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I understand the common Monty Hall Problem and why switching provides a 2/3 chance of winning, but I'm having trouble wrapping my head around how the probabilities work when multiple players are involved, as their probabilities seem to be contradictory.

Let's say we have two players and four doors. Each door has a 1/4 probability of hiding a prize. Let's say contestant 1 chooses door A and contestant 2 chooses door B. Door D is then revealed to be empty, leaving A, B, and C. From contestant 1's perspective the odds are 1/4 for door A, 3/8 for door B, and 3/8 for door C. But from contestant 2's perspective the odds are 3/8 for door A, 1/4 for door B, and 3/8 for door C.

What are the actual odds for each door hiding the prize, does it change depending on if they know each other's choices, and does it differ from the normal example of the problem? If so why, and if not, why not?

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    $\begingroup$ A and B should both stay 1/4 while C should be 1/2, at least under the usual assumptions about the hosts behavior. 1/2 the time the car will not be selected by either contestant, and in that situation you will win if you adopt an always-switch strategy. $\endgroup$
    – guy
    Jan 20 at 3:12
  • $\begingroup$ Is that assuming both players are aware of each other's choices? $\endgroup$
    – Adam Vale
    Jan 20 at 3:14
  • $\begingroup$ You might want to clarify whether the contestants are both aware of everything or not, especially if not. $\endgroup$
    – Trevor
    Jan 20 at 3:21
  • $\begingroup$ I added that to the question. $\endgroup$
    – Adam Vale
    Jan 20 at 3:30
  • $\begingroup$ For Monty, the probability is $1$ for one of the doors are $0$ for all others, which also is different from any contestant's probability calculations, but that usually doesn't bother us. $\endgroup$
    – David K
    Jan 20 at 4:14
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In the case where both contestants can see everything, it's more or less the same as a single contestant getting to select two doors. The probabilities under both doors effectively freeze, and the the $1/4$ for door D flows to door C, giving C a $1/2$ chance of having the prize, leaving A and B at their original $1/4$.

In the case where contestants can't see what door the other contestant picks, then you have the situation where their door has $1/4$ chance, and the other two remaining doors both have $3/8$. It's fine that both contestants have different probabilities for the doors; it results from both the contestants having different information about the probability distribution, since only they know which door they selected.

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  • $\begingroup$ Ok, let's see if I understand it now. So if there was then a 3rd person observer who knew both of their choices, then they would know that the probability in my original example would be 1/2 for C, and then from each of the player's perspectives the 3/4 chance of it not being the one they originally chose would be divided unevenly (without their knowledge) as 1/2 for C and 1/4 for the other one they didn't choose. But the 1/4 and 1/2 still adds up to 3/4, leaving the odds properly distributed. Is that correct? $\endgroup$
    – Adam Vale
    Jan 20 at 14:36
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If the players know what the other chose then the options are (and the actual probabilities that we know):

Prize Behind Door A: C is shown. $1$ in $8$. (Didn't happen)
Prize Behind Door A: D is shown. $1$ in $8$.
Prize Behind Door B: C is shown. $1$ in $8$. (Didn't happen)
Prize Behind Door B: D is shown. $1$ in $8$.
Prize Behind Door C: D is shown. $1$ in $4$.
Prize Behind Door D: C is shown. $1$ in $4$. (Didn't happen).

Of the ones that happened. Prize behind door A, or B are each $1/4$ and behind $C$ is $1/2$.

But if the players don't know what the other picked then from Player 1s perspective the probability as he knows is:

Prize behind Door A; other player choose A: D (one of 3 options) is shown: 1 in 48
Prize behind Door A; other player choose B: D (one of 2 options) is shown: 1 in 32
Prize behind Door A; other player choose C: D (one of 2) is shown: 1 in 32
Prize behind Door B; other player choose A; D (one of 2 options) is shown: 1 in 32
Prize behind Door B; other player choose B; D (one of 2 options) is shown:1 in 32
Prize behind Door B; other player choose D; D (only option) is shown: 1 in 16
Prize behind Door C; other player choose A; D (one of 2 options) is shown: 1 inn 32 Prize behind Door C; other player choose B; D (only option) is show: 1 in 16
Proze behind Door C; other chose C; D (one of two) is shown: 1 in 32

So the weighted options are A: 1 in 4
B: 3 in 8
C: 3 in 8

Why that differs from the actual probability is simply the player is lacking information we have.

Knowing information always changes probability.

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  • $\begingroup$ Ok, let's see if I understand it now. So if there was then a 3rd person observer who knew both of their choices, then they would know that the probability in my original example would be 1/2 for C, and then from each of the player's perspectives the 3/4 chance of it not being the one they originally chose would be divided unevenly (without their knowledge) as 1/2 for C and 1/4 for the other one they didn't choose. But the 1/4 and 1/2 still adds up to 3/4, leaving the odds properly distributed. Is that correct? $\endgroup$
    – Adam Vale
    Jan 20 at 14:31

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