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Game of luck (Posted on 2004-01-27) Difficulty: 2 of 5
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?

See The Solution Submitted by Tristan    
Rating: 3.4000 (5 votes)

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Solution my solution | Comment 10 of 22 |
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
  Posted by Ady TZIDON on 2004-01-30 11:54:53
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