The observable universe is not the entire universe. Our observable universe could be surrounded by antimatter.
Maybe the antimatter surrounding the observable universe is destroying the outer parts of our matter pocket and that is why our universe is expanding -- into it!
Opponent throws rock more than he wants to => You shouldn't throw scissors at all (assumption)
Let your optimal throw be according to probability { R: 1-x, P: x, S: 0 } Opponent doesn't want to throw R at all in this case; he should respond by matching optimal probabilities on what he can choose: { R: 1/2, P: (1-x)/2, S: x/2 } You continue to assume that you shouldn't throw any scissors; you respond by matching optimal probabilities on R and P: { R: x/(x+1), P: 1/(x+1), S: 0 }
Since your throw probabilities are optimal 1-x = x/(x+1) and x = 1/(x+1)
Guess what? x = 1/(x+1) is the definition of the golden ratio conjugate! (x = 0.618034) And it works in the first equation as well Therefore your optimal throw is according to probabilities { R: 0.381966, P: 0.618034, S: 0 } Therefore your opponent throws according to probabilities { R:.5, P: 0.190983, S: 0.309017 }
You will win (2x-x^2)/2 = 0.427051 of the time, lose (1-2x+2x^2)/2 = 0.263932 of the time, and draw (1-x^2)/2 = 0.309017 of the time. When draws are dropped you win 0.618034 (x) of the time.
I believe this is the optimal solution (but I still have to confirm the above assumption)
Slashdot Mirror
The observable universe is not the entire universe. Our observable universe could be surrounded by antimatter.
Maybe the antimatter surrounding the observable universe is destroying the outer parts of our matter pocket and that is why our universe is expanding -- into it!
Opponent throws rock more than he wants to => You shouldn't throw scissors at all (assumption)
Let your optimal throw be according to probability { R: 1-x, P: x, S: 0 }
Opponent doesn't want to throw R at all in this case; he should respond by matching optimal probabilities on what he can choose: { R: 1/2, P: (1-x)/2, S: x/2 }
You continue to assume that you shouldn't throw any scissors; you respond by matching optimal probabilities on R and P: { R: x/(x+1), P: 1/(x+1), S: 0 }
Since your throw probabilities are optimal 1-x = x/(x+1) and x = 1/(x+1)
Guess what? x = 1/(x+1) is the definition of the golden ratio conjugate! (x = 0.618034) And it works in the first equation as well .5, P: 0.190983, S: 0.309017 }
Therefore your optimal throw is according to probabilities { R: 0.381966, P: 0.618034, S: 0 }
Therefore your opponent throws according to probabilities { R:
You will win (2x-x^2)/2 = 0.427051 of the time, lose (1-2x+2x^2)/2 = 0.263932 of the time, and draw (1-x^2)/2 = 0.309017 of the time. When draws are dropped you win 0.618034 (x) of the time.
I believe this is the optimal solution (but I still have to confirm the above assumption)