Hi, I'd like to solve the following problem, a version of Rock-Papers-Scissors:
Rock beats Scissors 100% of the time
Scissors beats Paper 100% of the time
Paper beats Rock 99% of the time
What is the Nash equilibrium (for choosing what to play)?
I know that it will be a mixed strategy equilibrium, but I haven't found a place that will help solving 3x3s without simplification / reduction (they always find dominated strategies, for example).
I'm pretty sure I'm just forgetting something simple...
Thanks for any help.
Jul 1st 2012, 06:09 AM
Re: Game Theory 3x3
Rock beats paper 1% of the time?
The nash equilibrium can be written as:
Play rock with probability r.
Play scissors with probability s.
Play paper with probability 1-r-s.
Against a player with this strategy, it does not matter which strategy you choose. Therefore, playing 100% rock or 100% scissors or 100% paper is equivalent. This gives you 2 simple equations which you can use to solve for r and s.