Find the expected value of a dice sum If fair dodecahedron is rolled until at least $k$($k$ is fixed between 2 and 12) is gotten, and $X$ is the sum of all numbers appeared until the last time, what is $E(X)$?

The probability that any roll is greater than or equal to $k$ is 
$$
\frac{13-k}{12}
$$
so the expected number of rolls until a roll of $k$ or greater is 
$$
\frac{12}{13-k}.
$$
All but the last one of these rolls is less than $k$, so the sum of those rolls has an expected value of 
$$
\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2}.
$$
Add to this the expected value of the final roll
$$
\frac{k+12}{2}
$$
and so the expectation of the sum is
$$
\left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2} + \frac{k+12}{2} = \frac{78}{13-k}.
$$
