Question

Many of you probably played the game “Rock, Paper, Scissors” as a child. Consider the following variation of that game. Instead of two players, suppose three players play this game, and let us call these players A, B, and C. Each player selects one of these three items—Rock, Paper, or Scissors—independent of each other. Player A will win the game if all three players select the same item, for example, rock. Player B will win the game if exactly two of the three players select the same item and the third player selects a different item. Player C will win the game if every player selects a different item. If Player B wins the game, he or she will be paid $1. Assuming that the expected winnings should be the same for each player to make this a fair game, how much should Players A and C be paid if they win the game?

Answer 1

Let Xa, Xb, Xc be the amount paid if they win the game. Here Xa = 1

Each player can select Rock, Paper, or Scissors with equal probability.

so prob( A win ) = Prob( All three select same item )

= prob( All select Rock ) + Prob( all select papaer ) + prob ( all select scissor)

= 1 /3 x 1/3 x 1/3 + 1 /3 x 1/3 x 1/3 + 1 /3 x 1/3 x 1/3

= 1/9

Prob( C win ) = prob( A choose rock)xProb(B choose paper) x prob( c choose scissor) x 3!

= 1 /3 x 1/3 x 1/3 x 6

= 2/9

Prob( B win ) = Prob( two select same and one different)

=1 - prob( A win ) - Prob( B win )

= 1 - 1/9 - 2/9

= 2 /3

For a fair game the expected value should be same

so

1 x 1/9 = Xb x 2/3 = Xc x 2/9

Therefore Xb = 1/6 dollar

Xc = 1/2 dollar