Monty Hall bites again! While preparing a conference talk on this topic I noticed that all the versions of September 2007 or earlier need the following correction (now incorporated into the final, published version of the essay):
There's a variant of the 3-cards problem worth mentioning because it shows just how treacherous intuition can be. In interpreting the equation numbered (14) in the older versions, and in saying that everything had become obvious and calculation superfluous, I was using the symmetry of (11) without realizing it. But what if my personal knowledge of Player One were to tell me that (11) should be replaced by
P(R2| A1Z') = q , P(R3| A1Z') = 1 - q
where q differs from 1/2? For instance Player One might have a form of Tourette's syndrome that compels removal of the card as near as possible to the card I fingered, making q = 1. Then if R3 eventuates, I can be certain of A2. That is, P(A2| R3Z') = 1. For general q, it's another straightforward exercise from (6), (9), (10) and Boolean algebra to show that P(R3| Z') = P(R3| A1Z') P(A1| Z') + P(R3| A2Z') P(A2| Z') = (2 - q)/3 and hence that P(A2| R3Z') = 1/(2 - q). Setting q = 1/2 recovers the value 2/3 for the standard case, as well as demonstrating it more securely. The opposite-extreme case q = 0 is interesting as well.
(Had I followed my own advice and been twice as explicit as I felt necessary, I'd have picked this up much sooner!)