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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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 re: No Subject | Comment 5 of 12 |
(In reply to No Subject by Avin)

What you have shown is the probability of not having won by the round with n balls in play:

1*3  *  2*4  *  3*5  *  ...  * (n-1)*(n+1)
-----------------------------------------------
2*2  *  3*3  *  4*4  *  ...  *       n*n

as, for example,

1*3
----   = 3/4
2*2

is the probability that you will not win on the first round, with two balls in play.  This is why the probabilities multiply together, for successive non-wins.

Of course, ultimately 1/2 = 1 - 1/2, so coincidentally, the probability of winning in the long run is the same as the probability of not winning.

 Posted by Charlie on 2004-11-01 14:37:33

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