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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 solution | Comment 2 of 22 |
The probability that, during a given round, no one wins is 1 - 3/5 = 2/5. This is the product of the probabilities that the first player doesn't win, the 2nd player also doesn't win, and likewise the 3rd and 4th. As all of these are the same, x^4=2/5 where x is the probability that on one turn one player does not win.

The fourth root of 2/5 is .7952707287670506, so the probability of a win on a given turn is 1 - .7952707287670506 = .2047292712329494. So this is the probability of the first player's immediate win.

The probability that the 2nd player will win on his first turn is .7952707287670506 * .2047292712329494, representing the probability the first player does not win multiplied by the probability that the 2nd player then does win.

The 3rd player likewise has to wait for players 1 and 2 to lose before being given the opportunity to win, and therefore has probability .7952707287670506^2 * .2047292712329494 of winning in the first round.

The 4th player then has probability of winning of .7952707287670506^3 * .2047292712329494 of winning on the first round.

These come out to .2047292712329494, .1628151967333748, .1294821601605017 and .1029733718731742 for the 1st, 2nd, 3rd and 4th players, respectively. As a check, they do add to .6, the probability that one of the players will win within the round.

Some player will eventually win, and the probability will be in proportion to the individual probabilities within one round. As they add up to .6, we must divide each of the above individual-round probabilities by .6 to get the overall probability that that person will ultimately win, giving probabilities of .3412154520549157, .2713586612222914, .2158036002675029, .1716222864552903 respectively, or about 34%, 27%, 22% and 17% for the first, second, third and fourth players respectively.
  Posted by Charlie on 2004-01-27 13:41:23
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