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
my solution
