You start with a box holding one black and one white marble, and sample twice with replacement.

probability/marble.game New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

You start with a box holding one black and one white marble, and sample
twice with replacement.  If you get the white marble both times, you
win.  Otherwise you add another black marble and try again: on each
round you sample twice with replacement, winning if you get white twice
and otherwise adding a black marble and moving on.  What is the
probability that you eventually win?

Show solution

I found many different ways to show that it's 1/2.
Here is a very easy way to do it, without any advanced calculus or
whatever...only highschool algebra:

Call H.n the probability to lose all rounds up to 'n' inclusively.
H.1=3/4 obviously.

H.2=(Prob. to lose all prev. rounds=H.1)*(Pw*Pb+Pb)
     Where Pw=prob. to choose the white marble,Pb=Prob to choose black one.
hence...
H.2=H.1*(1/3*2/3 + 2/3).
Using this reasoning and remembering that Pw at n'th round is 1/(n+1)
since there are n+1 marbles now in the box ( 1 white,n black) we get:

H.n=H.(n-1) * (1/(1+n) * n/(n+1) + n/(n+1))

rewrite this as:

H.n = n(n+2)/(n+1)^2 * H.(n-1)

This is a simple recursive formula that readily gives:

H.n = 1/2 + 1/(n+1)

[ Editor: this formula gives H.1 = 1, not 3/4; the product of the
factors n(n+2)/(n+1)^2 is (n+2)/(2(n+1)) = 1/2 + 1/(2(n+1)).  The
limit is 1/2 either way. ]

Hence, the probability of never winning this game is as n goes to infinity
which gives 1/2. so the Probability of winning this game is:

1-1/2= 1/2 .

Tada, the answer without orthogonal functions and integration.

-- Stephane Baune <an462@FreeNet.Carleton.CA>

[ Editor: the puzzle was posted by Martin Eiger <mie@thumper.bellcore.com>
from Technology Review. ]