Expectation of total scores when rolling a die until the score is not 6

Question

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!

Answer 1

The expected sum after $k$ rolls is not $k$ but $6(k-1)+3$ indeed if we stop at first roll it means that we get $1,2,3,4,5$ and the average sum will the mean i.e. $3$, if we rolled twice means that the first roll we get $6$ and now one of the values $1,2,3,4,5$ and so on. It means that the exact formula is \begin{align} \mathbb{E}[X]&=\sum_{k=1}^{\infty}(6(k-1)+3)P(X=k)=\frac{5}{6}\sum_{k=1}^{\infty}(6(k-1)+3)(\frac{1}{6})^{k-1}=\\ & = 5\sum_{k=1}^{\infty}(k-1)(\frac{1}{6})^{k-1}+\frac{5}{2}\sum_{k=1}^{\infty}(\frac{1}{6})^{k-1}=\\ &=\frac{6}{5}+3=\frac{21}{5} \end{align}

Answer 2

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

Answer 3

Unless you roll a $1$ on your first roll which happens only $\frac16$ of the time, the total is going to be at least $2.$ So it should be clear that the expected value is going to be greater than $\frac65,$ which is just a little more than $1.$

In fact, $\frac56$ of the time your first roll will be less than $6,$ and the expected score (given that this happens) is $3.$ The other $\frac16$ of the time you roll a $6$ (so the final score is at least 6) and then you get to roll again. This implies that the expected value of the whole game is greater than $3.$

In fact, let's try it this way. You have a $\frac56$ chance to end on the first roll, with expected score $3$ in that case. You have a $\frac16$ chance to roll a $6,$ which means your final score will be $6$ plus whatever you get from the rolls afterward. But "the rolls afterward" are just like the original game, and they have the same expectation (provided that the expectation is finite, of course).

So as long as $\mathbb E[X]$ is finite, we will have

$$ \mathbb E[X] = \tfrac56(3) + \tfrac16(6 + \mathbb E[X]). $$

Solving for $\mathbb E[X],$ we get $\mathbb E[X] = \frac{21}{5}.$

Note that this is not really a valid method for finding the answer from first principles, only for computing the answer when you already know there is one (that is, you have to know already that $\mathbb E[X]$ is finite).