At MontyHallSolution, the following variation of the MontyHallProblem was posed:

	* There are 3 boxes, one  has the prize. '''No-one''' knows which (and you know that no-one knows).
	* Monty lets you choose one box. 
	* You choose a box.
	* Monty opens one of the other boxes - it's empty.
	* Monty tells you: keep the box you chose, or keep the other box.
	* What do you do?

Answer (a): It doesn't matter.

Answer (b): You should change your choice.

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The answer is (a). Note that Monty didn't say he knew which in the original problem, so the above only differs very subtly!

The key difference is that in the original MontyHallProblem, you choose one box, and then Monty picks a different box which is ''necessarily'' empty, regardless of your first choice.

In the variation, you choose one box, and then Monty picks another box which he cannot know to be empty. The fact that it turns out to be empty is a constraint on the whole problem, including your original choice of box. The right way to analyse this is to exclude all possibilities in which Monty's choice turns out to have the prize from the probability space, like this:

Assume without loss of generality that the prize is behind door #3.
* [1/3] You choose #1
** [1/2] Monty eliminates #2
*** stay with #1 loses; switch to #3 wins
** [1/2] Monty eliminates #3; '''excluded'''
* [1/3] You choose #2
** [1/2] Monty eliminates #1
*** stay with #2 loses; switch to #3 wins
** [1/2] Monty eliminates #3; '''excluded'''
* [1/3] You choose #3
** [1/2] Monty eliminates #1
*** stay with #3 wins (A); switch to #2 loses
** [1/2] Monty eliminates #2
*** stay with #3 wins (B); switch to #1 loses

To find the probability of winning by staying, we add the probabilities of A and B, and divide by the probability that the outcome was not excluded:
   (1/3 * 1/2 + 1/3 * 1/2) / (1 - (1/3 * 1/2) - (1/3 * 1/2))
 = (1/3) / (2/3)
 = 1/2

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''Probability of you winning this game overall: 2/3. If Monty does expose the correct door (1/3), you switch to it. If not (2/3), then your shot is (1/2).  1/3+(1/2*2/3) = 2/3.''

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'''An incorrect analysis'''

For this problem: Call the boxes A, B, and C. Without loss of generality, assume that you always pick A. Then the following are the possible games:

 A-prize B-empty C-empty; M picks B; stay->win; switch->loss
 A-prize B-empty C-empty; M picks C; stay->win; switch->loss
 A-empty B-prize C-empty; M picks B - excluded by hypothesis; never happens
 A-empty B-prize C-empty; M picks C; stay->loss; switch->win
 A-empty B-empty C-prize; M picks B; stay->loss; switch->win
 A-empty B-empty C-prize; M picks C - excluded by hypothesis; never happens

To see why this is different from Monty knowingly picking an empty box, consider the original problem tablified in the same format:

 A-prize B-empty C-empty; M picks B; stay->win; switch->loss
 A-prize B-empty C-empty; M picks C; stay->win; switch->loss
 A-empty B-prize C-empty; M picks B - excluded by hypothesis; never happens
 A-empty B-prize C-empty; M picks C; stay->loss; switch->win
 A-empty B-empty C-prize; M picks B; stay->loss; switch->win
 A-empty B-empty C-prize; M picks C - excluded by hypothesis; never happens

This of course is NOT a correct analysis, because when I try the problem (which I formerly thought I understood), I do get 2/3 wins with switching. What's wrong with it? I would guess that the first and second cases need to be counted as a single one, for some reason. The question is, why is this true if Monty knows the contents of the boxes, and false if he was guessing?  (If indeed it is.)  The information that the box he picked is empty is still available to you at the time that you decide to make the switch, regardless of whether it was available to Monty when he picked it.

''Because you pick a wrong box, his choice is predetermined by your choice. In other words you have correlated events, and not independent events. If you pick the good box, only then Monty's choice is independent of what you did.''

So?  Two possibilities if you pick the good box; one possibility for each of two bad boxes you might pick.  How does that answer the question?

''If Monty opens a box at random, and it's empty, that makes it more likely that you picked the right box to begin with.''

Contributors: AndrewMcGuinness, DavidSarahHopwood, anonymous

See also NotTheMontyHallProblem.
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CategoryMath