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According to the Monty Hall Problem, I select a door, say door number $1$. Monty then opens $3$ of the remaining $6$ doors that do not contain the prize. The question is should I switch.

Note that all the doors are said to have equal chance of containing the prize.

My attempt: $$P(1) = \frac{1}{7}$$

Once Monty opens three doors, we know that none of them contain the prize, so, we update our beliefs $$P(i|j, k, m) = \frac{1}{4}$$

Thus any of the closed doors has probability $\frac{1}{4}$ of containing the prize, so switching is pointless.

What am I doing wrong here?

SQB
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  • Assuming Monte know where the prize / goat is, and will always mess with you opening doors that do not have a goat behind them. If you don't switch you have a $\frac {1}{7}$ chance of winning. – Doug M Jan 06 '18 at 03:39
  • @DougM That is the prior. Why do I care about the prior when I have learned that the three doors Monty opened don't have the prize. –  Jan 06 '18 at 03:40
  • You shouldn't be updating your priors if Monty can open doors without revealing the location of the prize. – Doug M Jan 06 '18 at 03:41
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    @DougM Can you please elaborate a bit on this? If I have learned that three of the doors do not contain the prize, why should I not be updating my priors? –  Jan 06 '18 at 03:50
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    Possible duplicate of Monty hall problem extended., going by the accepted answer (mine), which provides for $n$ doors of which $k$ are revealed. – SQB Jan 28 '19 at 12:10
  • Suppose Monte opens 5 doors and you know he will do that regardless of your first pick, always revealing 5 losers. Knowing he will always do that, do you think you now have a 50/50 chance of being right in your first pick? Imagine it for 25 doors with him revealing 23 losers, etc. – Ned Jan 28 '19 at 15:08

1 Answers1

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Under your formulation of the problem, there are seven possible positions for the prize, each equally likely. Under only one of those positions was the prize behind the door you selected (the first door). Thus far, prior to revealing any doors, you should agree that the probability of winning is $1/7$.

Now, the opening of the other doors does not change the fact that in only one of these seven scenarios, you would win. For example, if the prize was behind door #3, the host opens any other three doors excluding #1 and #3, and you would lose by staying with #1. We can enumerate the outcomes after the non-winning doors are revealed. Let $P$ represent the location of the prize, $*$ represent an opened door (which has nothing behind it), and $E$ represent an empty but unopened door. Then the following list might represent the instructions to the host for opening the required doors--note that he never opens door #1 nor a door labeled $P$:

$$(P, *, *, *, E, E, E) \\ (E, P, *, *, *, E, E) \\ (E, *, P, *, *, E, E) \\ (E, *, E, P, *, E, *) \\ (E, E, *, *, P, *, E) \\ (E, *, E, *, *, P, E) \\ (E, *, *, E, E, *, P) \\ $$ So for example, if the prize is behind door #3, and you always choose door #1, the host would open doors #2, 4, and 5. It becomes clear that both before and after opening the doors, each of the outcomes is equally likely. If you stay with door #1, you win only $1$ out of $7$ times. If you switch to any other unopened door, you would lose with certainty $1$ out of $7$ times, and in the other $6$ cases where the door you originally chose did not have the prize, you now have a $1/3$ probability of selecting the correct door; thus the probability of winning by switching is $0(1/7) + (1/3)(6/7) = 2/7$.

heropup
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