Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total score on one turn.
My working:
$\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k).$
where, $P(X=k) = \frac{1}{6}^{k-1}\frac{5}{6}$
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{2}k\frac{1}{6}^{k-1} = \frac{5}{6}(1+2(\frac{1}{6})) = \frac{10}{9}$
EDITED
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{\infty}k\frac{1}{6}^{k-1} = \frac{5}{6}\frac{1}{(1-\frac{1}{6})^{2}} = \frac{6}{5}$
But on the answer sheet, it is $\frac{21}{5}$
I would appreciate if someone can point out my mistake. Thanks!

Let the expected value be X. If you throw a six, your expected value is 6+X, otherwise the score is whatever you threw. Therefore
X = (1 + 2 + 3 + 4 + 5 + (6+X)) / 6

(5/6) X = 21/6

X = 21/5

