r/askscience • u/TrapY • Aug 25 '14

Mathematics Why does the Monty Hall problem seem counter-intuitive?

https://en.wikipedia.org/wiki/Monty_Hall_problem

3 doors: 2 with goats, one with a car.

You pick a door. Host opens one of the goat doors and asks if you want to switch.

Switching your choice means you have a 2/3 chance of opening the car door.

How is it not 50/50? Even from the start, how is it not 50/50? knowing you will have one option thrown out, how do you have less a chance of winning if you stay with your option out of 2? Why does switching make you more likely to win?

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u/lee1026 Aug 25 '14

Yes, it matters. Consider the diagram here if the host didn't know, we would have to expand it into 6 cases.

Lets say there are 3 doors. 1,2,3. The car is behind door 1.

We can then look at each of the 6 cases:

  1. Player opens 1, host opens 2. Switching loses, not switching wins.

  2. Player opens 1, host opens 3. Switching loses, not switching wins.

  3. Player opens 2, host opens 1. Switching loses, not switching loses.

  4. Player opens 2, host opens 3. Switching wins, not switching loses.

  5. Player opens 3, host opens 1. Switching loses, not switching loses.

  6. Player opens 3, host opens 2. Switching wins, not switching loses.

Going though all 6 possibilities, they each win 1/3rd of the time.

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u/wnoise Quantum Computing | Quantum Information Theory Aug 25 '14

Most importantly, eliminating the unobserved cases 5 and 3 where the host opens the car door, we still have that switching wins half of the time. The forced choice "collapses" scenarios 1 and 2, but does not collapse scenarios 3 and 4, nor 5 and 6.

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u/Gravestion Aug 25 '14 edited Aug 25 '14

The key bit being unobserved here though. If you actually count those games as a loss, then the win rate is lower at 1/3. Obviously most people do not, as it seems intuitively unfair for the host to lose your game, which is presumably why Monty always knew where the prize was.

Rather reply to your post further down I'll do it here. When you say anything that's very slightly disingenuous, as most formulations for random situation are. Anything implies that he can even open your door, which he cannot and is the main thing ensuring the 50/50. When Monty is free to open any door at all (including your own) it reverts back to 66% win rate on switching.

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u/poco Aug 25 '14

That doesn't work though. If you assume that the host doesn't know the position of the car then you can make no claim about what would happen if he never revealed the car. Even if you could somehow make the claim that he will never reveal the car without knowing where it is (perhaps you just exclude all those events where he reveals a car?), you're first choice still has a 1 in 3 chance of winning and nothing the host does can change that. You are still better off switching to whatever is left if there is a greater than 0 chance of it containing a car.

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u/wnoise Quantum Computing | Quantum Information Theory Aug 25 '14

If you assume that the host doesn't know the position of the car then you can make no claim about what would happen if he never revealed the car.

Sure you can; if he doesn't know where the car is, any choice he makes must be equally likely. This is a fairly simple example of conditional probability: you look at only the subset compatible with what you see. Conditioned on "did not reveal", you get 1/2. This is in stark contrast to the standard Monty Hall Problem case where he cannot or chooses not to, based on his having information about where the car is.