very similar results (perhaps not identical if the rules say anything about "random" - but if we repeat the experiment enough the results will converge). Do you accept that there are (at least some) P values associated with (at least some) game rules that are in the realm of "what is", i.e. intrinsically related to the game rules in the same way the P of an ordinary coin coming up heads is intrinsically related to physical properties of the coin? -- Rick Block (talk) 01:17, 4 March 2009 (UTC)


 * Rick, if you are saying that a binary attribute, such as yes, no; on, off; exposed, covered; stationary, mobile; known, unknown; knowable, unknowable; included, excluded, should be considered "intrinsic", I'd say yes. And yes too, to some multifaceted states of "is". 216.153.214.89 (talk) 04:40, 4 March 2009 (UTC)

Just to make sure you noticed, in the above there was an example of a "game" (2 cards, ace in spot 1, other card in spot 2) where the "distribution ratio" does not match the intrinsic P value. I'd like to start using the word probability now if you'll let me. My contention is that given the rules of this game the probability (lower case p) of the ace being in spot 1 is 100% and the probability of it being in spot 2 is 0%, and these probabilities are a direct consequence of the problem rules (not because of any "overlay"). If you know the rules, and you pick slot 1 you'll find the ace there. If someone does not know the rules and they pick slot 1 they'll find the ace there (100%, guaranteed). We could tell someone who does not know all the rules that one of the cards is the ace of spades and the other is not, and let them pick one. They would likely think that because of the plainly obvious distribution ratio they have a 50-50 chance of finding it (and, in fact, they do), but this doesn't change the fact that the ace is under slot 1 with 100% certainty (because of the game rules). The probability and the distribution ratio are both intrinsic in this situation, and they're not the same. Do you agree with everything I've said here (yes/no)? -- Rick Block (talk) 05:07, 4 March 2009 (UTC)


 * I'm not sure. I think there's two ratios there - ace/cards and not-ace/cards, both of which match P depending on which you are looking for, so I am not sure. As for the p being zero, it's only zero for a particular search, a search for ace. If we were searching for "not ace" it's 1, not zero, right? So why isn't p the consequence of our own logic premise; ie; search parameters? And isn't our parameters a data set in and og itself? I am not razzing you here, I am sincerely listening. Please give me a more clear example of distribution ratio as you might use the term. Let's say I have a two bay garage with 1 car in 1 bay. Is my DR car/bays or 1/2 or not? This is a question seeking clarity, not a "gotcha" 216.153.214.89 (talk) 17:52, 4 March 2009 (UTC)
 * Let's say you have a two bay garage but you always park on the left (your boat is typically on the right). Your DR (as you've defined it - the ratio of of cars per door) is 1/2.  But this is not the probability of finding your car behind either door.  The probability of finding your car behind the left door is 100% (due to the rule you always follow) and the probability of finding it behind the right door is 0.  This doesn't ever change.  It is the underlying reality (in exactly the same sense that there is only one car and two doors).  We can ask questions about this situation that might have different answers, but these would be "overlays".  For example, what is the chance of someone (who doesn't know your rule of always parking on the left) finding your car?  This is a probability question, but it is not asking about the probability of where your car is - it is asking about the probability of a random "guess" being correct.  This probability is always the same as the DR (cars/door) - but this doesn't mean the probability of each door is the same as the cars/door ratio.  There are two different probabilities here - the actual probability of where the car is (100% on the left, 0% on the right) and the probability of a random guess being correct (50% - which is actually the ratio of correct guesses to guesses).  We'll get to more interesting scenarios soon, but is this much understandable? -- Rick Block (talk) 20:45, 4 March 2009 (UTC)

I agree with what you are saying and thank you for acknowleding my conception of DR, but... if my DR is a valid concept, why fight me when I say the DR when a door it removed drops to 1/2? Is it because by your rules, the removed door, impossible though it is to be a location, still counts as an element of the location set? If so, that's why people get this wrong, they don't think this door still counts. Can we get some clarity on this point please? Also, the reason why I've kept saying "information layer" is I'm thinking like a dutiful jurist. If the judge says "you will not give facts A and B any weight while you deliberate"; if facts A nd B are essential to support the theory of the case, then I'm bound to return a verdict of "not proved". Doesn't matter to me if there's a photo of Joe with the ax in his hand, if the rules of evidence prohibt the use of that data, I can't reach a conclusion. Do you understand me better now? Joe may have in fact did it, but I'm not allowed to find that to be so if the evidence is disallowed. If the 1/3 freeze fact is allowed to carryover to choice 2 and the 1/3 door exclusion fact is allowed carry over to choice 2, then yes the odds can be calculated and declared. But if those facts are not allowed to carryover, then it doesn't matter what is true and it doesn't matter what I want to say is true, as an honest court's finding would have to be "not proved". It's like I would have my fingers in my ears saying "I can't hear you" when I refuse to acknowledge the odds. Sort of like an inventor with a new idea is sometimes shut out for political reasons - people are too wedded to the old ways (old set size). It's pretty clear that linchpin to this problem catching people is that fact #2, the 0 probability state of the removed door is an allowed fact in the odds calculation. If I'm allowed that fact, I can write a formula that zeros exactly in on the true odds of 2/3. If not, I can't refine my calculation any closer than to show 50/50 - regardless of the true odds. That said, what's next? 216.153.214.89 (talk) 21:09, 4 March 2009 (UTC)

Moving on

 * The problem with your DR is that it has the form of a probability (which is the ratio of outcomes/trials) and it is sometimes the same as the probability, but it isn't always the same as the probability. In fact, the confusion between this and the probability is the primary reason people arrive at the wrong answer for the MHP.


 * Your DR is the probability if and only if the distribution is random. The random distribution can be cars over doors (like at the start of the MHP at the point the player is making her initial selection and the probability of the car being behind each door is 1/3) or winning choices over possible choices (like "guessing" whether to switch after the host has opened a door in the MHP in which case the probability of the guess being correct is 1/2 - note this is not the same as the probability of the car being behind either door).  In the normal version of the MHP the final choice is not a guess and thinking about your DR simply interferes with the actual solution.


 * If you want to skip ahead and talk about the MHP, I'm willing but first do you accept that the probabilities associated with the MHP are a consequence of the rules that are specified and are in the realm of "what is" (yes/no)? -- Rick Block (talk) 11:03, 5 March 2009 (UTC)

Reply coming soon - more thinking needed. 216.153.214.89 (talk) 16:50, 5 March 2009 (UTC)

Rick, I hope you know that I already agree you are correct in the math and have said as much. What I am having trouble with is being thwarted at each point when I try to express my understandings. As for your last question, if you are saying that a door is not a "door", but rather a door is "a door's 'is' in each problem is defined by that problem's rules" then yes, I agree. Then my question becomes, aren't then the calculations the "information overlay" (invariable though they may be) which we lay on a circumstance? In other words, we see some cows in a field and we draw block diagram of them on paper. That's layer 1. Then, we decide what we are solving for and with the invariable math of probability, we assign values to those blocks. Now probability law tells us that the values we assign are the true values, and are an immutable attribute of the cows as they exist in their current set, but isn't it true that if we flow-charted the steps we took to create our finished calculation it's 1) map objects 2) assign values? This is what I've ben trying to say. And if by assign, you want to say "recognize", I'm ok with that. We still can't recognize values, immutable pre-existing ones or not, unless we go through a discrete step of applying a discernable data layer (our crib notes) onto our block diagram. The fact is, we can't know the probability without the crib note layer, which I call the information layer. It's what broadcasts back to us our knowledge of what "is". That's why I was saying that without information we can't know the odds. Hence, I felt it was on the information layer. Perhaps we should parse our vernacular and say that Probability is the immutable aspect which we seek to know, but the completed steps we undertake to create our knowledge-base of the Probability, creates a data set we should call "Odds". This could allow us to distinguish between what "is" and "what we know". The Probability "is", the Odds are "what we know" about the Probability. How's that? With this vernacular in place, even if sloppy, it helps me accept your views without objection. And to take this further, it would then be that when our understanding of set's numbers is accurate and when we apply the right formula, the Probability and the Odds will correlate exactly. And I'll further suggest that under this vernacular, the player in the game, when looking at the 2nd choice faces 50/50 "Odds" only until he applies the right numbers and calculations to his thinking - then he sees that the Probability is 2/3 by switching. Ok with that? And along these lines, my DR ceases to correlate to the "Odds" as soon as the "Odds" are refined correctly. Then what happens is that the Odds and Probability correlate and the DR doesn't. People are getting stuck when they stay with the DR rather than refine the Odds to match the Probability, which by the rules of the game, they are allowed to do. In this vernacular here, the DR (at any given step) is fixed and the Probability is fixed, but the Odds are transient. The Odds are what we know about the Probability, so it's the Odds which we re-assign as we learn more. If I were modeling my thinking with this vernacular as explained here, I'd say that the DR and Probability are on layer 1 and the Odds are on Layer 2, the overlay. Also it seems that the ratio of car/remaining doors (my DR) is irrelevant to solving the Probability calculation, yes? 216.153.214.89 (talk) 17:00, 5 March 2009 (UTC)


 * Now I think we're getting very close. It sounds like you're agreeing we can talk about the Probabilities inherent in a situation, controlled by the rules that created that situation but you'd still prefer to think of any computation of Odds as an overlay.  This view seems similar to the Bayesian probability view where all probabilities are formally expressed as probabilities given certain background information (usually denoted as I).  This is perhaps similar to what you mean as an "overlay" - however probabilities in a Bayesian analysis aren't "overlaid" so much as "derived from" ("overlay" implies multiple different overlays are possible, while Bayesian probabilities are the direct result of a given I).  In the case of the MHP, the background information includes the rules we're imposing on the host - from which the probabilities can be mathematically derived.


 * And, yes, the ratio of car/remaining doors in the case of the MHP is irrelevant to solving the probability calculation.


 * I think we're perhaps finally ready to talk about the solution. Assuming it's sufficient to analyze the problem only in the case where the player initially picks door 1 (we can renumber the doors if necessary) there are only 4 different events (in both the English and event (probability theory) sense of the word - not a coincidence) that can happen.


 * 1a. The player has picked the car and the host opens door 2.
 * 1b. The player has picked the car and the host opens door 3.
 * 2. The player has not picked the car and the host opens door 3.
 * 3. The player has not picked the car and the host opens door 2.


 * Assuming the player has initially picked door 1, these are the only possible outcomes. The game rules (I) say the host never reveals the car, so in case 2 the car is behind door 2 and in case 3 the car is behind door 3.  There are no other possibilities so the probabilities of these outcomes must add up to 1.  If we call the probability of each of these p(i), then because the car was initially placed randomly behind the doors we know:
 * p(1a)+p(1b) = p(2) = p(3) = 1/3
 * The completely unambiguous version of the problem statement says the host randomly chooses which of door 2 or door 3 to open if the player initially picks the car, so p(1) = p(2) = 1/6.


 * This is exactly what is shown in the figure in the article.


 * Are you with me so far? -- Rick Block (talk) 20:34, 5 March 2009 (UTC)

Yes, but... I've been stretching my understanding to appreciate your vernacular, but you not as much to reach mine. By "overlay", I mean (as a metaphor) a clear piece of plastic "overlayed" onto another page - like an overhead projector acetate sheet, except that I use it for a different purpose than overheads. The black wax-crayon markings go on the clear sheet. These markings are my thinking, my knowledge, my perceptions, any math that I do etc. When I think about a circumstance, I visualize the information I develop regarding it as an "overlay". Let me give you an example: I excel to almost a mistake-free level at looking at a given qty of material and knowing by looking if it will fit in a given space. This talent comes in very handy when I move the contents of a family member's home in any particular vehicle. As I impliment this ability, I do so by looking at (or remembering) the stuff; sofas, tables, boxes, bags, etc and projecting onto them my understanding of the cargo capacity of the vehicle. I see the size of the cargo capacity by simply looking and mulling over how big it is, then my assessment of the fit I establish by "overlay" - it's a 3-D knowledge grid that lands on the stuff as I measure in my mind. Combine this with the fact that I am very good at assembling and disassembling things and there's almost no pile of items that I can't fit in to maxiumum carry just by looking and starting. I never get "stuck" with an item that only fits in part way. If I pick it up, it fits in the van. I look at the item, my overlay of knowledge says, yes or no and I either fit it in the van or it can't be fitted in. In fact, this ability is so uncanny that I've had others involved in a move remark on it. Indeed, just the other day, I drove 25 minutes to go pick up a table which I hadn't moved in 8 years, driving a mini-van which was not the vehicle I originally moved the table in the 1st time. And yet, there was no doubt in my mind that it would fit in the van, which it did, at a slight angle, by just over an inch. Now how is that possible, except that I must be spatialy calculating the volume and shapes of the items with a high degree of accuracy? It's the same thing I've been trying to explain to you. Something may always be there, but you can't see it if you don't have the perception/detection/math "overlay" on top of the circumstance. I'm willing to agree that Probability is always there. Are you willing to concede that it can't be preceived except via calcuated "Odds" as I described in my immediately prior post (above)? And I am saying "overlay" for "Odds" because there is a quantity to this knowledge and I can see it in my mind as a formula, written out on the clear sheet. I see both calculations on that sheet. I see the DR calculation of 1/2 and I see my crib note version of the Probability formula which is 1-1/3(removed)=2/3<>1/3(stay);=switching is better. I see two calculations on that sheet, only one of which is right. And when I compare them under the rules of Probability math, I see that the DR doesn't apply, so the P formula is right. But what I know is on written the clear sheet, not on the 1st layer which is a block diagram of the doors. What you know goes on top of what is. You can't perceive what is except by what you know and they are not the same thing. Do you understand me now? Just in case, I'll give you one more illustration - here's a good one: A long time ago, I was a foolhardy youth on a "travel the states" journey, which mostly transpired in FL and CA and included over a year of living out of an army issue back-pack duffel bag. Now, I've heard it said that the prospect of hanging wonderfully concentrates the mind and I assure you that being homeless and hungry for long periods of time does as well. Suffice it to say, one day in South FL, on State Road 84 in Ft. Lauderdale, just after sunset, I was walking past a Wendy's burger place and I saw a nice car running with the keys and unlocked. Not being a car thief, I quickly dismissed the thought of boosting that car as a more convenient trasnport than walking the next 12 miles or so where I was heading. Anyway, as an alternative, I ducked into the Wendy's for a cup of chili, which was all I could afford. Now while I was in there, I was fortunate enough to sit near a family and be presented with an unforgettable vignette. Because the sun had set, it was dark out and with the lights on in the store, the plate glass windows had a perfect mirror effect going. Right at this point, the family's precocious daughter (about age 5) said in a clear voice "they look like us, but they're not us". It took me a moment to recognize that she was talking about the mirror effect reflection of her family in the window. So anyway, looking in the dark reflection of the window, you could see the family perfectly. In my way of thinking, the family is level 1, the base facts, the "is", but the reflection is level 2, the "overlay", our knowledge of the facts. They exist on distinct planes and are not the same. As I was seeing this MHP problem, until I was persuaded that the two disputed datums of the removed door 1/3 and the fixed choice 1/3 were allowable, then to me, those datums are like vampires, I can't see their reflection in the mirror and so they can't be part of my answer. It's not that they aren't sitting in the restaurant, but rather, I wasn't convinced I was allowed to refer to them.216.153.214.89 (talk) 22:42, 5 March 2009 (UTC)


 * I think I've been bending over backwards to accommodate your vernacular! I understand what an overlay is and that probabilities aren't directly perceived.  What I'm saying (and you will apparently never agree with it) is that probabilities are the Truth™ just as much as any other math.  It appears that I am unable to explain this to you to your satisfaction, so OK you win.  If you have any math literate friends, I suggest you talk to them about this.  I will note that your "correct" equation above (1-1/3=2/3) is not exactly the right way to look at this problem either (this approach falls down if the question is changed to what is the probability if Monty opens one of the unpicked doors randomly and it happens to be a goat).  I'm sorry I can't explain this better but there doesn't seem to be any point in continuing trying, so I won't. -- Rick Block (talk) 03:31, 6 March 2009 (UTC)

Rick, your problem is that you are an over-educated snob. I've been trying to explain to you how it is that people misunderstand MHP - how people leave their watch on their dresser at home by mistake, but you are obsessed with detailing to me the schematic for the guts of the watch. We can't communicate because you think that only your aspect of the conversation is interesting. You want to talk about the guts of the bananna, but I find the peel interesting too. And regarding "I'm sorry I can't explain this better" - I never asked you to. You chose to speak with me and I was happy to explain to you my thinking. So quit the conversation; I fully understand your point and did before we started talking. What remains for me is to consider is whether your inelastic mind was caused by the study of formulas, or is what drew you to study them? Depending on which it is, I may decide that learning more of the math involved here is harmful to one's mind. In the meantime, I am content to understand MHP (as presented in this article) as 1-1/3(removed)=2/3(remaining)<>1/3(stay);=better to switch. And Rick just like a bananna has a center & peel as two different things, whether you care to accept it or not, there are two levels here, and the 2nd one is an overlay which is the information/knowledge level. No matter how much you desire it to be so, you mind can never interoperate directly with the layer 1. So lust as much as you want for Probability access, but you will never get any closer than the information level. Everytime you reach to touch the Probability, you can only do it with information gloves on. If you can't see that, you might be brain damaged. You should read this book: The Man Who Mistook His Wife for a Hat as it's a good gateway into an area of understanding that might help you stretch your mind. Also, your assertions of being an advocate for "the Truth™" puts you in good company with scores of other one-note-charlie fantatics who care only about their narrow focus and nothing that abuts it. Sort of like this fellow. There's a good chance that he was, by virtue of his reading material, in possession of "truth". But what did he get for it? Rick, what good is your truth if you mind is so weak you can't help others appreciate what you like about that truth? Ok, so you know math formulas - so what? What good does it do you? Have ever even once in your life benefited in day-to-day life from your esoteric superiority in math? I doubt it. The formula detailed in this article is "true", but was in the end, applied in such a way as to cause nothing but trouble. Did truth help those financial lemmings? So tell me Rick, how do problems like MHP which are good at fooling people help anyone benfit from math? If I ask someone like you how to calculate a tip, you give me all kinds of formulas and skoff and my shorthand of "take 10% of the bill and add half of that amount to the 10%, there's your 15%." Why is there even a problem like MHP? I would never foist this on someone without pointing them in the right direction by stating "before saying 50/50, ask yourself 'Am I using all the available information in my calculation?'". Apparently, you think it's all good fun to see people misunderstand things. I, on the other hand, want to know why they do misunderstand and help them to not do it in the future. It's not enough for me to tell them the directions to Dallas, I want to know why they keep driving to Atlanta by mistake, so I can give better directions next time. As far as I am concerned, if you don't understand why people get something wrong, you can't really help them get it right. They can mimic what you show them, but if they can't articluate it in their own words, it's all for naught. Rick, you are an excellent mathemetician, but a very poor linquist and only a so-so conceptual thinker. So scurry off now and have a nice time brown-nosing your "math literate friends". I suppose I should call a this for your "literate" taunt, but frankly what good would it do? You'd just close your mind that much more as a result. That said, thanks for the chat. 216.153.214.89 (talk) 05:28, 6 March 2009 (UTC)


 * I think you're mistaking me for someone else, perhaps that fellow in the plate glass window. All human knowledge is an overlay on reality (including both your DR and probability).  I get that and agree.  Is that what you're looking for from me? -- Rick Block (talk) 15:26, 6 March 2009 (UTC)

Rick, sorry if I offended. My point is that lots of people who don't know Probability math are nonetheless very good at following a premise through to it's conclusion. I know I am. However, lots of people, including me, are dummies in Pmath compared to a guy like you. Therefore, clarity on why and how they have misframed the premise -explained in non-math terms- is what will help the non-math mind understand how and why they misunderstood this. Directions to Dallas won't help you - you already know how to get to Dallas. I am interested in seeing the addition of more information to help people not go to Atlanta by mistake. That said, kudos to you for not just ignoring me - I'll give you credit for that. It makes you more of a man than an ordinary knowledge-expert and I appreciate you for that. 216.153.214.89 (talk) 16:28, 6 March 2009 (UTC)

Final contention
Here's my final ccontention regarding this problem and my final assertion about how 50/50 creeps in here. The so-called sample space of three doors, I'll call "the 3 doors se"t is a fraud. It's only a set of three from the game player's perspective. From the host's perspective, it's always a set of two, with decoy adulterant added in. Because the host already knows in advance which doors don't have the car, the host always is able to remove one door from the game without removing the car. What the host actually has is a 2nd state door-choice set with a DR of 1/2 and a transient supra option which he can use to modify the 1st choice set. The host already knew in advance that the final choice would be between two doors, a 1 of 2 choice. The host is banking on the likelyhood that the player won't know how to do a probability calculation and won't know the rules of allowable data. As I said (far) above, the player seems to be doing only legitimate things, but now I see it's the host who pulls a fast one and tries to mask a one of two choice with auxillary data to confuse the player. But, our player armed with superior skills can determine the true 2/3 odds by using data which was gleaned from the 1st state and carried over to the 2nd state. MHP is legitimate as posed in this article, but I stand by my original assertion that the word "guess", if used, changes that - because plain guessing prohibits definate knowledge and calculations. If these are prohibited, I am back to the juror analogy. Also, here's a question: If the removed door has 0 probability when shown to not be it, then that means this was/is fixed always at 0, right? Suffice it to say, if MHP were to ask the question: Can it ever be true that a 1 of 2 choice doesn't have 50/50 odds, a lot more people would look for the 2/3 solution sooner.216.153.214.89 (talk) 21:33, 4 March 2009 (UTC)

population density survey model proves my point
This principle applies to MHP thusly:


 * A) In MHP, we start with an initial population density of 1 car to 3 doors, 1/3 (State 1)

This is an indisputable fact.


 * >>>Comment: Population density is an existing term. It means the function p, defined on the population (= sample space), assigning probability to each of the population members (= possible outcomes). Hence you have to start with defining your sample space.Nijdam (talk) 20:22, 3 March 2009 (UTC)
 * >>>>Comment reply: Njidam; You sound like the old Stan Freberg advertising skit where the people in the fake business meeting had their own ad which disclaimed the use of the word "hunger" by anyone else... the line in the spoof disclaimer said "Hunger, a patented trademark of Food, Incorporated". The fact is, the car is the population, and the possible locations are the base across which the density is spread. At 1st it's 1/3, then it's 1/2. The density is not measured across the removed door - that door is removed. It's like 1 cat across three houses is 1/3 until 1 house burns down, then it's 1/2. Or better yet, it's car per unknowm door status. And of course, when one door is removed its status is not unknown any more. So you certainly can't argue with 1 car / 3 unconfirmed locations, becomes 1 car / 2 unconfirmed locations. When we confirm the status of a door by removing it, we reduce the number of unconfirmed locations, do you deny this? Also, please answer below and stop interpolating your comemnts into my text. 216.153.214.89 (talk) 22:54, 3 March 2009 (UTC)
 * >>>>>Never heard of the man, and better keep it this way. The point is, I think, you're not a probabilist, or even not a mathematician. No offense, but putting the car as the population is complete nonsense. Look at it: a population consisting of one element. Hardly interesting, and not much to experiment with. You might mean something else, but not able to put it in words. Believe me: the car is not the population. Even the doors are not.Nijdam (talk) 22:25, 4 March 2009 (UTC)


 * B) Next, we move to a population density of 1 car to 2 doors, 1/2 (State 2).

Also an indisputable fact.


 * >>>Comment: Such a statement is meaningless, unless you mean: conditional on the next situation (= event) the conditional population density is ... But the answer is not 1/2 for each remaining door. Nijdam (talk) 20:22, 3 March 2009 (UTC)

At the initial population state of 1/3 (State 1), during the game, two actions are taken - both of which provide us information:


 * 1) A door is chosen. This provides us the information that the odds for that door are 1/3.
 * 2) A door is removed. This provides us the information that this door's odds are 0.


 * -- Armed with this information, we now move to the secondary population state of 1/2.--
 * (I've drawn a line to indicate the before removal (above) and after (below)


 * Now we are at the secondary population state of 1/2 (State 2)
 * While in this state, do we take any actions which develop new information?
 * The answer is no, we do not.
 * However, we are still armed with information developed above the line.
 * And, armed with that information, we can calculate the odds for each door.
 * Original door is fixed at 1/3 (fixed)
 * Original door and remaining doors together = 100% of car
 * 100%-1/3 (removed) = 2/3 which <> fixed 1/3, so we switch


 * So far so good, but this doesn't answer the question: Is it legitimate to retain the information acquired in density state #1 to do a odds calculation on the population density of density state #2 in efforts to locate the car? I've decided that this question is irrelevant as we already have enough information to think things through.

Evidenced by what transpires in the game (and is therefore allowable, because the host either allows it or makes it happen), is the player allowed to know that the door which is removed does not have the car? The answer is yes. And by the rules of the game, is the player allowed to keep in mind which door he originally chose? The answer is yes. This resolves whether or not the player gets to carry-over the information from state 1 to state 2. The answer yes, he does.

This leaves us with a final question: Do the rules of the game, as played out in the game, correspond with the rules of probability?

Yes and no. If we have data from a sample space, provided that the continuity of our knowledge about that space is not broken along the way, should the size of the sample space change, we can definitely still apply a calculation to the re-sized sample space.

However, because odds calculations are extrapolations, we must corroborate that the change in the sample space doesn't also change the population density in an unknown manner. And the removed door satisfies this test in that it changes the sample space, but it's very removal, known as it is, also for sure establishes the new density of 1/2.

All of that said, this is still a trick question as population density is a statistics fact and probability calculations (aka "odds") are the end result of a formula. Both aspects are present at once, so people fall into the trap of stating the density (a statistics answer) instead of calculating the odds (probability).

And, as I stated above, the formula to calculate the odds, which is: 100%-1/3(removed)=2/3<>1/3(fixed)=Switch, can only be constructed if we acquire information above the line (state 1) and apply that information when we are below the line (state 2). To me, when I apply information to a circumstance, I think of it as an overlay, but that's my choice. Everyone is free to model their visualizations of information as they see fit. Because the circumstances (number of remaining doors) changes, and the information previously acquired is still held, I see myself as applying acquired information to a new fact set as an overlay. Personally, I still think this is a trick question and there's no prep given to help the layman to distinguish between statistics and probability. So whooopee, there we have it, this question is effective in eliciting a statistics answer from most people, when it's asking for a probability answer. Oh super, aren't we clever, we can trick people. That said, I have no more complaints here, other than the distinction between statistics and probability is not mentioned in the article. All that said, as I contended above, the odds do indeed exist solely on the information "layer" (which I liken to an overlay) that's carried over from above the line to below the line. Without the discovered information acquired above the line, we can't, when below the line, possess the composed formula and the odds can't be calculated. Those odds can be assinged as a probability, but without the two informational elements, they could never have been calcuated and therefore would remain unknown to us.

This opens an entirely new debate - does probability exist beyond our knowledge? The answer to that is yes, it does, but it's an asinine excercize to try to conceive the unknowable. For in the end, the only probabilies which matter to mankind are those which we can understand by knowing. And this is why I insist on modeling my probabilities as being on my information layer. I am willing to admit that my knowledge has a finite quantity to it and all of it is connected back at some point to information. That which is not me, is an external fact set. That which I know is an information layer. When learning something new, I apply what know to what I am trying to learn, in multiple iterations, blending-in the increased knowledge each time, until I am satisified with the results. If you can't know it, you can't use it and unsuable knowledge is by definition, useless. I'm not interested to squander my limited thinking capacity attempting to know the unknowable. I prefer to apply multiple iterations of what I know works to get the results I seek. And I suppose, most people do the same. And since most of us -the ingorant unwashed- can't distinguish between probabilities from statistics, well, we end up with MHP so we can amuse ourselves guffawing over each other's ignorance. 216.153.214.89 (talk) 12:24, 3 March 2009 (UTC)


 * There are many things wrong with what you say here and we'll be getting to them in the thread above if you continue to play. The point of this response is not to pick a fight, but just to avoid the impression that a lack of response indicates agreement.  We're talking about all the same issues above, so I'd rather continue there.  -- Rick Block (talk) 01:57, 4 March 2009 (UTC)