# Probability question regarding Monty Hall problem

I understand how switching is the best strategy and it doubles one’s probability of winning, but what would the probability be if Monty randomly opened a door and it happened to reveal goat?

In that case, I’m guessing switching doesn’t matter. Whether you switch or not, the probability of winning is 1/2. Is this correct?

In this case, you have a 1/3 probability that you will win by switching, a 1/3 possibility that you will lose by switching, and a 1/3 probability that Monty accidentally reveals the car. The overall probability of getting the car then depends on what happens next. If you get the car when it’s revealed, then your overall probability of winning will obviously be 2/3. If Monty makes you do “do-overs” until such time as a car is not revealed by the door opening, then the overall probability of winning approaches 1/2.

The question you seem to be asking is, “Of the games in which Monty flips a coin and reveals a goat, what’s the probability of winning by switching?” This would be 1/3 (the number of games you win by switching) divided by 2/3 (the number of games in which Monty randomly reveals a goat), or 1/2 as you expect.

Finally, if Monty’s choice is truly random, then the strategy of staying with the original door is just as good (in either scenario) as switching.

I just spent about five minutes going over all of that, and Ididn’t see where my question was addressed.

Huh? What does “If you get the car when it’s revealed” mean?

Why approaches 1/2? Isn’t it exactly 1/2?

If Monty accidentally reveals the car by showing you a different door (that is a possibility, if he is truly guessing randomly), do you win, lose, or get a “do-over”? If a win, then you have a 2/3 chance of winning the game. If it’s a do-over, the probability of winning approaches 1/2

That’s mathematical language designed to account for the fact that the game could theoretically go on forever in an infinite series of do-overs (if Monty were extremely unlucky enough to always end up revealing the car). If, more practically, we just quit the game after N do-overs, you must account for this quitting probability as one of the outcomes in confirming the probabilities all sum to 1, e.g. if N=10, probability of winning =.4995, probability of losing =.4995, probability of quitting = .001. No matter how high N is, its probability will still be non-zero.

Ahh, got it.

Interesting. Is there a way to calculate how close to 1/2 the probability would be if the rules stated the game would go on until Monty didn’t randomly reveal a car (and they are immortal, the Sun will last forever, etc.)?

Isn’t this analogous to .999… = 1 ? .999… Doesn’t approach 1, it is one.

The probability doesn’t approach 1/2, it is 1/2. No?

No. The probability is some function of the number of trials, n, and as n becomes arbitrarily large, the probability becomes very close to 1/2. It’s analogous to how .9999…9 approaches 1 as the length of the … increases.

So, I’m guessing there’s no way to calculate how close to 1/2 the probability is, because “for infinity” is not a number?

No, you can work out the expression for the probability in terms of the number of trials, even if I don’t feel like doing it right now.

This seems a bit odd… the OP presumably doesn’t want the probability that one wins the car in N trials (over some variable N); he wants the probability that one wins the car ever. And this is exactly 1/2; granted, there is some chance that trials will continue indefinitely without resolution (probability 0, but still possible), but that needn’t interfere with our ability to answer the OP’s question.

I mean, one doesn’t generally restrict oneself to saying a flipped coin has probability merely approaching 1 of eventually landing on heads, or merely approaching 1/3 of first landing on heads after an odd number of tails, or whatever. One says these probabilities are exactly 1 and exactly 1/3; granted, this is because “What is the probability that a flipped coin will land on heads within N trials?” approaches 1, and so on, but all the same, “…ever?” and “…within N trials?” are different questions, with different appropriate answers.

Yes, this is correct, given that you want to analyze the game as not stoppable after any finite number of trials except via a car being discovered, one way or another. [Which is sensible; the introduction of the idea of quitting after N trials no matter what, for some variable N, has been an unfortunately distracting and needlessly obfuscating red herring)

I’m confused. How can the probability be 0 and still be possible? How can it be possible that the trials will continue indefinitely without resolution and still be exactly 1/2?

I’ll address your first question in a second. As for the second question, I’m confused by what you mean by it. It’s possible that the trials will continue up until a point at which you pick the wrong door and don’t get the car. It’s possible that the trials will continue up until a point at which you pick the right door and do get the car. It’s possible (on the orthodox view) that the trials will keep going forever with Monty constantly fucking up and having to institute a do-over, thus preventing you from ever even getingt to pick a door at all [which is another way for you to not get the car]. Each of these three possible outcomes has some probability of occurring. Your original question was about the probability of the middle one; as it happens, that is 1/2.

As for how events of probability 0 can be possible, the orthodox mathematical formulation of probability does not equate “probability 0” with “impossible”; all impossible events have probability 0, but not all events of probability 0 are impossible. That is, more things have probability 0 than just the impossible ones. For example: take a sphere and think of it as giant infinite-sided die, the faces of which are points on the sphere. We’re going to toss it up and let it land on one of them. All the points on the sphere are equally likely to be the one landed on; therefore, for any particular point, the probability that it is landed on is less than 1/2 (since there are more than 2 points), and less than 1/3, and less than 1/4, etc. The only nonnegative real number satisfying those constraints is 0; thus, every particular point on the sphere has probability 0 of being landed on. All the same, some particular point will be landed on. An event of probability 0 will happen. Not all events of probability 0 are impossible.

Granted, this means the conventional mathematical formulation of probability can assign the same probability of 0 both to possible and impossible events, but this coarseness isn’t generally considered problematic. Probability isn’t about addressing questions of what is and isn’t possible, it’s about addressing questions of how probable things are. If you want to analyze the former, you need to go get another tool (or, what amounts to the same thing, you could modify the conventional mathematical formulation of probability to better address your needs/desires).