PDA

View Full Version : Question for all you aspiring game theorists...

jayheaps
08-02-2005, 08:35 PM
For all you game theorists out there.

Here is a simple game with two players, A and B. A is given \$1000 and is able
to split it between he and B any way he chooses. B has the option to accept or
reject the offer. If he accepts the offer, the money is divided according to
A's split. If B rejects the offer, both players get nothing.

At face value, this game isn't interesting if played one time. B should always
accept. However, if this game is played 1000 times. it gets more interesting. Now consider that there will be 10 games going on at the same time with 10 different As and 10 different Bs. For the competition, the As are competing against the other As, not their actual opponent in the game.

What is A's optimal offer strategy? What is B's optimal strategy?

spaminator101
08-02-2005, 10:01 PM
Are the B's competing against the other B's as well as this changes the strategy of A as well.

emp1346
08-02-2005, 11:13 PM
[ QUOTE ]
Are the B's competing against the other B's as well as this changes the strategy of A as well.

[/ QUOTE ]

while this may not be the same question, i'm curious if yeah, it's a team competition...

otherwise, it would seem to me that if the benefit of A and B (on the same team) or independent, then A should always offer to split the money with B... *shrugs*

spaminator101
08-02-2005, 11:55 PM
if a team compatition then this would have been obvious al they both would split 50/50 everytime

passion
08-03-2005, 02:20 AM
[ QUOTE ]
For all you game theorists out there.

Here is a simple game with two players, A and B. A is given \$1000 and is able
to split it between he and B any way he chooses. B has the option to accept or
reject the offer. If he accepts the offer, the money is divided according to
A's split. If B rejects the offer, both players get nothing.

At face value, this game isn't interesting if played one time. B should always

accept. However, if this game is played 1000 times. it gets more interesting. Now consider that there will be 10 games going on at the same time with 10 different As and 10 different Bs. For the competition, the As are competing against the other As, not their actual opponent in the game.

What is A's optimal offer strategy? What is B's optimal strategy?

[/ QUOTE ]

I'm not sure the actual team competition matters to the game or at least you have not formalized how it matters. I am going to proceed by assuming that all As and Bs care about is their playoffs.

To solve these types of games it makes sense to start at the back and work forward.

What would happen in the 1000th period? A would offer B a the smallest possible share and B would accept.

What about the 999th period? Both players know that A will offer B the smallest share in the 1000th period and that B will accept. Knowing this, there is no reason for A to offer B anything but the smallest share in the 999th period.

What about the 998th period? Same thing - both players know A will offer B the smallest share in the last two periods and the B will accept - so in the 998th period A offers B the smallest share and B accepts.

Actually I think its the same thing for all prior periods.

When do you get someting different? If the game was played for an infinite (or indefinite) number of periods you would get different equillibrium strategies.

Passion

SumZero
08-03-2005, 04:21 AM
[ QUOTE ]
[ QUOTE ]
For all you game theorists out there.
Here is a simple game with two players, A and B. A is given \$1000 and is able
to split it between he and B any way he chooses. B has the option to accept or
reject the offer. If he accepts the offer, the money is divided according to
A's split. If B rejects the offer, both players get nothing.
At face value, this game isn't interesting if played one time. B should always
accept. However, if this game is played 1000 times. it gets more interesting. Now consider that there will be 10 games going on at the same time with 10 different As and 10 different Bs. For the competition, the As are competing against the other As, not their actual opponent in the game.
What is A's optimal offer strategy? What is B's optimal strategy?

[/ QUOTE ]
I'm not sure the actual team competition matters to the game or at least you have not formalized how it matters. I am going to proceed by assuming that all As and Bs care about is their playoffs.
To solve these types of games it makes sense to start at the back and work forward.
What would happen in the 1000th period? A would offer B a the smallest possible share and B would accept.
What about the 999th period? Both players know that A will offer B the smallest share in the 1000th period and that B will accept. Knowing this, there is no reason for A to offer B anything but the smallest share in the 999th period.
What about the 998th period? Same thing - both players know A will offer B the smallest share in the last two periods and the B will accept - so in the 998th period A offers B the smallest share and B accepts.
Actually I think its the same thing for all prior periods.
When do you get someting different? If the game was played for an infinite (or indefinite) number of periods you would get different equillibrium strategies.
Passion

[/ QUOTE ]

Well this is a case where your theoretical optimal strategy under the assumption that both players are super rational actors is potentially flawed. If you run this game as an experiment many B's will reject offers that they don't feel are fair. This seems may seem irrational to people who only value the units and don't realize that people are not rational actors and that money isn't always a "good" or isn't always the only "good". Some people will turn the money down on a matter of principle.

Also, even if people are relatively rational, B may be able to tell A and convince him that he'll reject any offer that is less than X% of the prize pool. If B follows through on it the first few times rejecting the offer will A continue to offer a 999-1 split? Should he? If B knows that A will change his behavior then B should have this spite strategy. But if A counters and says he'll always offer 999-1 no matter what and if B trusts this and if all B cares about is maximizing his own money and no outside values (like spite or fairness or whatnot) than the 999-1 split is the theoretical optimal. So again this is a case of theoretical optimal versus experimental optimal.

Our economics class ran this game (as a game, with no real money just play money) and iirc tended to find that nearly everyone would accept a 50-50 split, most people would accept a 60-40 split, about half would accept a 2/3-1/3 split, and about a third would accept any split where they got any amount of money. So with that population an offer of somewhere around 60-40 was best.

passion
08-03-2005, 07:54 AM
[ QUOTE ]
Well this is a case where your theoretical optimal strategy under the assumption that both players are super rational actors is potentially flawed. If you run this game as an experiment many B's will reject offers that they don't feel are fair. This seems may seem irrational to people who only value the units and don't realize that people are not rational actors and that money isn't always a "good" or isn't always the only "good". Some people will turn the money down on a matter of principle.

[/ QUOTE ]

Of course I agree with you. If you run this as an experiment you you would not get the As offering Bs the smallest possible shares. That said, it is game theory so we are operating under the assumption that all the players care about is their payoffs. If all players care about is their payoffs, the equillibrium strategies for A to offer B the smallest share in every period and for B to accept.

[ QUOTE ]
Also, even if people are relatively rational, B may be able to tell A and convince him that he'll reject any offer that is less than X% of the prize pool. If B follows through on it the first few times rejecting the offer will A continue to offer a 999-1 split? Should he? If B knows that A will change his behavior then B should have this spite strategy. But if A counters and says he'll always offer 999-1 no matter what and if B trusts this and if all B cares about is maximizing his own money and no outside values (like spite or fairness or whatnot) than the 999-1 split is the theoretical optimal. So again this is a case of theoretical optimal versus experimental optimal.

[/ QUOTE ]

Again its game theory so we must assume, in absence of other information, that all players care about is their discounted stream of payoffs. In the case where the game is played a known and finite number of periods Bs threats to not accept payouts are not credible because both players can clearly see what is going to happen in the last, second to last, third to last, ....., periods ect.

If the game is played an indefinite number or infinite number of periods then Bs threats will have some bite and you will get equillibrium strategies that involve the Bs getting offered a higher share.

My sense is that in an actual experiment (in the finite period game) the outcome would depend on the total amount of money to be divided in each period. If it where \$1 then A, then the Bs are not likely to accept a 1% share. If it were say \$1,000,000 then the Bs would gleefully accept a 1% share. It is easy to stand up for whats just and fair when it doesn't cost you anything.

Passion

08-03-2005, 01:28 PM
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. Theoretically, any amount is good for B. For assuming he had nothing to begin with it's all positive expected value. "A" on the other hand is competing against other "A"'s so any loss to him may cost him the whole competition. If I were "A" I would push for a 80/20 split each time, hoping for a 75/25 negotiation and banking that no one else is keeping 75% over the long run. However, if I were "B" I would realize this and go for more by threatening to reject offers. "B" would have to reject about 25% of the offers to get a 60/40 split about 20% of the time. The math is loose but the idea seems solid to me.

passion
08-03-2005, 02:02 PM
[ QUOTE ]
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.

[/ QUOTE ]

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

SumZero
08-03-2005, 09:45 PM
[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

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

[/ QUOTE ]

And, iirc, this type of game has been simulated. Where a group of people were supposed to devise rules for how their programs would act as A or B knowing that they'd get to play 100 times as A and 100 times as B against every other program with their goal being to have the most money possible. That is a slight wrinkle on the game because if there were 10 programs then accepting any split is better for you against 8 of the programs not involved in the split but may be worse against the other player you are playing against.

08-04-2005, 12:54 PM
[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

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

[/ QUOTE ]

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

bobman0330
08-04-2005, 01:19 PM
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 &gt;= X/K when X=N-1.

(N-N+1)(1-K)/K = (N-1)/K
1-K = N-1
N &lt; 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
08-04-2005, 02:05 PM
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 &gt;= N-1
1/K - 1 &gt;= N - 1
1/K &gt;= 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.