You're playing a game with a standard 52 card deck, high card wins (ace is high) and if you both pick the same card it's a push. Player A picks a card. Player A now has the option to keep this card, or discard it (it does not go back into the deck) and pick a new card. If player A keeps his first card, player B picks a card and must keep it. If player A discards his first card and picks a second card, player A must keep the second card (even if the first picked card is higher) and player B picks a card and must keep it.
I will put my thoughts into a spoiler so others can work on this unencumbered if they wish.
Spoiler:
As always, I give the standard disclaimer that mistakes in what is presented below are possible and, in any event, people should approach the question and answers in however they so desire.
Clearly there are a variety of strategies A can employ for his keep/discard cutoff. I think it is fairly obvious what is A's optimal cutoff strategy (see below).
Once you have a well-defined strategy for A, then it is merely a matter of turning the crank on dealing out every possibility and seeing how often A beats B or B beats A.
These can be written straightforwardly as sums using integers 1-52 for the cards and greatest integer functions for the ranks. The evaluation of such sums is probably also straightforward but I took the easy way out and coded this up on a computer.
I coded up any cutoff strategy for A and then dealt out all possible 3-tuples of 52 cards (X,Y,Z) for the possible outcomes. The ultimate cards that get shown down for A and B may be different combinations of X,Y,Z depending upon whether the first card drawn (X) exceeds the cutoff.
Anyway, I used the program to loop over all possible cutoff strategies A could employ. The computer found that the optimal cutoff card is an 8; player A should keep the first card if it is higher than an 8. Under that strategy, the computer finds that A wins 79,200 3-tuples, B wins 45,600 3-tuples, and they tie 7,800 3-tuples.
This works out to be an edge of 33,600 / 132,600 = 25.3393665%. For what it is worth, this fraction simplifies to 56/221 which is probably derivable via direct methods but I leave that to others who may be interested in that approach.
If it was an infinite deck then half the time A would be somewhere between 50% and 100% for an average of 75%. The other half he would be even money. That's an average of 62.5 % or an edge of 25%.
Nate Silver asked the more interesting question for the infinite deck. They each get a card. A can choose to replace it and after seeing what he does B has the same choice.
