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?
OK, so let's try again. The product for n=1 to ∞ of (1-(z/n)²) is sin(πz)/(πz) so the product I mentioned in the previous message is the limit, as z→1, of 1/(1-z²) times sin(πz)/(πz), which comes out to ½.
Edited on November 1, 2004, 11:24 pm
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