Modify this puzzle in the following way: Suppose that the encoding is in trinary, with 0’s, 1’s and 2’s, and that 0 represents white and 1 represents black (as in the original) and that 2 represents red. Suppose further that each digit is a 0 with probability 40%, 1 with probability 40%, and 2 with probability 20%. Then, the new question is: What is the expected number of digits between 2 adjacent non-white (i.e. red or black) bars, each with width at least 20 bits?
($20 to the anyone the first person who solves it)
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