A six-sided die is tossed once and its outcome is noted. It is then tossed repeatedly and independently until an outcome is observed that is at least as large as that on the first toss. Compute the expected number of tosses needed, not including the first one. Hint - condition on the outcome of the first toss.
So you'd use a formula like: E(X) = sum of E(X|N=n)*P(N=n) , where N is the outcome of the first toss. So I believe P(N=n) would be 1/6 for each one, but I'm kind of confused on the E(X|N=n) part ... I was thinking that if you roll a 1, the expected number of tosses would be 1 (b/c the prob is 6/6 that it's >=1). If you roll a 2, the expected number would be: 1*(5/6) + 2*(1/6)*(5/6) + 3*(1/6)^2*(5/6) + ... , and similarly for the rest ... 3: 1*(4/6) + 2*(2/6)*(4/6) + 3*(2/6)^2*(4/6) + ... etc. Is this correct or am I thinking about it the wrong way? If it is correct, how do you find the exact values of those sums (I think they're a form of the geometric series but I'm not sure).