Marion,
I don't mind you asking. I'm professor of psychology and the work I mentioned was a recent diploma thesis by one of my students. She explicitely told the subjects that she knew where the prize was and would use this knowledge to make sure that she only opened cups with no prize under them. If she hadn't told this before the experiment the subjects might have been right assuming that they could as well stick to their first choice. Why?

I try to make it understandable without mathematics (but you could calculate it by using the tables of events as has been done above). Often it works if you make the case more salient by using more extrem numbers. Let's take Jim Dixon's example from above:
Instead of 3 cups, suppose there are 1,000 cups. You choose number 483 at random. Then the house turns over all the cups EXCEPT numbers 483 and 722, showing that all 998 cups are empty. Now, would you keep number 483? Or switch to number 722? To make it completely unambiguous you have to assume that the house (and you) know before the test that 998 cups (known to the house to contain no prize!) will be turned over. Then you should switch by all means.
But imagine you are in a lottery with 1000 tickets, one single prize and you have ticket 483. Now, for the sake of suspense, the house slowly calls out tickets in random order that have not won. With each call you could know that you have lost, but fortune is with you and 998 tickets have been called losers with your ticket and ticket 722 left. Wouldn't you feel now being a bit nearer to a win than you were before? Would you switch? Perhaps not, but it doesn't matter, for in this situation the chance of winning for both tickets left is 50%.

The very same problem arises in your prisoners example (the most often and best studied problem in conditioned probabilities). You have not told us enough to make the response unambiguous. First, the decision which prisoner has to die has not just to be random but also random with equal probabilities (how goes random with unequal probabilities? E.g., you'd have an urn with 100 red beads (you die), 700 black beads (coprisoner B dies) and 200 green beads (coprisoner C dies)). The two responses below are only correct under that assumption, otherwise it is much more tricky. But that's only a minor quibble.

The major is this. You (as prisoner) have to know beforehand that the next day definitely not you but another prisoner will be released to come to the response you (Marion) think of, namely, you should not be worried. If, however, you know that the next day one of the three prisoners (not necessarily you, but possibly you) will be released and the one released is not you, you should be worried for now the probability of you dying goes up from 1/3 to 1/2.

Wolfgang