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I have a model of the Monty Hall problem that as far as I know is standard: three doors, the contestant chooses one at random, then if Monty has a choice (i.e., the contestant has chosen the door with a car) he chooses his door to open at random, and events are generally independent.

For the contestant, a strategy of always switching doors gives a $2/3$ probability of winning, and a strategy of never switching doors has a $1/3$ probability of winning. Intuitively the switching strategy is better because in order to win, you need to have choosen a door with a goat behind it; for the non-switching strategy, in order to win you need to have choosen the door with a car behind it. Since you're more likely to choose a goat door than the car door, you're better off switching.

I have two further questions about this:

  1. I think it's clear that the contestant's "gain" from adopting a switching strategy is 2, in the sense that over many trials, the contestant is twice as likely to win by using the switching strategy. However, is there a way to compute a "gain" given that the contestant will only play the game once?
  2. How would we prove that the strategy of switching is optimal? I have searched but not found anything. Obviously if you only consider the two strategies, then computing the $2/3$ and $1/3$ probabilities above constitutes the proof. However, you could open up the model to allow for weird strategies like "Always switch when selecting door 1 and Monty opens door 3, but switch only half the time when Monty selects door 2", and so on. I am not sure how to consider these strategies given my model of the problem.
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  • $\begingroup$ We already know that probability of winning if we switch is $\frac{2}{3}$ and if we don’t is $\frac{1}{3}$. Say we switch if event $X$ occurs and don’t switch otherwise. The probability of winning is then $\frac{1}{3}\leq P(X)\cdot\frac{2}{3}+(1-P(X))\cdot\frac{1}{3}\leq\frac{2}{3}$ $\endgroup$ Sep 15 '21 at 1:45
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    $\begingroup$ For your first question, as you know, the probability of winning is 2/3 when switching. You need to define "gain" for your question. $\endgroup$ Sep 15 '21 at 2:15
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    $\begingroup$ Be careful. Monty is not choosing randomly. He has to know where the prize is or you would end up throwing a third of the games away. In that case, the probability would be $\frac{1}{2}$, not $\frac{2}{3}$. When the contestant's first guess is correect, Monty has a choice of two doors he can open and that can be random. $\endgroup$
    – John Douma
    Sep 15 '21 at 3:12
  • $\begingroup$ Monty has been the subject of at least a dozen questions on this site, so by now I suspect every possible question about his doors has been asked and answered here. Please do a search for one that answers your questions. $\endgroup$ Sep 15 '21 at 3:41
  • $\begingroup$ You might be interested in looking into the notion of "expected value". It allows you to formalize the expectation of the always switch strategy, and of the not-switching strategy (and also of other strategies that are more peculiar, whose value will be between the other two). About your first question: there are lots of paradoxes that come out of the fact that, as you mention, expected value is really about what you expect on average if you run the experiment many times, which may not be possible in practice. (Though no such paradox appears here.) $\endgroup$ Sep 15 '21 at 4:07
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However, you could open up the model to allow for weird strategies like "Always switch when selecting door 1 and Monty opens door 3, but switch only half the time when Monty selects door 2", and so on. I am not sure how to consider these strategies given my model of the problem.

Don't.

Strategies should only be considered when they are effective: ie their condition informs your expectations.

You do not have any information that Monty prefers not to select door 2 unless he has to. So you should expect that always switching when this happens is exactly as effective as always switching when it does not.

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OP: How would we prove that the strategy of switching is optimal?

Here's my visual representation of the classic Monty Hall game. It extends the traditional probability tree by framing the contestant's winning decision as the third trial of the probability experiment.

Here—without loss of generality—the contestant has chosen Door $1.$ From the diagram:

  • the sample space is $\{12c,121,13c,131,23c,231,32c,321\}$
  • $P(\text{wins by sticking to Door $1$})=P(121,131)=\frac13$
  • $P(\text{switching to Door $2$ or $3$})=P(23c,32c)=\frac23.$

Alternatively, frame the game as a four-trial experiment, with the first trial being the contestant's initial door choice and the subsequent trials as above. Then there are $24$ outcomes, but $$P(\text{wins by sticking to initial door choice})$$ and $$P(\text{wins by switching door})$$ remain the same as above.


OP: and events are generally independent

You must be mistaken: the events (subsets of the sample space) associated with this game are not generally independent, and the trials are mutually dependent (however, the contestant's initial door choice and the car location are pairwise independent).

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