To end the week, a problem very similar to the famous "two envelope paradox," except that while the two envelope version continues to be a source of contentious debate, the "three envelope problem" has a definite answer.
I've adapted this from Thomas Povey's "Professor Povey's Perplexing Problems," a volume I'll say more about in a Sunday posting over at MathTango:
You're handed an envelope, which upon opening, has X number of dollars in it. The presenter now places (out of your view) 2X dollars into another envelope and X/2 dollars in a third indistinguishable envelope (i.e. the values could be 100, 200, and 50). Now you are asked if you wish to hold onto your current envelope with X dollars or swap for either of the other two envelopes. Should you swap???
This sounds very similar to the two-envelope situation but is subtly different. In the two-envelope case, the X value used must simultaneously or ambiguously be viewed as potentially the largest or the smallest value when computing the various probable outcomes. In the three envelope case we have 3 distinctive and fixed values, X/2, X, 2X. As a result it turns out that the computed "expected value" of switching is more definitively 5X/4, and thus worth doing (i.e., 5X/4 is greater than X).
[ 5X/4, by the way, is one of the solutions to the two-envelope paradox as well, obviously arguing for swapping; the problem is that alternative calculations are logically possible that lead to a don't-swap conclusion -- and the back-and-forth arguments, based on small nuances, could give you a migraine! ;-)]