Ok, I'll have to think awhile about what it would mean if the house's choice is random. Thanks for the leads.

Is anybody going to help me out and argue with Eddie over the red/white cards one? I think I still have to concentrate on my little cups.

But about those prisoners... I'll put here my thinking on the problem, and why I think it doesn't contradict my stance on the cups, although at first analysis (that's you, Eddie!) it seems to.

You might think that there is a 1/3 probability that you are doomed, and therefore a 2/3 probability that you aren't. Since the doomed person has already been chosen by the jailer, as I specified above, the release of a prisoner shouldn't change your chances, so the other prisoner is stuck with the 2/3 chance of being executed. The problem is, he's thinking the same about you. As Eddie says, there is a contradiction here.

However, I would say that the real issue is in Eddie's 5th statement: "We each figure we have a 1/3 chance of being doomed." In fact, if the jailer has already decided who to execute, then each person's probability of being executed is either 1, or 0, not 1/3. Because the prisoners have no way of knowing whether they're a 1 or a 0, they perceive it as randomness, so they think their chances start at 1/3 then go to 1/2. But the execution will not be random, the victim has already been chosen. The probability that they perceive is different from the probability that is actually there.

So psychologically, the two prisoners left are likely to be frightened by the third prisoner's release. But logically, if they know that the doomed one has already been chosen, they should consider it indifferent news.

Marion