This is a problem from AoPS's Introduction to Counting & Probability book. It does not ask you to prove the general case: just compare the probability of the better team of winning a best-of-5-games, a best-of-7-games, and a best-of-9-games. I am interested in solving the general case:
Assume the better game has probability of $p>\frac 12$ of winning a single game against its opponent. In a best-of-$(2n-1)$-games series, the probability of the better team winning the series is $f(n)$. Prove that $f(n)$ increases when $n$ increases.
My approach: It's easy to show that $$f(n) = \sum_{k=0}^{n-1}\binom{n+k}{k}p^n(1-p)^k \tag1$$ and with the help of Wolfram Alpha we conjecture that $$f(n+1)-f(n) = \frac{\binom{2n}{n} p^n(1-p)^n((2n+1)p-n)}{n+1} \tag 2$$
After some lengthy and ugly calculation I believe I proved $(2)$ is true. However it's very likely this has been asked in the past (though I wasn't able to find it) so I don't want to post the not-so-good proof.
If you have a simple way to prove $(2)$, or a simple argument to show $f(n+1) > f(n)$ without using $(1)$ please post your answer. Thank you.
Update: I found two similar questions and answers to show $f(n+1)>f(n)$ without using $(1)$. Now it remains to find a simple proof for $(2)$.
