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- Variation on the Monty Hall Problem 2 answers
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?