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I was wondering how we could think about the following variant of the classic Monty hall theorem,

Suppose now that the host actually does not remember what is behind any of the doors, once you choose one of the three doors, he will open at random one of the other two doors. If he reveals the prize the show ends, if he does not reveal the prize, should you still switch doors?

I was trying to formulate it but I am just not sure if I am making correct assumptions.

For example,

say we choose door $A$, then the probability that he opens door $B$ is equal to the probability he chooses door $C$ is equal to $0.5$.

The probability that the prize is behind door A, B and C is 1/3

the probability that Monty ends up opening any given door is also $1/3$

So should one just condition on something else now?

Looking forward to hearing any opinions on this,

Thanks

Quality
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3 Answers3

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Suppose you choose door 1. And Monty reveals door 2, but Monty has no prior information.

Then the conditional probability that the prize is in fact behind door one, given that it is not behind door 2 is $50\%$

If Money has no information, then not enough information gained by the reveal, to justify switching.

Doug M
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  • I don't see why this is downvoted. It looks correct to me. Anyone care to explain? – Patrick Stevens Feb 27 '17 at 23:14
  • The information of what is behind the door is still conveyed to the player irregardless of Montys prior knowledge (as long as it is clear that 1 and only one has a car). – mathreadler Feb 27 '17 at 23:21
  • $P(1|\neg 2) = P(\neg 2|1) \frac {P(1)}{P(\neg 2)} = \frac {1}{2} = P(3|\neg 2)$ – Doug M Feb 27 '17 at 23:24
  • @mathreadler The classical Monty Hall problem depends upon the condition that Monty has information that the player does not have. If Monty has no information, he imparts no ADDITIONAL information. There is enough information to say that the odd you have the right door have increased from $\frac 13$ to $\frac 12$ but not any information to suggest that $P(3) > P(1)$ – Doug M Feb 27 '17 at 23:28
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    Using undefined symbols does not really help much. (Also I did not do the downvote). – mathreadler Feb 27 '17 at 23:54
  • @DougM Hi, I also did not down vote, in fact, this seems to make sense to me. Because the possible occurrences that result in it being non beneficial seems to be the same as those where it is. Also you say not enough information to justify switching but would it also be true that it does not matter if you switch or if you dont – Quality Feb 28 '17 at 04:34
  • @Quality That is correct. – Doug M Feb 28 '17 at 17:18
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  1. We pick right at first (happens 1/3 of the time).

If that happens, then Monty will for sure show a goat and we will win 100% if we stay, 0% if we switch.

  1. If we pick wrong at first (happens 2/3 of the time).

    1. Monty now ends the game with 50% chance and neither loss or win.
    2. The other 50% we will benefit switching 100% of the time.

So with 1/3 we will lose for sure on switching if a goat is shown. With 2/3 if a goat is shown we win by switching. If a goat is not shown there is nothing for us to choose.

There is 1/3 stolen away if Monty shows the car. But that is nothing we can do anything about. The only thing we can consider is our initial 1/3 vs 2/3 and then given that a goat is shown, we will lose or win 100% respectively, so the conditionals are clear.

mathreadler
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  • Interesting that your interpretation gives differing results then some of the other answer . What do you think it is that is resulting in these differences ? Let me try to make sure I understand your method , In 1/3 cases we win by staying ( initial choice correct). In 2/3 cases we chose the initial wrong and in half of these , he will reveal a goat and we would want to switch to get the correct door – Quality Feb 28 '17 at 13:49
  • The method starts by analyzing the probability that the first picks (made by the player) are right or not. Yes so 1/3 to lose if switching comes from the branch where we pick correct initially and then a sheep is shown. But given that we picked wrong from start (which is 2/3), and added information from seeing one sheep will make us right for switching 100% of those cases. – mathreadler Feb 28 '17 at 13:52
  • Hmm , so what do you think is wrong with the answers that give that there is no benifit/loss to switch or not ? – Quality Feb 28 '17 at 13:55
  • As in almost all probability questions, there is a lack of carefulness in specifying the event space. Everyone else assume the game keeps going (with either default loss or default win) if Monty shows a car. Why would it? – mathreadler Feb 28 '17 at 13:57
  • But in the case that we dont lose, are there not equal amount of times that it is better to switch as it is worst to switch? – Quality Feb 28 '17 at 15:03
  • In the case that we don't lose? I don't understand. I think what makes the confusion here is that it is unclear in the question what happens and what we allowed to do or not in the case Monty exposes a car. The 2/3 chance of winning on switching which I try to explain is only if we consider the game to restart and "reshuffle" the doors whenever he shows a car. – mathreadler Feb 28 '17 at 16:00
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There is a long discussion of this on The Straight Dope by Cecil Adams. I will add the link here:

http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3

victoria
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