Players A and B, each starting with a score of zero, continuously roll a d6 until a 6 is rolled, with player A starting first. Each player’s rolls are added to their individual scores (including the 6). What are the expected values of a) the total scores of A and B, b) A’s score, c) B’s score?
What I know is that the total score is independent of what A and B roll, and can be treated as an individual player continuously rolling until a 6 is obtained. The recursion relation is $5\mathbb{E}(X)+21=6\mathbb{E}(X)$ so $\mathbb{E}(X)=21$. I also know that by linearity of expectation the sum of the expected values of A and B’s scores should be 21. However I am unsure of how to obtain A’s score. I do know the answer is $126/11$ but I do not know what my recursion relation looks like. Help is greatly appreciated!
Let $v_A$ and $v_B$ be the expected value of the game for Player $A$ and $B$, respectively. Let $$ v\equiv\begin{pmatrix}v_{A}\\ v_{B} \end{pmatrix}. $$ A turn is comprised of two rolls, one from Player $A$ and one from Player $B$. On each turn, one of three things can occur: (i) the game can stop because Player $A$ rolled a six, (ii) the game can stop because Player $B$ rolled a six, or (iii) none of the above, in which case the game continues. Therefore, the expected value can be represented as a sum of three terms: $$ v=\frac{1}{6}\begin{pmatrix}6\\ 0 \end{pmatrix}+\frac{1}{36}\sum_{1\leq i\leq 5}\begin{pmatrix}i\\ 6 \end{pmatrix}+\frac{1}{36}\sum_{1\leq i,j\leq 5}\left(\begin{pmatrix}i\\ j \end{pmatrix}+v\right). $$ Simplifying, $$ \left(1-\frac{25}{36}\right)v=\begin{pmatrix}7/2\\ 35/12 \end{pmatrix}. $$ Dividing both sides by the coefficient of $v$ above, $$ v=\begin{pmatrix}126/11\\ 105/11 \end{pmatrix}. $$ From the above, $v_A = 126/11$, $v_B=105/11$, and $v_A+v_B=21$.