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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.

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Alison already gave you a reference, but note that the same question (and reference) appeared on MO at… – Joe Silverman May 10 '11 at 20:23
@Joe. I formed my question by myself, but it is natural enough to bridge players that it comes independently to many people. – Denis Serre May 11 '11 at 5:33
up vote 17 down vote accepted

I believe the game you describe is two-person single suit whist and was solved by Johan Wastlund in this paper.

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Amazing! Such an accurate answer given within 15'. MO is really is nice toll. Thanks. – Denis Serre May 10 '11 at 19:46
And some probabilistic results that had to be left out of the EJC-version can be found at . In particular, asymptotically half the time, not having the lead will be worth a trick. To summarize the paper within the limits of a comment: (1) The game becomes much nicer if "ducking" is forbidden. (2) For large n, such a rule changes almost nothing. – Johan Wästlund May 10 '11 at 19:52
By the way, the question was asked before:… – Johan Wästlund May 10 '11 at 20:44

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