Re: Riddle -- Probability of Expectation
Sorry, this doesn't work.
It's true that the chance four randomly picked letters will be some permutation of "hues" is (4*3*2*1)/(26^4). But it's not true that the chance it will be a permutation of either "hues" or "hews" will be twice that. If that were true, then C(26,4)*(4*3*2*1)/(26^4) would have to equal 1. In fact it equals 3.93. The problem is that some of the random 4 letter combinations have some letters the same.
You have a similar problem with your next step. If I know that the first four letters of the line are "hues" then the probability that the first three letters plus the fifth contain "hues" is 1/26, not 120/26^4.
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