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The Monty Hall Problem - Am I stupid?

UPDATE: I get it now. Thanks to MHW, Jan-Helge and Wikipedia :O) I got way too hung up on isolating each choice. If the number of choices is expanded to, say, 1 million doors, then it’s easier to not overlook the probability of the first choice. If you have 1 million doors, there’s only a 1/1,000,000 chance of getting the right door the first time, and then the game show host removes all but 2 doors, so that makes it quite intuitive that changing doors gives you a good chance of choosing the right one. For some reason this is quite easy to overlook when there are only 3 doors in the first place. Some kind of mental misdirection going on there.

ORIGINAL POST:

The Monty Hall Problem is fascinating, and frustrating. I get the math, I just don’t agree that it solves the particular problem at hand. But I see that it actually does solve it. But it really doesn’t. Confused? Read on…

When presented with the option to switch doors, you have in practice re-started the game with two doors - one of which hides a car and one which hides a goat. This gives a 50/50 percent chance of choosing the door hiding the car.

I fail to see the practical relevance of whether the host (of the game show) knows which door the car is behind. No matter what happens, you end up with a choice between two doors, one of which hides a goat and one which hides a car. There is no other possible final choice in this “game”.

So, in practice, this is a simple probability without replacement problem. You have a jar with 2 white marbles and 1 black marble. The probability of choosing a white marble is 2/3 and 1/3 for a black marble. The host removes a white marble for you. You’re left with a 1/2 chance of a white marble and 1/2 chance of a black marble.

The Wikipedia article points out that it’s never disadvantageous for the player to switch, because the probability of getting the car is always at least 1/2. I simply mean to say that, as far as I can tell, the probability is always, in practice,  exactly 1/2.

This whole thing comes down to how you define the problem. Yes, if you choose one door, then the host chooses another door, then you switch, there is a 2/3 chance that switching will get you the car - theoretically speaking, because you can’t (and wouldn’t want to) choose the door the host chose, and that door is still defined as being in the game. However, keeping the door the host chose in the game makes no sense precisely because it’s already opened to reveal a goat at the time you are to make your final choice and thus gives you 2/3 chance of getting the car if you don’t switch and a 2/3 chance if you do switch. That obviously doesn’t add up. So the game is, effectively, like I noted above, reset and becomes a new game with only two doors.

This is starting to look like one of those “magic” quantum theory things, where the right answer is that there is a 2/3 AND a 1/2 chance of getting the car (NOTE: I do realize this doesn’t actually have anything what-so-ever to do with quantum theory).

For fun I whipped together a little Java app to simulate 1 million rounds of this game, using the theoretical version, and sure enough, switching doors results in winning 2/3 of the time. So even though both definitions of the game, and their corresponding answers, can be said to be correct, it seems only the theoretical and least intuitive one is available to us in practice - which makes no sense to me.

Hey, I never promised you’d be less confused when done reading…

If you happen to know why my assertion that this problem can also correctly be defined as a simple probability without replacement problem is false, please feel free to explain it in the comment section below. Because that’s either the case, or this problem does indeed have two mutually exclusive correct answers - which, well, my rationality-loving mind would have a hard time accepting.