Hi, This is a solution to problem 2 in the the latest Tech review puzzle corner. It was a nice problem; for a while I could not come up with a better winning probability than 1/2 even demanding just *at least 1* correct guess and no incorrect guesses, much less the real problem which requires an infinite number of correct guesses. The justification of the solution may be a bit long; the solution itself comprises just the first 2 paragraphs. Regards, Mark Fischler --- Problem 2 (Wizards in Hats) Assign numbers 1, 2, ... to some (countably infinite) set of the wizards. Now let's define a sequence of "cabals", where cabal number N is a sequence of length L= 10*2^N-1 starting at wizard k, obeying the following rules: a) The hats on wizards k-3, k-2, k-1 are black, black white respectively, and on wizards k+10*2^N-1 through k+10*2^N+1 are white, black black. b) At most one hat among the wizards in cabal N is black. c) Cabal N occurs after cabal N-1 (except for number 1, which is the first sequence of 19 hats, at most one of which is white, surrounded by the required BBW ... WBB hats). Most wizards will be able to determine that they are not part of any cabal; others will be unsure (they may be part of a cabal if their own hat is white, but not if it is black); yet others will know for sure that they are part of a cabal. For example, if hats 1 through 25 are BBW ... WBB there the ... stands for 19 hats at least 18 of which are white, then any wizard among those 19 who can see 18 other white hats will know she is in cabal 1. The agreed plan is that any wizard who knows he is in a cabal will guess his hat is black, and all others will abstain. This plan guarantees (shown below) an infinite number of guesses, and a total probability of any wrong guess of less than ten percent. Let's see what happens when this plan is followed. First, there will be an infinite number of cabals, since if there were no cabal after cabal N-1, that would mean that some specific finite sequence of hats *never* will occur after the end of cabal N-1, which is impossible given an infinite sequence of randome hat assignments. Second, each cabal will contain at least one wizard who is sure she is in that cabal, because at least one will see the other 10*2^N-2 white hats. Thus there will be an infinite number of wizards guessing their hat color. Now let's define a cabal as "evil" if any wizard in that cabal will guess incorrectly -- if any cabal is evil, the the wizards lose their contest. Let's look at the expected number of evil cabals (which must, of course, be greater than or equal to the probability of one or more cabals being evil). The probability of cabal 1 being evil is 1/20 -- there are 19 configurations of one black and 18 white hats, and only one with all white hats. Similarly, the probability of cabal N being evil is 1/(10*2^N). The expected number of evil cabals is thus 1/20 + 1/40 + 1/80 + ... = 1/10, and the probability of at least one wrong guess is less than that. So this plan solves the puzzle. Note that the expected number of wrong guesses is still infinite: If one member of a cabal guesses wrong, it will be because the entire cabal has white hats, so all the members will guess wrong. Thus the expected number of wrong guesses is 19/20 + 39/40 + 79/80..., which of course diverges. The agreed plan has so strongly grouped the cases of multiple wrong guesses, that the probability of no wrong guesses at all is left non-zero, and in fact above 90%.