4

In a match between two players A and B, the first to win $n$ games wins the match. Player A wins a game with probability $p$, the outcomes of each game are independent. (The probability of player B winning a game is $q=1-p$.) I've found that the probability that A wins the match with a scoreline of $n:k$ is $${n+k-1\choose k}p^nq^k\ .$$ Here is my problem: how to prove that the total probability is $1$, i.e. $$p^n\bigg[1+{n\choose 1}q+{n+1\choose2}q^2+\cdots+{2n-2\choose n-1}q^{n-1}\bigg]+q^n\bigg[1+{n\choose 1}p+{n+1\choose2}p^2+\cdots+{2n-2\choose n-1}p^{n-1}\bigg]=1?$$ I've tried induction but it seems impenetrable.

Teddy38
  • 3,309
  • 2
    I think the first term is also $\sum \limits_{i=0}^{n-1} {2n-1 \choose i} p^{2n-1-i}q^i$ and the second $\sum \limits_{j=0}^{n-1} {2n-1 \choose j} q^{2n-1-j}p^i=\sum \limits_{i=n}^{2n-1} {2n-1 \choose i} p^{2n-1-i}q^i$. I would not expect each of these to have simple forms but they add to $1$ as a binomial expansion – Henry Nov 11 '18 at 22:40
  • @Henry Thanks for your comment. I didn't believe it until I tried a few cases. I'll attempt to prove it. – Teddy38 Nov 13 '18 at 09:51
  • 1
    Teddy38: My two expressions are equivalent to playing $2n-1$ games with the match winner being the person who wins most games, i.e. at least $n$ – Henry Nov 13 '18 at 10:02
  • @Henry Ok, that makes sense. Many thanks! – Teddy38 Nov 13 '18 at 10:16

0 Answers0