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?

(In reply to solution by Charlie)

I may have been reading the puzzle wrong. If the idea is that each player have the same ultimate chance of winning, then within each round, each player must have a different chance of getting an immediate win.

We want each player to have an equal chance of winning within a round, and the total chance is .6, so each one is to have a 15% chance of winning in each round, including the first round.

Thus the first player must be given a 15% chance of an immediate win. Then, as the second player is to have an overall probability of a first round win of 15%, his turn probability must be .15/(1-.15) = 3/17. The third player must then be given .15/(1-.3) chance of winning or 3/14. The 4th player is then given probability of .15/(1-.45) = 3/11.

Thus if the randomizing procedure for use by player 1 gives him 3/20 probability of a win (when he shoots, draws or whatever procedure is used to randomize), that for player 2 gives him a 3/17 probability of a win, that for player 3 a 3/14 win probability and for player 4 a 3/11 chance of a win, then in each round each will have an overall 15% probability of a win given that he has to await a failure by his predecessors. Then each has an even chance of winning the game as a whole.

Posted by Charlie on 2004-01-27 14:14:29