There are many solutions to this one. I like the following:

1. What is the probability of seeing a number ? 1/6 2. 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/p 4. Therefore we have, that for the k tentative the expectation is 6 / (n-k+1) 5. Since the expectation is linear k=1 => 6/6 k=2 => 5/6 k=3 => 4/6 k=4 => 3/6 k=5 => 2/6 k=6 => 1/6 This 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.

There are many solutions to this one. I like the following:

ReplyDelete1. What is the probability of seeing a number ? 1/6

2. 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/p

4. Therefore we have, that for the k tentative the expectation is 6 / (n-k+1)

5. Since the expectation is linear

k=1 => 6/6

k=2 => 5/6

k=3 => 4/6

k=4 => 3/6

k=5 => 2/6

k=6 => 1/6

This 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.