Expected number of steps and probability

Question

I have a problem that I am not quite sure how to solve using my elementary knowledge of probability. My question is this: suppose a friend and I are playing a game. We both start at 0 points, and whoever gets to 9 points first wins. There is no way of subtracting points. My friend has a 20% chance of scoring in each step, and I have a 15% chance of scoring in each step. What are the expected number of steps until I reach 9 points? And what is the probability that I will win? If you could show me how to derive these answers that would be great, thank you.

Answer 1

The general strategy is to condition on the first result of the game.

For instance, to compute the probability that you win, one introduces the probability $u(x,y)$ that you win starting from partial scores of $x$ games won by you and $y$ games won by your opponent.

Then, for every $0\leqslant x,y\leqslant8$, $u(9,y)=1$, $u(x,9)=0$, and, $$u(x,y)=15\%\cdot u(x+1,y)+20\%\cdot u(x,y+1)+65\%\cdot u(x,y).$$ This linear system determines uniquely $u(0,0)$, the probability that you win before the game starts. Note that the system can be rewritten as $$ 7u(x,y)=3u(x+1,y)+4u(x,y+1). $$ Can you solve this?

What are the expected number of steps until I reach 9 points?

In which sense? How does one count the cases when you never reach 9 points because your opponent did first and the game stopped? Please explain.

Edit: Unless I am mistaken, considering $a=\frac37$ and $b=1-a=\frac47$, one finds $$ u(9-i,9-j)=a^i\sum_{k=0}^{j-1}b^k{i+k-1\choose i-1}, $$ in particular, $$ u(0,0)=\left(\frac37\right)^9\sum_{k=0}^{8}\left(\frac47\right)^k{8+k\choose 8}\approx27.3975\%. $$ Edit: In the second question, it appears finally that one plays alone with probability $15\%$ of success at each step, hence the mean number of steps necessary to win $9$ times is $9$ times the mean time for the first success in a heads or tails game, thus, $9$ times $1/15\%$ (numerically, $60$).