4 people play a game of chance. They each take turns until everyone has taken a turn, then they begin a new round. They stay in the same order every round. Every time a player takes a turn, they have a certain chance of winning. When someone wins, the game ends. They all have even odds of winning a game. The chance of someone winning in any given round is 3/5.

What is the probability for each person to win during their turns?

let pa pb pc pd denote probability of winning within the 1st round by players A B C D respectively. Let x denote probability of winning the game per se regardless of player's place in the round.

CLEARLY:

pa=x

pb=x*( 1-x)

pc=x*( 1-x)*(1-pb)

pd=x*( 1-x)*(1-pb)*(1-pc)

and pa+pb+pc+pd=3/5

substituting and solving we get

x=pa=19.5%

pb=15.7%

pc=13.2%

pd=11.5%

to get the values of probabilities for the whole game (multiple rounds) we multiply those values

by 5/3 ( infinite geometric series) to get:

PA = 32.6 %

PB= 26.18 %

PC= 22.0 %

PD= 19.17 %

ady

*Edited on ***January 30, 2004, 11:57 am**