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 Marble Game (Posted on 2004-11-01)
You're playing a game. You start with a box with one black marble and one white marble, and you sample twice with replacement. If you select the white marble both times, you win. If you select the black marble either time, you add another black marble and try again. On each round, you sample twice with replacement, winning if you select the white marble twice, otherwise adding another black marble and moving on to the next round.

What is the probability that you eventually win? Equivalently, if P(n) is the probability that you win on or before round n, what is the limit of P(n) as n -> infinity?

 See The Solution Submitted by Brian Smith Rating: 4.2000 (5 votes)

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 Fancy Answer | Comment 6 of 12 |
The odds of winning the first time are 1/2² -- the odds of getting the white ball are 1/2, and you need to manage it twice; thus, P(1)=1/2². The odds of winning the second time are 1/3², so P(2)=1/2²+1/3². In general, you find P(n)=1/2²+1/3²+...+1/(n+1)². The limit is ζ(2)-1 (ζ() is the Riemann function) about 64.5%.
 Posted by Federico Kereki on 2004-11-01 15:17:34

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