Game theory in Rock-Paper-Scissor
Given that the current chapter of the class in is on game theory, I wanted to apply this concept on a simple game like rock paper scissor. Everyone knows the game and it seems like a fairly balanced game. The players have 3 options, play Rock, Paper or Scissor: Rock crushes Scissor, Scissor cuts Paper, Paper covers Rock and if both player pick the same it’a a tie. If we play first to one, there is no real strategy, no Nash equilibrium both player have a probability of 1/3 to win.
However, it gets much more interesting when we go for multiple rounds. That’s when the some sort of strategy comes in. Let’s say player A’s strategy is to always pick Paper. Player B’s best response is going to be play Scissor all the time. Player A is going to adapt and his counter strategy is to play Rock. It can go on indefinitely. There is no pure strategy that makes you win all the time. The sum of your payoff is going to be 0.
| A | B | R | P | S |
| R | 0,0 | -1,1 | 1,-1 |
| P | 1,-1 | 0,0 | -1,1 |
| S | -1,1 | 1,-1 | 0,0 |
To win more often than your opponent, you need to add “randomness” into your game. For that you have to adopt a mixed strategy. Player A new strategy is to play 2/4 Scissor, 1/4 Rock, 1/4 Paper and player B’s strategy is to play Paper all the time. Player A is going to win half of the time and player B only one quarter of the time.
| A | B | R | P | S |
| R | 0 | 1/4 | 0 |
| P | 0 | 1/4 | 0 |
| S | 0 | 2/4 | 0 |
| A | B | R | P | S |
| R | 2/16 | 1/16 | 1/16 |
| P | 2/16 | 1/16 | 1/16 |
| S | 4/16 | 2/16 | 2/16 |
| A | B | R | P | S |
| R | 1/9 | 1/9 | 1/9 |
| P | 1/9 | 1/9 | 1/9 |
| S | 1/9 | 1/9 | 1/9 |
Neither player can improve their strategy, the Nash equilibrium is reached. The draw is the most common outcome.
The best strategy is to play randomly.
Source: https://www.inverse.com/mind-body/rock-paper-scissors-not-sports