Variation of Monty Hall problem

Question

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?

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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?

Answer 1

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

Answer 2

$\frac{1}{2}$ of course. The uncertainty is resolved before the beginning of the game (choosing a door) unlike the ordinary Monty Hall problem.

Answer 3

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.

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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 $\frac23$.