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Let's say that you have 3 doors and you pick door #1. At that point the host says "Ok, I'm going to open one of the other doors at random and then give you the

by jiggy2011 13y ago

Let's say that you have 3 doors and you pick door #1. At that point the host says "Ok, I'm going to open one of the other doors at random and then give you the chance to change your choice". At this point 2 things could happen. Either he opens the door with the car in which case you will definitely switch to that door and have a 100% chance of winning. Or he opens the door with a goat. There is a 1/3 chance that you picked correctly in the first place, therefore there is a 2/3 chance that you picked wrongly, so it's safer to assume that you picked wrongly. You know that he didn't show you the car, so either you picked correctly to being with (1/3) or you picked wrongly (2/3), but if you picked wrongly then it must be behind the door that the host did not open. So you still have better odds from switching even though the host did not know.

Jtsummers 13y ago
You are correct, as long as the host doesn't reveal the prize the odds of winning by switching remain better than the odds of winning with your initial selection. EDIT: And in the case that he reveals the prize your odds have gone up. I suspect (but haven't calculated) that a version where you can switch to the winning prize when it's revealed would give you much better odds of success and a bankrupt game show would result.
- Jtsummers 13y ago
  Crunched the numbers, I'm wrong. If the host shows a goat (when the host doesn't know where the car is), switch or stay doesn't matter as you have the same odds either way. There are 4 outcomes to the selection phase of this game: 1/3 - both pick goats 1/3 - you pick the goat, host picks the car 1/3 - you pick the car, host picks the goat 0 - you pick the car, host picks the car So given the host picks a goat, switching or staying doesn't matter it's 50/50. EDIT: An aside - why we shouldn't trust our intuitions, but instead actually work out these problems.
  - jiggy2011 13y ago
    You're right. When doors are opened randomly you can not tell whether the host was "lucky" (to not reveal the car) or whether you picked the right door to begin with since these are both equally likely.
maxerickson 13y ago
The host doesn't open a door at random. He always shows a goat. You are discussing some other game show.
- jiggy2011 13y ago
  Let's assume that both of these game shows are showing on different channels and you were invited to participate in one of them. But once you get to the studio you have a case of nerves and forget which show you are on. Should that change how you make your decision?
  - maxerickson 13y ago
    No, the odds for switching at the real show are 66-33 and the odds at the fake show are 50-50, you might as well switch after the reveal (what you are calling 'better odds' boils down to 'sometimes you can see the car so you switch to it'. When you see a goat, you have equal information for the remaining doors.).

Jtsummers 13y ago

You are correct, as long as the host doesn't reveal the prize the odds of winning by switching remain better than the odds of winning with your initial selection. EDIT: And in the case that he reveals the prize your odds have gone up. I suspect (but haven't calculated) that a version where you can switch to the winning prize when it's revealed would give you much better odds of success and a bankrupt game show would result.

Jtsummers 13y ago
Crunched the numbers, I'm wrong. If the host shows a goat (when the host doesn't know where the car is), switch or stay doesn't matter as you have the same odds either way. There are 4 outcomes to the selection phase of this game: 1/3 - both pick goats 1/3 - you pick the goat, host picks the car 1/3 - you pick the car, host picks the goat 0 - you pick the car, host picks the car So given the host picks a goat, switching or staying doesn't matter it's 50/50. EDIT: An aside - why we shouldn't trust our intuitions, but instead actually work out these problems.
- jiggy2011 13y ago
  You're right. When doors are opened randomly you can not tell whether the host was "lucky" (to not reveal the car) or whether you picked the right door to begin with since these are both equally likely.

maxerickson 13y ago

The host doesn't open a door at random. He always shows a goat. You are discussing some other game show.

jiggy2011 13y ago
Let's assume that both of these game shows are showing on different channels and you were invited to participate in one of them. But once you get to the studio you have a case of nerves and forget which show you are on. Should that change how you make your decision?
- maxerickson 13y ago
  No, the odds for switching at the real show are 66-33 and the odds at the fake show are 50-50, you might as well switch after the reveal (what you are calling 'better odds' boils down to 'sometimes you can see the car so you switch to it'. When you see a goat, you have equal information for the remaining doors.).