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I have read two sources where the outcomes of the Monty Hall problem are triples of the form:

(initial guess, he curtain Monty opens, the curtain where the car is)

However, I think it should be:

(initial guess, the curtain Monty opens, switched curtain, the curtain where the car is)

Is the sample space consisting of these outcomes wrong? It seems that it is because I am getting that it doesn't matter if one switches curtains.

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2 Answers 2

up vote 1 down vote accepted

Your two statements are equivalent as the initial guess and the curtain (or door) Monty opens determine the remaining curtain (or door) in the standard version of the problem.

But the real issue is that not all the possible events you would list are of equal probability: if your initial guess is the same as the car's location then Monty has a choice of two; if your initial guess is not the same as the car's location then Monty does not have a choice, at least in the standard version of the problem.

The events where Monty has had to make a choice have half the probability of those where he does not, assuming he has made a random choice.

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If the contestant chooses at random,$2/3$ of the time s/he will choose a goat. Then Monty will show a goat in another door, meaning changing doors will give you the car. Only in $1/3$ of the time, will a contestant select the door containing the car; in this last case, the contestant loses by changing the choice of doors. This means $2/3$ of the time, you're better-off by changing doors.

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