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On a game show, the Monty Hall problem is being played. The contestant is told to pick a door, and he does, but just before being able to tell the host which door he picked, one of the doors that the contestant had not chosen gets knocked over by a poorly-hung light post, revealing a goat.

The host decides to continue the game, and asks the player to pick a door.

Does the player now have a 1/3 chance, a 2/3 chance, or a 1/2 chance of winning if he switches?

I don't understand why the chance after the event would be 1/2. Imagine that two Monty hall games are being played at once: one where the host picks randomly (chance of winning before entry of 1/2), and one where the host picks a door with a goat that the player has not picked (chance of winning before entry of 2/3).

In the random game, the contestant is asked to pick a door, and he picks door A where there's a goat. In the logical game, the contestant picks door A too, where there's another goat.

In the random game, the host (or the accident) randomly opens door B, revealing a goat. In the logical game, the host opens door B because he knows the other goat is there.

Now, both players are asked to choose a door. Why does the random game's player have a different chance of winning than the logical game's player?

SomeoneElse
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  • I think you misunderstand because you are trying to focus on only a single event. Imagine both games played out 1000 times. Any time the light post reveals the car, the contestant wins. – Ryan Aug 31 '14 at 23:14
  • Actually, I take back my last sentence. Any time the light post reveals the car, we consider the game to be "invalid." The most important thing to realize here is that the game host has knowledge of the car's location, while the light post does not. – Ryan Aug 31 '14 at 23:23

3 Answers3

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$\frac12$ of course. Contrary to the original Monty Hall problem, the accident revealed a goat by chance (it might have revealed the car).

Hagen von Eitzen
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  • I am asking what the chance is AFTER the accident happened. – SomeoneElse Aug 31 '14 at 22:01
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    Yes, and I am answering that. Before the accident, there was a $\frac13$ chance of picking (or having pocked) the right door. If the accident had revealed the car, switching or not switching would both give you a goat, by the way – Hagen von Eitzen Aug 31 '14 at 22:08
  • Read what I added in my question. I agree that a person entering this game show would have a 1/2 chance of winning, but technically, everything went according to plan out of chance in this case. – SomeoneElse Aug 31 '14 at 22:10
  • @SomeoneElse It has nothing to do the host's intention. It is about when the information about the door with the goat is revealed to us. – Calculon Aug 31 '14 at 22:18
  • @L'universo: The information is revealed after the contestant has picked the door. The only difference is that it was revealed out of chance. – SomeoneElse Aug 31 '14 at 22:21
  • @SomeoneElse Nope that is not the only difference. He did choose the door but he didn't disclose his choice to the host (according to your variation of the problem). – Calculon Aug 31 '14 at 22:24
  • @L'universo: Why would that be relevant? – SomeoneElse Aug 31 '14 at 22:24
  • @SomeoneElse If he were to disclose his choice to the host, then the host would be forced to pick a goat-door out of the two doors he was left with. In your variation, a goat-door is revealed arbitrarily. The host/poorly-hung light post is then not restricted. – Calculon Aug 31 '14 at 22:28
  • @L'universo: What I'm asking is why. The host picked a door at random, and it happened to be the same one that a host respecting the rules would have picked. I really don't understand why the right door being picked at random is not the same as the right door being picked intentionally. – SomeoneElse Aug 31 '14 at 22:31
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$\frac{1}{2}$ of course. The uncertainty is resolved before the beginning of the game (choosing a door) unlike the ordinary Monty Hall problem.

Calculon
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The chance of someone who has not started the game show, but who plans on switching is $\frac13$:

If the contestant picked the one with the car (chance of $\frac13$), he has a $(\frac130)+(\frac23*0) = 0$ chance of winning:

  • There is a 1/3 chance of his door being knocked over, leaving him to pick between the two doors without the car, giving him a 0/1 chance of winning
  • There is a 2/3 chance one of the doors with the goats got knocked over, and he switches and loses

If the contestant picked a door without a car (chance of $\frac23$), he has a $(\frac13*\frac12)+(\frac13*1)+(\frac13*0) = 1/2$ chance of winning:

  • There is a 1/3 chance his door gets knocked over, leaving him with a 1/2 chance of winning
  • There is a 1/3 chance the door with the car gets knocked over, leaving him with a 0/1 chance of winning
  • There is a 1/3 chance the other door with the goat gets knocked over, and he switches and wins.

This gives him a $\frac23*\frac12=\frac12$ chance of winning.

After the accident revealed a door with a goat which the contestant has not picked (chance of $(\frac13*\frac13)+(\frac23*\frac23)=13/9$), the chance of him winning if he switches is $\frac23$.

SomeoneElse
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