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?
(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.
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Posted by Charlie
on 2004-11-01 14:37:33 |
re: No Subject
