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Jan 31, 2012 at 23:45 comment added Barry Cipra @Zsban, if A holds 1,2,3 and B holds 1,2,4 (and you're starting the game at 0), then A can win by playing 2, but loses if he plays 3.
Jan 31, 2012 at 18:49 comment added Zsbán Ambrus Is it ever worth not to play the highest card you have when the rules allow you that?
Jan 31, 2012 at 3:28 comment added user20948 @Barry Cipra: I only meant that the game is finite. @Guilaume Brunerie: I think that it's not obvious that if the player has cards to play, the decision of pass will be worse. I'll rewrite the problem more formal.
Jan 30, 2012 at 16:34 comment added Guillaume Brunerie @Barry Well, yes B should play his 2, but passing is not better than playing 1 (if B pass or play 1, A will always be able to play 2 and win). I’m not saying that you should always play your smallest card, but that it cannot be worse than passing.
Jan 30, 2012 at 14:57 comment added Barry Cipra @Guillaume, it's not quite true that it's "always" better to respond to a pass by playing your smallest card. Suppose A and B are both holding a 1 and a 2. If A (foolishly) passes, then B should play his 2, not his 1. But I think the basic principle is correct: You should always play something (if you can) rather than nothing. (@Frank, I wasn't suggesting a scenario where one player always passes no matter what the other player does. I was suggesting the conceivability of a scenario where the first player to make an active move loses.)
Jan 29, 2012 at 15:26 comment added Guillaume Brunerie @Barry Cipra : If one player pass, it is always better for the other player to play its smallest card than to pass
Jan 29, 2012 at 2:29 comment added user20948 Thanks. Because it is finite with perfect information, the game is determined. I think that it is a mathematical game. Not like chess, which is also a finite game with perfect information, my game seems to be easier. I want a analysis of the game to find out an effective algorithm to the game, as Nim is solved (en.wikipedia.org/wiki/Nim). There's another case that we can prove it to be NPC for $n$. (en.wikipedia.org/wiki/…) Barray Cipra: If one player wants to throw "pass" forever, the other one can win.
Jan 28, 2012 at 13:02 comment added Barry Cipra It's not a hundred percent clear (to me) that this game doesn't admit situations where each player's only way of avoiding a loss is to throw the game to the other player -- i.e., situations where the game toggles back and forth between $w(0,A,B)$ and $w(0,B,A)$ and never ends.
Jan 28, 2012 at 12:41 comment added Guillaume Brunerie I updated my answer.
Jan 28, 2012 at 12:41 history edited Guillaume Brunerie CC BY-SA 3.0
Added the case when both player know both hands
Jan 28, 2012 at 8:08 comment added user20948 Thanks. I meant that when the player received a situation, he not only knew the set $A$ which meant his cards but also knew the set $B$ which meant the other player's cards. As I've written in the postscript in the text, $B$ is known to the player. In other words, either player always knows the cards in his hand and in the other player's hand. Thank you for your comment.
Jan 28, 2012 at 7:37 history answered Guillaume Brunerie CC BY-SA 3.0