User talk:Martin Hogbin/Monty Hall problem/dissenters

Please discuss questions and answers here.

K & W Statement is conditional
The unambiguous formulation given by Krauss and Wang must be treated conditionally ''Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?''


 * Martin, I'm sorry to find you here under "disagree", because it is impossible to find a solution not based on some conditional probability. But, then, please prove it to me. Nijdam (talk) 12:54, 28 December 2009 (UTC)


 * Still no one has yet given me a definition of what constitutes a condition but let me, for the moment, accept that there is a condition in that the host must choose one door or the other.


 * So I agree (for the moment) that the problem is, strictly speaking, conditional but I assert that it is not necessary to treat it this way. The host choice is defined to be random thus his action cannot transmit any information to the player and thus change the odds that she originally chose the car.  We can therefore safely ignore the condition in the interests of clarity of explanation, even though it might be considered strictly necessary.  As you know, I would be happy to have a footnote in any simple unconditional explanation, stating that some people consider the problem one which must be answered using conditional probability and referring them to such a solution. This would seem a good compromise to me.  Would you accept it?  Martin Hogbin (talk) 15:25, 28 December 2009 (UTC)

One last effort: possible ways of the game with chosen door 1:


 * unconditional:
 * chosen door 1 car behind 1 opened door 2 chance 1/6
 * chosen door 1 car behind 1 opened door 3 chance 1/6
 * chosen door 1 car behind 2 opened door 3 chance 1/3
 * chosen door 1 car behind 3 opened door 2 chance 1/3

after door 3 has been opened:
 * conditional
 * chosen door 1 car behind 1 opened door 3 chance 1/3
 * chosen door 1 car behind 2 opened door 3 chance 2/3

Do you notice the actual reduction? Please do try to understand? Nijdam (talk) 15:42, 28 December 2009 (UTC)


 * I understand your point perfectly. The sample set has been reduced by the host's choice of door thus the problem is conditional, but I am asking for some leeway to initially present a solution that avoids this complication.  So we might have


 * chosen door 1 car behind 1 opened door either 2 or 3, it makes no difference which  chance 1/3
 * chosen door 1 car behind 2 opened door 3                                            chance 1/3
 * chosen door 1 car behind 3 opened door 2                                            chance 1/3


 * I agree that we may not be able to easily represent this in 'proper' notation but it simplifies the problem enough that most people should have a chance of understanding it. You have forgotten how difficult and confusing t his problem is for people who have never seen it before.  So my proposed solution is not rigorous but I intend to cover that by having a footnote to a more rigorous one for those (probably very few) who are interested.


 * What you want to do may be strictly correct but it is pointless, as most readers will not be able to understand it, even though you find it quite simple. We are writing for both the statistics purist and the average person.  Martin Hogbin (talk) 16:24, 28 December 2009 (UTC)


 * It is very common, indeed necessary, to vary the degree of rigour to suit the intended audience. Let me quote from one of my old mathematics (for physicists) text books: The degree of rigor to which we have aspired is that customary in careful scientific demonstrations, not the lofty heights accessible to the pure mathematician.  For this we make no apology; if the history of the exact sciences teaches us anything it is that emphasis on extreme rigor often engenders sterility.  We trust, of course, that our efforts to avoid rigor mortis has not brought us dangerously close to sloppy reasoning.  In other words the rigour has been toned down to suit the needs of the intended audience, in this case physicists.  Our audience is the general public and we need to ensure that the solutions and explanations are clear and convincing to them.  In our case we have the advantage that we can present a more rigourous solution later. Martin Hogbin (talk) 20:49, 28 December 2009 (UTC)


 * If you understand the situation perfectly, why then insisting on the "unconditional solution". There is none. The only point is how to bring this without being formally technical. Nor me or Rick insist on mentioning even the word "conditional". We only will make clear that something like the "combined doors solution" is not complete. Or that the following reasoning: chosen door has chance 1/3 to hide the car, hence(?) the remaining door must have chance 2/3, is not complete, but should read like: chosen door has chance 1/3 to hide the car, opening of the goat door does not affect this chance, hence ... the remaining door must have chance 2/3. Nijdam (talk) 08:15, 29 December 2009 (UTC)


 * I did already (and also in an earlier stage) made a suggestion on the "construction" page.Nijdam (talk) 08:19, 29 December 2009 (UTC)


 * I am perfectly happy with your suggestion. All I want to do is not complicate the first part of the solution with mention of 'conditional probability' or 'host door choice', this includes the diagrams. Martin Hogbin (talk) 11:01, 29 December 2009 (UTC)
 * Just to explain something. I have been using the non-standard term 'non-conditional' to mean 'without specific mention of the issues of conditionality', or 'where the conditional nature of the problem has been ignored or circumvented', just as you have done above.  I do not mean 'unconditional' when I use this term.  Martin Hogbin (talk) 11:15, 29 December 2009 (UTC)


 * Rick nor me has ever insisted of the mentioning of the word "conditional". We only want to emphasize in the article that the simple solution, as formulated by some authors, is not complete, which is mentioned by more than just one or a few of the sources. And in the discussion we base this on the need of conditional probability.Nijdam (talk) 12:40, 29 December 2009 (UTC)


 * I am OK with that, provided that it is done in a low-key way that does not detract from the clarity of the solution. Maybe it could be mentioned only at the end of the section or perhaps by means of footnotes.  It must also be pointed out that only some formulations of the problem are conditional.  For example this, perfectly reasonable, interpretation of what Whitaker's meant to ask, You will be offered the choice of three doors, and after you chose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch is described by Morgan as the unconditional problem.  There is also the case where no door numbers are given, where there is a dispute as to whether this case is conditional.


 * We might also point out that even in the, agreed strictly conditional, K&W formulation there is a simple and obvious symmetry which means that the conditional case must give the same answer as the unconditional case. Martin Hogbin (talk) 12:58, 29 December 2009 (UTC)

Pardon The Intrusion (Once Again)
Celebrate the simple solutions in the article! Without them, there is no joyous paradox!

Explain that without the symmetry (& total law of prob) & 100% condition there's *is* no "2/3 & 1/3 vs 1/2 & 1/2".

Carlton's tree is a great visual tool for emphasizing the "peaceful co-existence", don't you think? Glkanter (talk) 19:31, 23 January 2011 (UTC)