Random permutations of S



Initially we start with one of the random permutations of S. There are N! permutations of S(all possible). But of these, the probability for the 2nd output byte of the result to be 0 is 2/N and not 1/N(as obviously but incorrectly seen from the fact that the probability of a byte to take a particular value is 1 divided by total different values possible).

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