The game consists of a sequence of independent plays for each of the two players. Each round I have probability $p$ of making a point and my opponent has probability $1-p$ of making a point. (so each round only involves one player taking their chances and each player alternates, there has to be an even number of games). What are my chances of winning?

So my chances of winning a game with $2n$ trials is: $\sum_{n+1}^{2n} {{2n}\choose{x}} p^x (1-p)^{2n-x}$

In the case of $p=0.5$ for two rounds, my probability of winning is 1/4, then 4 rounds my probability is 5/16 which I obtained by manually counting outcomes. Can I obtain a general formula for 2n games which uses counting outcomes?


  • 3
    $\begingroup$ Are you looking to continue the $p=\frac{1}{2}$ case? If so it seems like there is a symmetry that you could exploit here. Namely that $p=1-p$ in $\sum_{i=0}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i}= 1$ and that $\binom{2n}{i} = \binom{2n}{2n-i}$. Unfortunately the second identity isn't much help if $p \neq 1/2$ $\endgroup$ – Kitter Catter Aug 22 at 17:29
  • $\begingroup$ @KitterCatter I agree with your identity, and I am focused on the p=1/2 case. I am curious is there is a combinatorial argument for evaluating the probability of getting more points. The probability should converge to 1/2 as the trials get larger. $\endgroup$ – rannoudanames Aug 22 at 17:41

From the fact that you have to have scored some number of points we have: $$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} = 1-\sum_{i=0}^{n} \binom{2n}{i} p^i (1-p)^{2n-i}$$ Using the fact that there are just as many ways to distribute $i$ points as to distribute $2n-i$ points when there are $2n$ points available: $$ 1-\sum_{i=0}^{n} \binom{2n}{i} p^i (1-p)^{2n-i}=1-\sum_{i=0}^{n} \binom{2n}{2n-i} p^i (1-p)^{2n-i}$$ letting $j=2n-i$ to bring the expression closer to the thing we are solving for: $$1-\sum_{i=0}^{n} \binom{2n}{2n-i} p^i (1-p)^{2n-i} =1-\sum_{j=n}^{2n} \binom{2n}{j} p^{2n-j} (1-p)^{j} $$

For $p=1/2$ we have that $p=1-p$ and thus they are exchangeable: $$1-\sum_{j=n}^{2n} \binom{2n}{j} p^{2n-j} (1-p)^{j}=1-\sum_{j=n}^{2n} \binom{2n}{j} p^{j} (1-p)^{2n-j} $$

splitting out the $j=n$ term: $$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} =1-\binom{2n}{n}p^n(1-p)^n-\sum_{j=n+1}^{2n} \binom{2n}{j} p^{j} (1-p)^{2n-j} $$

Converging to: $$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} = \frac{1}{2} - \frac{1}{2^{2n+1}}\binom{2n}{n}$$

Essentially if the probability of winning a point is 1/2 then you are as likely as your opponent to score some number of points. Your probability of winning is thus $(1- P_{tie})/2$. The probability of a tie is $2n$ games pick $n$ points multiplied by the probability of getting exactly that number of points for both sides.

  • $\begingroup$ it's $1/2^{2n+1}$. thank you though. I am mainly curious about a combinatorial argument to reach this result. $\endgroup$ – rannoudanames Aug 22 at 18:54
  • $\begingroup$ Essentially fifty except for the case of a tie. Thanks for the correction. 😀 $\endgroup$ – Kitter Catter Aug 22 at 19:35
  • $\begingroup$ If the answer doesn't fit what you want then you don't need to accept it. You should leave the question open. $\endgroup$ – Kitter Catter Aug 22 at 21:12

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