# A good problem in probability

Two players play a series a games. In each game, each one of the two player wins a point
(i.e. there is no draw), the first player have probability $p$, and the second have probability $q=1-p$, that are known for the players. The number of plays is even. To win is needed to achieve more than the half of the points. If $p<q$ and the first player can choose the number of plays. Which is the best election?

• I think I have a partial solution, but I'm nor sure. I used "Negative Binomial Distribution". If the number of plays is even, suppose $2r$ with $r \in \mathbb{N}$ the probability of $r+1$ successes in $2r$ trias is: $P(X=2r)= {2r-1 \choose k}p^{r+1}(1-p)^{r-1}.$ Where X is ra r.v the count the number of trials. So if K is the r.v that count the number of wins we need to calculate $P(K>r)$ this is: $$P(K>r)=\sum_{k=r}^{2r-1}{2r-1 \choose k}p^{k+1}(1-p)^{2r-1-k}$$. Apr 25, 2014 at 0:13
• But I don't know who to calculate the best n=2r, where n is the election that the first player have to make, the numer of plays that they are goin to play Apr 25, 2014 at 0:21
• I assumed the goal was for player 1 to win more times than player 2. Is this correct? Ties are neutral? Apr 25, 2014 at 0:48
• I think is because player 1 wants to increment his probabilities to won the game. Apr 25, 2014 at 1:08

In a pair of plays, the chance of player 1 winning both is $p^2$, player 2 winning both is $q^2$, and of each winning a point is $2pq$. Since player 1 has lesser odds of winning, every pair of plays makes it more likely for player 2 to get ahead than player 1. As such player 1 wants to try to end the game as quickly as possible, before player 2 has an insurmountable lead.