January 2006 Puzzle Solutions

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1. If punishment were to increase to -9, then the upper right corner would give Alice (4 * 0.9) + ((-9) * 0.1). The bottom right corner would give an expected gain of 0. That yields the matrix:

¿¿	Alice High	Alice Low
Bob High	3, 3	0, 2.7
Bob Low	2.7, 0	0, 0

In this case, if Alice cheats, Bob receives the same gain whether he cheats or not. So, the lower right and the upper right are the same from Bob's point of view. If Bob decides not to cheat, then the game state will move to the upper right and from there to the upper left. To encourage this, you could make the punishment even slightly more negative than -9 (e.g., -9.01).

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2. Suppose punishment is reduced to -5 and the probability of punishment is raised to 17 percent. Then the expected gain for Alice from cheating is (0.83 * 1) + (0.17 * -5) = -0.02 in the bottom right state. In the upper right corner, the expected gain will be (0.83 * 4) + (0.17 * -5) = 2.47.

¿¿	Alice High	Alice Low
Bob High	3, 3	0, 2.47
Bob Low	2.47, 0	-0.02, -0.02

So the only Nash equilibrium is the upper left.

3. Assuming a 10-percent likelihood of being caught, let us say the punishment is -9.01. Then the lower right has an effective value of (-9.01 * 0.1) + (1 * 0.9) = -0.001. The upper right has (for Alice) a value of (-9.01 * 0.1) + (4 * 0.9) = 2.699, which we'll approximate to 2.7.

This yields a game matrix of:

¿¿	Alice High	Alice Low
Bob High	5, 2	0, 2.7
Bob Low	2.7, 0	-0.001, -0.001

So, Alice will take the game to the upper right corner. Bob can rail about immorality, but Alice sees it all as a game.

4. On the other hand, if punishment increases to, say, -17 (which might constitute cutting off the hand of a thief who steals bread) then the game matrix becomes:

¿¿	Alice High	Alice Low
Bob High	5, 2	0, 1.9
Bob Low	1.9, 0	-0.8, -0.8

The only Nash equilibrium is the upper left.

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