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There are many solutions to this one. I like the following:1. What is the probability of seeing a number ? 1/62. On n trials, what is the probability of seeing the k-th number, provided that you have seen (k-1)th before ? there are n-(k-1) numbers to see and so the probability is p=(n-k+1)/6 3. What is the average number of trial before a success? This is a classical question: \sum_i(i p (1-p)^(i-1)). It's a success after i-1 insuccess and that series converge to 1/p4. Therefore we have, that for the k tentative the expectation is 6 / (n-k+1)5. Since the expectation is lineark=1 => 6/6k=2 => 5/6k=3 => 4/6k=4 => 3/6k=5 => 2/6k=6 => 1/6This last passage is very intuitive and can be given directly. The result comes from the sum of all the above fractions since expectation is linear.