Two players, A and B, are each given 10000 dollars. They'll play a 10 round game. In each round both have to gamble some of this money (can be zero); in the 10th round, they both must use all their remaining money. The winner of each round is the one who gambles less. He gets both sums gambled in that round; these winnings are put away and not added to the money available for gambling. The winner of the game is the one who won more money.

What is the best strategy to win this kind of game?

Example for a 3 round-game: round 1, A gambles 5000, B gambles 4000 --> B gets 9000. Round 2, A 5000, B 4000 --> B gets 9000. Round 3, A 0, B 2000 --> A gets 2000, B wins getting 18000.

This seems like a zero-sum game theory, but there's something different, any help will be appreciated.

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    $\begingroup$ You may be interested in this: en.wikipedia.org/wiki/Blotto_games, though it's not quite the same. $\endgroup$ – joriki Dec 13 '11 at 16:19
  • $\begingroup$ This is also related: math.stackexchange.com/questions/76095/…. I thought about that one for quite a while and found that the analysis quickly becomes surprisingly complicated if the game has more than two rounds; I believe the same will be the case here. $\endgroup$ – joriki Dec 13 '11 at 16:24
  • $\begingroup$ Note also that, from a game-theoretric viewpoint (i.e. ignoring issues of psychology and cognitive limitations), the question to be asked about a symmetric game like this is not what's the best strategy to win, since there can't be one, but what is the optimal stragegy in the sense of a Nash equilibrium. Note also that the existence of a Nash equilibrium is only guaranteed for games with a finite strategy set; there are examples of continuous games like this one without a Nash equilibrium. $\endgroup$ – joriki Dec 14 '11 at 10:25
  • $\begingroup$ @joriki: like one in the prisoner's dilema, you mean? $\endgroup$ – zfm Dec 14 '11 at 11:08
  • $\begingroup$ No. You should probably read up on the prisoner's dilemma and/or on Nash equilibria if you thought that. I couldn't have meant the prisoner's dilemma both because it has a finite strategy set and because it has a straightforward (pure) Nash equilibrium, in which both players always defect. I saw an example a while ago; I'll post a link if I find it again. $\endgroup$ – joriki Dec 14 '11 at 13:42

You don't say what happens in case of a tie: I'll suppose they both keep their money.

If they both must gamble all their money in the 10th round, the winner will be the one with the least money after the 9th round. So in the 9th round, the players are trying to lose all their money: obviously they will gamble all their money. The winner is the one with the most money after the 8th round. In the 8th round, the players want to get as much money as possible, so the optimal strategy is to gamble 0 and win whatever your opponent is foolish enough to gamble. Similarly in each of rounds 1 to 7, the optimal strategy is to gamble 0.

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    $\begingroup$ don't really get this... everything we used (and we got, if won) in a round before can't be used again... $\endgroup$ – zfm Dec 14 '11 at 8:52
  • $\begingroup$ I initially misunderstood the rules in the same way (and came to the same conclusion); they're rather confusingly stated. $\endgroup$ – joriki Dec 14 '11 at 9:30
  • $\begingroup$ I've tried to clarify the statement of the rules. $\endgroup$ – joriki Dec 14 '11 at 9:36
  • $\begingroup$ @joriki: thanks! $\endgroup$ – zfm Dec 14 '11 at 9:48
  • $\begingroup$ To the downvoter: This is a correct solution to the game one might easily have taken the initial question to describe. $\endgroup$ – joriki Dec 14 '11 at 9:56

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