Game Strategy

## Computer Undercut: Game Strategy
For students who know some algebra and some probability, here is one way to arrive at what would be a "best strategy" using the idea of expected value. Assume that you selected your number randomly with the following probabilities. number chosen probability 1 a 2 b 3 c 4 d 5 e Remember that since these are probabilities, a + b + c + d + e = 1. Next you have to construct a payoff matrix, which gives the net payoffs for each set of move. For example, if I select 1 and my opponent selects 1, my net payoff equals my winnings minus his/her winnings which equals 1 - 1 = 0. If I selected 4 and my opponent selected 1 then my net payoff equals 4 - 1 = 3. If I selected 5 and my oppponent selected 4, my net payoff would be -9. Here is the payoff matrix: my choice 1 2 3 4 5 1 0 -3 2 3 4 2 3 0 -5 2 3 opponent's choice 3 -2 5 0 -7 2 4 -3 -2 7 0 -9 5 -4 -3 -2 9 0 For this to be a fair game, my payoff should be 0. This means that 0a -3b + 2c + 3d + 4e = 0 3a + 0b - 5c + 2d + 3e = 0 -2a + 5b + 0c - 7d + 2e = 0 -3a - 2b + 7c + 0d - 9e = 0 - 4a - 3b - 2c + 9d + 0e = 0 Remember that a + b + c + d + e + f = 1. This gives us 6 equations in 5 unknowns which turns out to be solvable. (This is a great problem for students who are just learning to solve systems of equations). It turns out that the original system has an infinite number of solutions, but if you add the requirement that a + b + c + d + e + f = 1 a= 10/66 b= 26/66 c= 13/66 d= 16/66 e= 1/66 This says that if you were to play randomly and play . 1 with probability 10/66 . 2 with probability 26/66 . 3 with probability 13/66 . 4 with probability 16/66 . 5 with probability 1/66 then in the long run you would break even. Since you are rational, this is the strategy you would choose or so say the gurus of game theory. For more on this see Douglas Hofstadter's Metamagical Themas. -- Jonathan Choate |