Two people in a greed therapy group are playing a game. There is a pot of 6 dollars and each person, while isolated, is asked the question "Do you want to be greedy and take all of it?" The money goes to the person who is greedy and if they answer the same they share it. To punish their greed, a player must pay 1 dollar of what he won if he were greedy and an extra dollar if both players were greedy.

You \ Other | Not Greedy | Greedy |
Not Greedy  |     3\3    |   0\5  |
Greedy      |     5\0    |   1\1  |

A) Which option should you choose the first time you play?

B) If you continue playing an unknown finite number of games, what strategy should you use to maximize the amount of money you can win? (Assume that your opponent doesn't necessarily use the same strategy as you.)

Note: Both players are trying to get as much money as possible, and neither needs to get a certain amount of money at all costs.

If we consider whole strategies rather than single games, we can create a new chart with the average amount of $ per game when different strategies go against each other.

I'm only going to consider 4 basic strategies, for now...

Strategy 1: Always be greedy.
Strategy 2: Never be greedy.
Strategy 3: copy last person's move, start out greedy.
Strategy 4: copy last person's move, start out not greedy.

Other possible strategies include a forgiving strategy, which occasionally forgives the other person's greed.  Perhaps what would more realistically occur is a learning strategy that attempts to recognize the other person's strategy and adapt.  I'm not going to consider these other strategies in this comment.

In this chart, the different columns represent the other person's strategy, while the different strategies "I" use are represented in the different rows.  Each number in the chart is the average money "I" get per game over a large number of games.

   1  2   3   4
1  1  5   1   1
2  0  3   3   3
3  1  3   1 2.5
4  1  3 2.5   3

I think it's pretty clear that if we limit ourselves to these four strategies, number 4 is always better than 3.  Eliminating number 3 from our table... (I eliminate it from both columns and rows because I assume, as given, that both players try to get as much money as possible)

   1  2  4
1  1  5  1
2  0  3  3
4  1  3  3

Now, it's pretty clear that 4 is always better than 2.  Furthermore, when we eliminate 2, we can eliminate 1 as well, since 4 is always better.

In conclusion, if these were our only available strategies, strategy number 4 is undoubtedly without any argument the only strategy that should be used.

However, this begs the question.  Surely adding other possible strategies to choose from will make it not so clean-cut.  Probably.

Posted by Tristan on 2005-10-07 01:40:21