Problem
N urns contain respectively 1 black n-1 white, 2 black n-2 white,..., k black n-k white,...
and finally n black and 0 white balls. An urn is chosen at random and a ball is
withdrawn. In case the ball is black, what is the probability that a
second ball drawn will also be black?
Solution
The probability that the next ball will be black is the weighted average
P1*0 + P2*1/(n-1) + P3*2/(n-1) +...+
Pn*(n-1)/(n-1)
where P1 is the probability that
the currently selected urn is urn1, P2 is the probability
that it is urn2, ..., and Pn is the probability that it is
urnn. Then letting T = n(n+1)/2,
P1 = 1/T
P2 = 2/T ...
Pk = k/T ...
Pn = n/T
(See the PostScript for proof, although it may be easier to believe if you just
think about it a while.)
Then weighted average is the sum from k=1 to n of
S Pk * (k-1)/(n-1) =
S k/T * (k-1)/(n-1) = 1/(T(n-1)) * (Sk2 -
Sk ) =
n(n+1)(2n+1) n(n+1)
= (1/T) * ------------ - (1/T)* ------
6(n-1) 2(n-1)
2n + 1 1
= 2 * ------ - 2 * ------
6(n-1) 2(n-1)
2n + 1 1 2n + 1 - 3
= ------ - --- = -----------
3(n-1) n-1 3(n-1)
2n - 2
= ------ = 2/3
3n - 3
Remarkably, in this formulation the answer is EXACTLY 2/3 for any choice of n > 1.
Lee
P.S.
To prove that Pi equals i/T where T is n(n+1)/2, let A be the event
that a black ball has been drawn and Bi be the probability that the ith
urn has been selected. Then by Bayes' formula
P(A|Bi)P(Bi)
P(Bi|A) = -------------
S P(A|Bi)P(Bi)
i 1
--- * ---
n n i
= ------------------------------------ = --------
1 2 n 1 n(n+1)/2
( - + - + ... + - ) * -
n n n n