The mathematics of prediction: [Avinash Q. Dixit and Barry J. Nalebluff; Thinking Stategically]
Let's say you have a tennis player receiving and his opponent is serving. If he correctly guesses his opponent will serve to his forehand, he will return it at 90% success. If he correctly guesses his opponent will serve to his backhand, he will return at 60%. If he predicts forehand, but his opponent serves backhand, he will return at only 20%. It is slightly better if he guesses backhand and opponent serves forehand, with a 30% return rate.
What is the best strategy for server and returner?
The equilibrium point is found by graphing the following:
On the x-axis, you have "percentage of times the server aims to forehand", ranging from 0 to 100. On the y-axis, you have "percentage of successful returns". Plot the line "anticipate backhand" between the endpoints (0, 60) and (100, 30). Plot the line "anticipate forehand" between the points (0, 20) and (100, 90).
Now, the SERVER wants to find the x-value where his opponent cannot change strategy to get a better result; i.e. the highest minimum value of these two curves. This occurs at the crossing point of the two curves.
Similarly, the RECEIVER wants to find the y-value where his opponent doesn't reduce the line by changing his percentage of forehand/backhand. The crossing point is the place this occurs.
The crossing point is the value (40, 48) in this scenario.
This same logic can be applied to the equilibrium point of prediction. You have to weigh up what the success value is for making 100% prediction one way or 100% prediction the other way, graph the lines of values for your opponents' different options, and find the equilibirum point. This tells you what your optimum percentage spread for making the decision will be.
Note, you have to be random. If you got 33%, you shouldn't alternate between Choice 1-Choice 2-Choice 2, Choice 1-Choice 2-Choice 2, etc. You should roll a die before each decision, and if you get 1 or 2, make Choice 1. If you get 3-6, you make Choice 2.
Let's say you have a tennis player receiving and his opponent is serving. If he correctly guesses his opponent will serve to his forehand, he will return it at 90% success. If he correctly guesses his opponent will serve to his backhand, he will return at 60%. If he predicts forehand, but his opponent serves backhand, he will return at only 20%. It is slightly better if he guesses backhand and opponent serves forehand, with a 30% return rate.
What is the best strategy for server and returner?
The equilibrium point is found by graphing the following:
On the x-axis, you have "percentage of times the server aims to forehand", ranging from 0 to 100. On the y-axis, you have "percentage of successful returns". Plot the line "anticipate backhand" between the endpoints (0, 60) and (100, 30). Plot the line "anticipate forehand" between the points (0, 20) and (100, 90).
Now, the SERVER wants to find the x-value where his opponent cannot change strategy to get a better result; i.e. the highest minimum value of these two curves. This occurs at the crossing point of the two curves.
Similarly, the RECEIVER wants to find the y-value where his opponent doesn't reduce the line by changing his percentage of forehand/backhand. The crossing point is the place this occurs.
The crossing point is the value (40, 48) in this scenario.
This same logic can be applied to the equilibrium point of prediction. You have to weigh up what the success value is for making 100% prediction one way or 100% prediction the other way, graph the lines of values for your opponents' different options, and find the equilibirum point. This tells you what your optimum percentage spread for making the decision will be.
Note, you have to be random. If you got 33%, you shouldn't alternate between Choice 1-Choice 2-Choice 2, Choice 1-Choice 2-Choice 2, etc. You should roll a die before each decision, and if you get 1 or 2, make Choice 1. If you get 3-6, you make Choice 2.













