# We Got A Problem #11: Rock Rock Paper Scissors

December 19, 2012 6 Comments

Most people know Rock-Paper-Scissors (or its more fun cousin, Ninja-Cowboy-Bear). Kate and Ogden are locked in a long battle of online Rock-Paper-Scissors, with a winner scoring 1 point, and a tie scoring nothing for either player. Game’s to 100.

Ogden won the first series 100-77 by constantly looking for patterns in Kate’s choices. For the second series, Kate tries something interesting: she has no pattern. She randomly selects rock, paper, or scissors with equal probability. She even tells Ogden that she is doing this, and he can do nothing about it but sigh and accept that he cannot beat this “strategy”. Kate wins the second series 100-95 by luck.

For the third series, Ogden changes a rule: if rock beats scissors, the winner scores 2 points instead of 1. Other victories (scissors over paper, paper over rock) are still worth 1.

Now Kate’s panicked: if she still randomly selects, is she beatable in the long run, or not? Can she alter the probabilities to make a new “strategy” that Ogden can’t beat? It seems like it would be bad to play scissors so often, and she should probably play more rocks… or not?

Let us know whether you find any ways that Kate can play to force Ogden into equilibrium.

If Ogden does rock every time and Kate continues her random choices, Ogden will win in the long run, since 1/3 times Kate will throw a scissors and he will beat her winning 2 points instead of 1. This part appears to be obvious, so Kate needs to do something to change her strategy.

it seems like all she’d have to do is just throw nothing but rock and paper every time, randomly of course between them. Then Ogden will never get the 2 point advantage, will not want to play scissors in fear of a 50% of it getting beaten, so Ogden will play nothing but rock and paper as well. Seems like then they will be back to a 50/50 luck based game.

If Ogden sees that Kate is throwing nothing but rock and paper every time, what will he do?

Throw paper every time…. He’ll never get beat since she isn’t throwing scissors….

Yup. So, Kate will lose if she doesn’t throw any scissors… but she’ll lose if she throws 1/3 scissors, too. Try something else!

This throwing scissors business is dangerous. I’ll have none of it.

We’d like Kate’s mixed strategy to make Ogden indifferent among rock, paper, and scissors, which means we need his E(playing rock) = E(playing paper) = E(playing scissors). Let Kate’s probabilities of throwing rock, paper, and scissors be r, p, and s, respectively. Then, for Odgen:

E(playing rock) = 2s … since he gets 2 points if Kate throws scissors, and 0 otherwise

E(playing paper) = r … since he gets 1 point if Kate throws rock, and 0 otherwise

E(playing scissors) = p … since he gets 1 point if Kate throws paper, and 0 otherwise

To find Ogden’s indifference point, we set up a system of equations:

2s = r

2s = p

r + p + s = 1

When you solve that, you get r = p = 2/5, and s = 1/5. So Kate should play scissors only 1/5 of the time (to keep Ogen honest), and otherwise play rock and scissors equally often. The situation is symmetric, so Ogden’s best response to Kate is to play the same strategy.

Right? I think. Been out of the game theory game for a while now.