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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(6): another remark to SK 'S solution | Comment 20 of 22 |
(In reply to re(5): another remark to SK 'S solution by Ady TZIDON)

re:
"However you did not comment about the sum of the probabilities not being equal to 1 (100%). "

Actually I think the question had been whether
pa+pb+pc+pd=3/5
and this is true per your interpretation of the game, but, even then, not given the way they were calculated in
pa=x
pb=x*( 1-x)
pc=x*( 1-x)*(1-pb)
pd=x*( 1-x)*(1-pb)*(1-pc)
due to the use of pb and pc in the formulae for pc and pd instead of x.

And again, this is in your current and my original interpretation of the puzzle, which is not the intent of the poser, who wanted the situation to be considered:
"When someone wins, the game ends. They all have even odds of winning a game." That is, the whole set of possibly multiple rounds is one game, and the author wants each player to have the same chance of winning that game.

The only problem I see is in the singular construal of the word "they" in "Every time a player takes a turn, they have a certain chance of winning." "They have" actually means "he or she has". We can see this from the wording of the puzzle as a whole--specifically the two sentences I quoted before.

 Posted by Charlie on 2004-02-01 12:06:36

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