Many of us know the [Monty Hall Problem]
But the other day I was asked a variation of this riddle.
The answer of the original question is, of course, $ 66\% $ in favor of changing doors, but this is based on the fact that the game show host knows where the prize is.
Suppose he does not know where the prize is, and after you make your pick, he opens one of the other two doors and it happens to be a goat. Is it still better to change doors when he asks?
I believe it is. (After all it still leaves us with two chances instead of one.) But some of my friends think otherwise. We are not mathematicians, just a couple of riddle-likers, so we are not sure of the correct answer. So I thought to post it here.
I read this on wikipedia and if I understand it correctly it seems to support my answer:
Morgan et al. (1991) and Gillman (1992) both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the well known statement of the problem in Parade despite the author's disclaimers. Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. If the player picks Door 1 and the host's preference for Door 3 is q, then in the case where the host opens Door 3 switching wins with probability 1/3 if the car is behind Door 2 and loses with probability (1/3)q if the car is behind Door 1. The conditional probability of winning by switching given the host opens Door 3 is therefore (1/3)/(1/3 + (1/3)q) which simplifies to 1/(1+q). Since q can vary between 0 and 1 this conditional probability can vary between 1/2 and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However, it is important to note that neither source suggests the player knows what the value of q is, so the player cannot attribute a probability other than the 2/3 that vos Savant assumed was implicit.
What are your thoughts?