This game looks like bridge, but 1- there are only two players Alice and Bob, 2- there is only one suit, whose cards are numbered $1, 2,\ldots,2n$. One deals each player $n$ cards. Therefore Alice knows Bob's cards and conversely; once the cards are dealt, there is no randomness.


Alice play a card, then Bob. The highest card wins the trick. The winner of the trick leads a card and so on. At the end Alice has got $p$ tricks and Bob $n-p$ tricks. The goal for each player is to get as much tricks as possible with the cards (s)he was dealt.


My question is about the strategy of play and the number of tricks you expect to win in a given layout. The answer should not be obvious. Let me give an example with $n=3$. If Alice is dealt $6,4,1$, she gets two tricks by leading first the $1$. If she has instead $6,3,2$, she leads the $3$ (equivalently the $2$). I see easily the way to get as many tricks as possible if $n$ is small, say $n\le6$, but I don't see a generalization.


Of course, the number of expected tricks with a given hand differs whether you begin or you opponent does.



Bridge game with only one suit: strategy
I believe the game you describe is two-person single suit whist and was solved by Johan Wastlund in [this paper](http://www.combinatorics.org/Volume_12/PDF/v12i1r43.pdf).


