Question for all you aspiring game theorists...

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Well first, I wonder if there is a "right" answer to this question because "optimal strategy" would essentially be relative to each player's idea of what is and is not a good deal.

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The optimal straties in game theoretic context reffered to as a Nash Equillibrium (NE). A NE is a set of strategies such that player A is reacting optimally (i.e getting the highest payoff) given Bs strategy and player B is reacting optimally given As strategy -i.e. both players are acting optimally given the strategies of the other player. Using this concept of an equillibrium in a finite period game what is fair or what is a good deal doesn not matter.

As was pointed out above there may be reasons to expect typical players to deviate from NE strategies in this sort of game (in the real world).

Passion

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Hmmm, learned something new right there. I've seen "NE" before but hadnt quite grasped what it means. Thanks man.

[bobman0330]

NE isn't ideally suited to this game. Firstly, if it's a finitely repeated game, the only NE is for B to get only 1 every round. He might play some head games, but none of his threats are strictly credible.

If this game is played for a randomly-determined, unknown to the players length of time, then a different concept comes into play, what is called an "Evolutionarily Stable Strategy" (ESS). The principle behind this is that player B will refuse to accept certain bargains. His threat is given credibility because there is no final round where he has to revert to the accepting 1 strategy.

What makes B willing to reject on any given round, an immediately -EV move, is his expectation that he will get a better deal on future rounds. Future rounds have to be discounted according to the probability that they will not occur. For ease of calculation, assume the rule is that each round there's a K% chance that the game terminates. B's future benefit for refusing to agree to anything less than N, assume the offer is X is (N-X)(1-K) + (N-X)(1-K)^2 + ... = (N-X)(1-K)/K. His loss is X + X(1-K) + X (1-k)^2 = X / K.

Now, the downside of B's strategy is that he refuses to accept any payoff less than N, so he has to have a strong enough future incentive to refuse an immediate payoff of X=N-1 in order for his strategy to be ESS. (otherwise A could offer X=N-1 and "break his will.")

So, (N-X)(1-K)/K >= X/K when X=N-1.

(N-N+1)(1-K)/K = (N-1)/K
1-K = N-1
N < 2-K
N=1.

Now, that seems like it can't possibly be correct, but I can't find the problem in my math/logic. Anyone?

[bobman0330]

OK, my problem was that X alone needs to be compared to the future gain from refusing and insisting on N.

So, when X=N-1
(N-N+1)(1-K)/K >= N-1
1/K - 1 >= N - 1
1/K >= N

So, if the game has a 1% chance of ending each round, A should offer $100 and B should accept.

Note that the amount of money in the pot is irrelevant to the solution. If N turns out to be a large portion of the total, then I think we have to look at the situation from A's perspective as well.