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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?

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  • $\begingroup$ 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: $$P(K>r)=\sum_{k=r}^{2r-1}{2r-1 \choose k}p^{k+1}(1-p)^{2r-1-k}$$. $\endgroup$
    – ViKaN
    Apr 25, 2014 at 0:13
  • $\begingroup$ 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 $\endgroup$
    – ViKaN
    Apr 25, 2014 at 0:21
  • $\begingroup$ I assumed the goal was for player 1 to win more times than player 2. Is this correct? Ties are neutral? $\endgroup$
    – RandomUser
    Apr 25, 2014 at 0:48
  • $\begingroup$ I think is because player 1 wants to increment his probabilities to won the game. $\endgroup$
    – ViKaN
    Apr 25, 2014 at 1:08

1 Answer 1

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The ideal number is 2.

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.

EDIT: This answer assumes player 1's goal is to win more times than player 2. As such, ties are factored out. If the goal is to win as much as possible, that is a different answer.

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