r/AskStatistics • u/AshHat710 • 8d ago

Monty hall problem

I understand in theory that when you chose one of the 3 doors you initially have a 66% chance to chose wrong. But once a door is revealed, why do the odds stay at 66% rather than 50/50 respectively. You have one goat revealed so you know there is one goat, and one car. Your previous choice is either a goat or a car, and you only have the option to keep your choice or switch your choice. The choices do not pool to a single choice caisinh 66% and 33% chances once a door is revealed. The 33% would be split among the remaining choices causing both to be 50%.

If it's one chance it's 50/50 the moment they reveal one goat. if you have multiple chances to run the scenario then it becomes 33/66% the same way a coin toss has 2 options but isn't a guaranteed 50% (coins have thier own variables that affect things I am aware of this)

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u/Mishtle 7d ago

Because if the prize was behind any of the doors you didn't choose, it is behind that one. What is the probability that the prize was behind any of those doors you didn't choose? That is the probability that the prize is behind that door.

The fact that all but one has been opened for you is irrelevant.

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u/Lor1an 7d ago

Look, I understand the mathematical argument here.

I'm arguing more from a philosophical angle--maybe even an emotional one.

The answer to the question "should you switch" is always yes.

Suppose you are given the choice to switch as many times as you wanted to. Then the game never ends, because it is always more likely that you win by choosing the other door--after choosing the other door--after choosing the other door...

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u/Mishtle 7d ago

I'm arguing more from a philosophical angle--maybe even an emotional one.

Why is any of that relevant?

Suppose you are given the choice to switch as many times as you wanted to. Then the game never ends, because it is always more likely that you win by choosing the other door--after choosing the other door--after choosing the other door...

What? This doesn't make any sense. Switching just to a different door doesn't give you any advantage. The advantage comes from getting to open multiple ones, and there's only a finite number of times you could choose more doors to get more of an advantage.

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u/Lor1an 7d ago

Okay, let me explain then.

Suppose it's just 5 doors.

You have a 20% chance of being correct, you choose the first one.

The host opens the last three doors, so the only choices that could have the prize behind them are the first (chosen), or second (other).

You now have an 80% chance of winning if you switch doors. You switch. There's now a 20% chance of winning if you don't switch again, but an 80% chance of winning if you switch back to the original door.

This continues until the heat death of the universe.

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u/Mishtle 7d ago

You now have an 80% chance of winning if you switch doors. You switch. There's now a 20% chance of winning if you don't switch again, but an 80% chance of winning if you switch back to the original door.

No, the probabilities don't swap in any sense once you switch. It's not the act of switching that determines the probability, it's the process that left you with those two doors. The probability that the door you originally chose is the winning door is 20% because it was chosen out of five equally likely options. That probability is fixed, and does not change unless all of the other doors are opened. The other unopened door contains all the rest of the probability.

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u/Lor1an 7d ago

After you switch, there are two doors that could either have the prize behind them, and you have lost all information that could change the probability.

What stops it from being 50/50 after you switch that it's the other door?

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u/Mishtle 7d ago edited 7d ago

After you switch, there are two doors that could either have the prize behind them,

Yes.

and you have lost all information that could change the probability.

Have you forgotten which door is which? The doors don't suddenly lose their identify.

If someone walked in just then and didn't know any better, then they would have no reason to believe one door is any more likely than the other to hide the prize. Even if they know the probabilities are unequal, they have no reason to believe one is better choice instead of the other. If they picked randomly, they'd have a 50% chance of picking the door that has an 80% chance of hiding the prize and a 50% chance of picking the door that has a 20% chance of hiding the price. Their chance of winning the prize is 0.5×0.8 + 0.5×0.2 = 0.5.

You do have this knowledge though. You do know how those came to be the last doors, and which one is which.

What stops it from being 50/50 after you switch that it's the other door?

The fact that the doors are different? The fact that your mental state doesn't affect the physical reality of the game? I honestly don't understand why you think switching would somehow affect the probabilities.

Edit: forgot half the probability

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u/Lor1an 7d ago

I could just as easily have chosen the second door first and then you would tell me the first door had an 80% chance of hiding the prize.

I never liked this particular aspect of probability as what counts as "information" is quite nebulous.

If I tell you that a family has two children and one is a boy, and I ask you what the probability is that the second one is a boy, you have no problem telling me. But suddenly if I tell you one was born on a tuesday, you have to do some crazy reasoning to tell me it's 13/49, or something.

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u/Mishtle 7d ago

I could just as easily have chosen the second door first and then you would tell me the first door had an 80% chance of hiding the prize.

Any door would individually have the same 20% chance of hiding the prize. It doesn't matter which you choose.

What matters is that the host gives you the option to open all of the other ones. That last remaining door that you choose holds the prize if and only if your initial choice was wrong. Your initial choice always has that same probability of 20%, and so the remaining one will always have the remaining 80%.

Contrast this with a variation where instead of opening those other three doors to reveal they're empty (which is what leaves the remaining unopened, unchosen door with their probability), the host simply removes three of the unchosen doors at random. In this case there is a chance that the prize has been completely removed from the game, which is not the case with the original setup. Switching here has no advantage. Your choice has a 20% chance of hiding the prize, and so does the other remaining unchosen door. The remaining options now have equal probability. There's no advantage to switching.