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 Game of luck (Posted on 2004-01-27)
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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 re: another remark to SK 'S solution | Comment 14 of 22 |
(In reply to another remark to SK 'S solution by Ady TZIDON)

It was initially my misinterpretation also that a "game" was one player's turn during one round. But that is not the case, as the puzzle states that "When someone wins, the game ends." Before someone wins the same game is still on. The idea is to adjust the probabilities for each player so that each has the same chance of winning the total game.

But even with the idea that there is a constant x, which is the probability of winning on a given turn, assuming you get to your turn, the set of equations you originally give is wrong:

The probability that c would win on a given turn, taking into consideration the previous probabilities is given by you as pc = x*(1-x)*(1-pb)

However, pb is the probability that B would win already reduced by the chance that A did not win. The actual probability that C would be given a chance is (1-x)*(1-x), and so the probability pc is x*(1-x)*(1-x), rather than x*(1-x)*(1-pb), as we are assuming an equal chance of any given turn producing a winner.

Look at my original solution to see how this (actually incorrect) assumption should work out. For the actual way Tristan intended, see SK's solution and my correction to my own post.
 Posted by Charlie on 2004-01-31 11:50:33

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