Two game-set inequalities

Here are a couple of curious related results about a generalized 2-player 1-set tennis game: the winner of the set is the first player to win $n$ games, and the winner of each game is the first player to win $k$ points. Let $p>1/2$ be the probability that the stronger player wins a point, and let $P(n,k,p)$ be the probability that the stronger player wins the set. Then

• $P(n,k,p)>P(nk,1,p)=P(1,nk,p)$ if $n,k>1$, and
• $P(n,k,p)>P(k,n,p)$ if $n>k>1$.

The way the paper goes about proving this is pretty clever, but I'm still wondering if there is a more direct combinatorial proof. Say, for a rational $p$, after clearing denominators, we are counting more of something on the left than on the right because of such-and-such an injection.

• Well, direct or indirect, an explicit expression is to the point anyway. According to the paper, $P(n,k,p)=P(n,P(k,p))$ and$$P(k,p)=p^k\sum_{i=0}^{k-1}\binom{k-1+i}{k-1}(1-p)^i$$(Lemma 1). I believe the rest must be easy after that. – მამუკა ჯიბლაძე Nov 18 '17 at 8:31
• @მამუკაჯიბლაძე Yes, it may be straightforward to some degree, but not very revealing. – Alexander Burstein Nov 18 '17 at 8:38
• Yes, I agree - I just wanted to say that maybe the central question must be how to obtain that explicit expression combinatorially. – მამუკა ჯიბლაძე Nov 18 '17 at 8:39
• @მამუკაჯიბლაძე Maybe, and it's also interesting to figure out exactly where the overage comes from. – Alexander Burstein Nov 18 '17 at 8:51