I came across this problem online and it has been in the back of my mind ever since. I'm currently taking a probability course so please forgive my limited knowledge if I'm missing a key concept.
Problem. Let's play a dice game with two players. Both players roll a fair dice. The winner gains a number of points equal to the difference of the two dice (if same number are rolled, the round is tied and no points are won/lost). E.g. player 1 rolls a $5$, player 2 rolls a $3$, player 1 wins, and gains $(5-3=2)$ points.
The game ends if a player reaches 20 points or more. If after 10 rounds and no one has 20 points, no one wins.
Question. What is the expected number of points gained in each round by the winning player?
My Work So Far. Intuitively, I wanted to take the conditional probability of gaining $1-5$ points and multiply those probabilities with their associated points and add these products which should give me the expected number of points gained in each round.
Let event $A_i$ be the event that the winning player gains $i$ points. The chance of winning a round is $\frac{15}{36}$, so:
$A_i=P(\text{gaining i points}|\text{winning})$
So my answer is:
$\sum_{i=1}^{5}i\times A_i=\sum_{i=1}^{5}i\times\frac{(6-i)/36}{15/36}=\frac{7}{3}$
This is sadly not the answer, I feel like I'm missing a key insight or I'm tackling this problem with the wrong concepts.
Clarification It seems like my work was to just calculate a single round. I feel very lost in an analytical solution for an entire game. By "winner" I meant the winner of the entire game. The winner is determined if they ever reach 20 points or more.
Best Answer
Probably this problem has a nice solution since it was placed online but I did not find it. Therefore I will give here the brute force solution.
Let $$ p_k(n_1,n_2) $$ be the probability, that after $k$ rounds the players have $n_1$ and $n_2$ points, respectively, where we assume a non-stop game.
Then we have: $$ p_k(n_1,n_2)=\frac16p_{k-1}(n_1,n_2)+\sum_{i=1}^5\big[p_{k-1}(n_1-i,n_2)+p_{k-1}(n_1,n_2-i)\big]\frac{6-i}{36}. $$ This recurrence together with the condition: $$ p_0(n_1,n_2)=\begin{cases} 1,&n_1=n_2=0\\ 0,&\text{otherwise} \end{cases} $$ allows us to calculate the probability for any $k,n_1,n_2$.
Due to symmetry, the probability that a player will as first surpass in the $k$-th round a total of $t$ points and, in doing so, accumulate $n$ points, can be expressed as follows: $$ P_k(n,t)=\sum_{i=n-t}^{5}\frac{6-i}{36}\sum_{n'=0}^{t} p_{k-1}(n-i,n'), $$ where $n=t+1\dots t+5$.
Now we are ready to compute the expected number of points gained in each round by the winning player: $$ \left[\frac nk\right]=\frac{\displaystyle\sum_{k=1}^{10}\sum_{n=20}^{24} P_k(n,19)\frac{n}k}{\displaystyle\sum_{k=1}^{10}\sum_{n=20}^{24} P_k(n,19)}. $$
My calculation gives the result: $$ \left[\frac nk\right]=\frac{6480877805871607}{2825235479368820} \approx2.29392. $$
The probability that the game will be finished in 10 rounds is only about $4.4\%$.