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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 4 of 22 |
Dang... Brian beat me to it... but my solution agrees with his... a nifty problem:
_______________________________________________

The four probabilities are:
3/20
3/17
3/14
3/11

respectively.
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Proof:

Let the respective probabilities for winning on each person's turn be represented by w, x, y, and z respectively.

"The chance of winning in any given round is 3/5" means that the change of NOBODY winning in a round is 2/5. This is equivalent to:
[1]: (1-w)(1-x)(1-y)(1-z) = 2/5

Now, because they all have equal odds of winning a game, we must take position into account.... because the first person has the advantage of position. Therefore, the first person must be least likely to win on his given turn. This gives the following equality:
w = (1-w)x = (1-w)(1-x)y = (1-x)(1-x)(1-y)z

which leads to the following equations...

[2]: x = w/(1-w)
[3]: y = w/(1-2w)
[4]: z = w/(1-3w)

Now we have 4 equations and 4 unknowns.

Plug equations 2, 3, and 4 into equation 1, and we see that
w = 3/20

Plugging this value into equations 2, 3 and 4, we get
x = 3/17
y = 3/14
z = 3/11

  Posted by SilverKnight on 2004-01-27 14:04:28
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