My approach: (I don't remember results. So had to derive this 1/p result.)
The problem can be reduced to heads and tails as follows. Say on the first roll, you get x. Essentially from the next roll, you are solving this problem: Expected number of tosses before an head comes where Pr(head) = (7-x/6). Because this is nothing but success to you. So each situation boils down to a heads and tails toss situation. So say you get a 3 on the first roll. Now getting {3, 4, 5, 6} is success to you. Pr(getting 3,4,5 or 6) = 4/6. So E[tosses before a head comes] where the Pr(head) = 4/6 is 6/4.
Since we know that expected number of tosses before an head comes = 1/p where Pr(head) = p, we just have to calculate the summation.
E[N] = Summation ( 6/(7-x) * 1/6 ) where (1/6) is nothing but the probability of x coming on the first roll. x goes from 1 to 6.
E[N] = Summation( (1/7-x) ) where x ranges from 1 to 6
Correction in solution : In the second line, it should be greater than or equal to 1 with probability 6/6
My approach: (I don't remember results. So had to derive this 1/p result.)
The problem can be reduced to heads and tails as follows. Say on the first roll, you get x. Essentially from the next roll, you are solving this problem: Expected number of tosses before an head comes where Pr(head) = (7-x/6). Because this is nothing but success to you. So each situation boils down to a heads and tails toss situation. So say you get a 3 on the first roll. Now getting {3, 4, 5, 6} is success to you. Pr(getting 3,4,5 or 6) = 4/6. So E[tosses before a head comes] where the Pr(head) = 4/6 is 6/4.
Since we know that expected number of tosses before an head comes = 1/p where Pr(head) = p, we just have to calculate the summation.
E[N] = Summation ( 6/(7-x) * 1/6 ) where (1/6) is nothing but the probability of x coming on the first roll. x goes from 1 to 6.
E[N] = Summation( (1/7-x) ) where x ranges from 1 to 6
E[N] = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6