User talk:Tweedledee2011/martin

invite to martin
Thank you.

From what I have seen you are misusing sample spaces. You can set up a sample space so that every element has an equal probability, and this is often a good idea but just enumerating the possible outcomes does not guarantee this fact. In you second sample space you assume, without proof, that all outcomes have equal probability. That is not the case. Martin Hogbin (talk) 11:25, 1 January 2014 (UTC)


 * Martin - You have landed on the crux of the issue. If the game player is allowed to use the previously acquired information, then it's possible to deduce that of the 2 doors, the previously not chosen one has better odds. And this is so for all the sound reasons the article gives. However, that still misses my point, so let me try again:

1) The first sample space contains all the possibilities which are 15, if you distinguish between the goats:
 * D1C D2Ga D3Gb
 * -- D2Gb D3Ga
 * D1Ga D2C D3Gb
 * D1Gb -- D3Ga
 * D1Ga D2Gb D3C
 * D1Gb D2Ga --

Note that I do not double-count the Car when the Goats switch positions - because the car doesn't move, so its possibility didn't change. So it's not 18 to start, but only 15.

And, if you do not distinguish between the goats, then the sample space has 9 possibilities:
 * D1C D2G D3G
 * D1G D2C D3G
 * D1G D2G D3C

2) The second sample space does not contain 15 or 9 possibilities. Rather, it contains only 4:


 * D1C D2G
 * D1G D2C

After a door is opened, it's definitely NOT this:


 * D1C D2Ga
 * -- D2Gb
 * D1Ga D2C
 * D2Gb --

That's because one of the goats has been removed, so it's not possible that either is still there. At this point, it's one or the other.

So then, to reason this through, let's start with the perspective of a new player walking onto the game. If a new player who has not made a choice #1 were to be given the choice #2, he would not have enough information to calculate that switching is better. For this player, the best calculation he could make would be 50/50. However, even though he could not calculate it (due to lack of information) switching would still be better, so if player #2 directed player #1 to switch, that would indeed win 2/3 times.

Now, let's equate the idea of a player #2 with a person trying to answer this problem. When a person tries to answer this problem, they might not be aware of the fact that they can use any information on the table. Rather, they might think that they have to pick as if they were a player #2 just walking onto the set. And if they do, they won't look at anything other than 2 doors, 1 car and conclude "50/50". This is why so many people get this question wrong.

Now, here is what I am saying:

Probability has formal rules of logic embedded in the definition of the terms:


 * 1) The sample space of an experiment is the set of all possible outcomes or results of that experiment.
 * 2) An outcome is a possible result of an experiment.
 * 3) An experiment is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space.
 * 4) An event is a set of outcomes of an experiment (a subset of the sample space) to which a probability is assigned.

If we stick to the literal meaning of those terms, because it's not a possible result of the 2nd choice to choose the removed goat or find the car behind the opened door, that door is not part of the second experiment. And if it's not part of the second experiment, then that raises this question: Is the second experiment discrete? If it is, then the information acquired from the opening of the door can not be used to calculate the odds of switching. Not if one is going to strictly adhere to the definition of experiment. There is not only one "well defined set of possible outcomes". Rather, there are two sets. The first sample space and the second. Either the definition of experiment is wrong on Wikipedia, or the experiments are discrete because they do not share the same set of possible outcomes.

Now do you understand what I am trying to say?

Tweedledee2011 (talk) 12:07, 1 January 2014 (UTC)


 * If you are saying what I think you are then you are quite right.


 * You are invited to answer the question from the point of view of the player. Thus you know everything that the player would be expected to know throughout the game.  The player knows that there are three doors, two goats, and a car.  The player is assumed to know the rules of the game which are that the host always opens and unchosen door to reveal a goat and always offers the swap.


 * With this information, the probability that the player will win by switching is 2/3. Do you agree?


 * Vos Savant discussed a different scenario in which a little green man appears on the stage after the host has opened a door. All this person knows is that one unopened door hides a goat and the other hides a car.  In that circumstance (with no information on what has gone before) the only assumption that the green man can make is that each door is equally likely to hide the door.  In this (unreal) case the probability that any chosen door hides the car is 1/2.  Is this your point? Martin Hogbin (talk) 12:30, 1 January 2014 (UTC)
 * We are getting close. If you've ben following this article as long as I have, then you would know that Massimo Palmarini Piattelli wrote about this very question back in 1991 In the March/April edition of Bostonia magazine. He also wrote a book, largely I think, based on feedback he got from the article. At the time of the article, I was one who wrote the magazine and in turn, was invited by Massimo to comment further. Recently, I was in touch with both Massimo and BU via email and I have in my possession a copy of the original article (get it here: https://www.transferbigfiles.com/download/file/92967483-c9dc-458a-982d-5e538b104ba4?rt=c1Kr42z--UzujIf4-hnAuQ2). If you read Massimo's article, in sidebar C he explains this problem, but he says "If, in your second and final choice, you have guessed correctly, you win...". But please note: The word guess means "to form an opinion or give an answer about something when you do not know much or anything about it". And this is why people get this question wrong: They are encouraged to guess, not calculate. However, this raises the question of what is the honest basis of calculation. I do not deny that switching gets 2/3 if the calculation is made using all the available information. However, I do question the admixture of the game example with probability theory. If this is a guessing game, then one does not calculate, one guesses - and it's 50/50 for the second choice. However, if this is a test of our ability to correctly calculate odds, then the definitions must be precisely applied. And as I stated above, it seems to me that the very definition of the terms used in probability prelude allowing us to use in our calculation, the information obtained from the open door. Unless the first choice and the second choice are partial steps in a single experiment, then it's illicit to use that information. That's why this is a trick question. Tweedledee2011 (talk) 12:54, 1 January 2014 (UTC)
 * It is certainly a very unintuitive question, probably the most unintuitive simple question in the world, but it is not a trick question. It is a rather vague question.  The most famous version of the question starts, 'Suppose you're on a game show...', and ends, 'Is it to your advantage to switch your choice?'.  It does not actually even ask for a probability.  In what is by far the most well known answer, vos Savant (who actually reworded the original question from Whitaker) says that that if you switch you have a 2/3 probability of winning the car and if you stick you have a 1/3 probability of winning.  In this she is totally correct.  Given the player's state of knowledge and what are generally taken to be the standard, but unstated, rules of the game (that the host always opens and unchosen door to reveal a goat and always offers the swap) the chances of winning by switching are exactly 2/3.


 * You seem to be addressing the question of what is the probability of winning the car for a player who chooses randomly between the two available doors. This is indeed 1/2 but I cannot see why you or Piattelli think that this is the question being asked.  It is certainly not how most people understand the question.


 * There are many possible variations of the game set-up, the game rules, and the assumed question that can be used to get out of the standard 2/3 answer. In my experience, most of these are used to try to recover some pride by people who got the answer wrong .  The facts are that most people assume the standard set-up, game rules and question (see vos Savant and Krauss and Wang) and that the answer to this question is 2/3. Martin Hogbin (talk) 14:51, 1 January 2014 (UTC)
 * Martin - Please re-read my comments, I am not (nor did Massimo) suggest what you are now saying. Rather, because the Bostonia article by Massimo (which I linked to - see above) clearly uses the word "guess" and an honest understanding of the word "guess" means to choose without making calculations, I am trying to be as precise as possible, hence the perspective that I've tried to add. But it's just that, perspective, not an alternative framing. The correct way to play this game is to make an educated guess - which is what one is doing when they make a choice bolstered by probability theory. Again, think of the impact of the word "guess" and what it means to people who take it literally - as I did when I first read Massimo's article. As for the question under discussion in this article, just 2 weeks ago, as a returning student, I passed my Quant Statistics (probability) class at a MA state university with a B+, so I am at least moderately versed in the basics here. And as such, I assure you that I am not denying that probability theory can be used to prove that the odds of switching are 2/3. However, what I am saying is that the verbal logic inherent in the definition of the terms (which I've explained above) suggests to me that the points I am raising, if examined more carefully, will reveal a) why this is a trick question and b) that the revealed reason(s) have more to do with a dishonest framing of the question than the inability of people to see the right answer. In other words, I am suspicious that this problem is more akin to a carny trick than a true example of erroneous reliance on intuitiveness. And if that's correct, then the article should have some information about those points (if and when I am able to make those points clear). That said, I need to think how better to state what I am trying to say as you keep fixing on the wrong points of my posts. Tweedledee2011 (talk) 18:01, 1 January 2014 (UTC)
 * I cannot understand why you, Piattelli, or anyone else are using the word 'guess'. It does not feature in any well known statement of the problem.  I have just quoted from what is generally accepted to be the standard problem statement, that of Whitaker/vos Savant.  There is no mention of the word 'guess' anywhere; you are asked simply 'Is it to your advantage to switch your choice?'.  In her famous reply, vos Savant confirms that it is to your advantage to switch and goes on to say that you will win the car with a probability of 2/3 if you do so.  She does not use the word 'guess' either. Martin Hogbin (talk) 18:37, 1 January 2014 (UTC)
 * Martin - Did you download and read the article form the link I supplied? I am not debating the merits of using the word "guess". Rather, I am only stating the fact that Massimo used it in his very notable article on the subject (a copy of which, you can get at the link I supplied). And as a result of his using that word, my initiation to this problem included my questioning the use of that word. In fact, that's what I wrote him about in 1991. I asserted that "guess" skews the premise and tricks people. That said, I am also concerned that this wiki article is also tricking people, but for a different reason: The failure to explain the change in the sample space, which I think is significant - but I have been unable to make you see that. Tweedledee2011 (talk) 20:11, 1 January 2014 (UTC)
 * You were quite right to question the use of the word 'guess' by Piattelli. It is the wrong word and is indeed misleading however, most people do not learn about the problem from his paper and the word is not used in the article so I do not see the problem.


 * You have not yet tried to explain why you think the wiki article is tricking people. The question is stated in its most well known form and the right answer to that question is given. Martin Hogbin (talk) 23:29, 1 January 2014 (UTC)
 * Let me think through my position more clearly and get back to you soon as I can Tweedledee2011 (talk) 00:28, 2 January 2014 (UTC)

Asymmetry in conditional probability
Yes and no. Tweedledee2011, your second "sample space" isn't addressing the famous paradox. Btw this pardox is a good way to illustrate "conditional probability". Your "second sample space" shows randomness and it does't depict the famous paradox, but it shows symmetry where there never is symmetry in that stage of the famous paradox. Your second "sample space" does not pay any attention to the fact that – as soon as the host has opened one of the two unselected doors – there is no longer randomness nor symmetry. From this point on there is asymmetry, because in 1/3 of cases the door offered to switch on contains the prize for sure, in the remaining 2/3 with a chance of 1/2, and in any case with probability 2/3 (as to the famous paradox). From another perspective it is zero in 1/3, and in the remaining 2/3 it's 1/1 each. But is never 50/50. See Leonard Mlodinow The Monty Hall Problem on youtube, min. 25:00 – 28:00 of total 51:35 min and The Drunkard's Walk: How Randomness Rules our Lives (ISBN 0-375-42404-0) deals with randomness and people's inability to take it into account in their daily lives (2008) chapter 3 p. 53–56. Please follow this fact in your mathematical considerations. Regards, Gerhardvalentin (talk) 13:20, 1 January 2014 (UTC)
 * Gerhard - thank you for your aid on the issue of "brain teaser". by keying in on that point, you are closest to getting what I am trying to say. Here's what I ask: Please use the convention of D1C, D2Ga, etc. which I have used above, and list for me below, all the possible outcomes of the second choice. Those possible outcomes will be the "sample space" of the 2nd choice, right? So then, if I am displaying the sample space of the 2nd choice wrong, please help me see it the right way. Thanks Tweedledee2011 (talk) 18:07, 1 January 2014 (UTC)