0
$\begingroup$

An urn contains B black and W white balls. What is the expected number of drawings required to get a ball of each color ? With or without replacement, up to you.

With replacement, am I right to think that it is

$$2bw+3b^2w+4b^3w\cdots+2wb+3w^2b+4w^3b\cdots=bw\sum_{k=0}^\infty (k+2)(b^k+w^k)=bw\left(\frac{1+w}{w^2}+\frac{1+b}{b^2}\right)=1+\frac WB+\frac BW$$ where $b$ and $w$ are the fractions of both colors ?

$\endgroup$
7
  • $\begingroup$ Without replacement, the number of draws cannot exceed $\max\{B,W\} + 1$, no? $\endgroup$ – Dilip Sarwate Nov 17 '15 at 16:27
  • $\begingroup$ Ooops, sorry, I mean with. $\endgroup$ – Yves Daoust Nov 17 '15 at 16:29
  • $\begingroup$ This is the coupon collector problem. The search turns up many threads that provide explicit formulas. $\endgroup$ – whuber Nov 17 '15 at 17:00
  • $\begingroup$ @whuber: all entries I found are about equiprobable coupons. For two coupons, the expectation is $2H_2=3$, which matches my solution. $\endgroup$ – Yves Daoust Nov 17 '15 at 17:06
  • $\begingroup$ Many of the methods described in the answers apply with almost no change to varying probabilities. $\endgroup$ – whuber Nov 17 '15 at 18:14
1
$\begingroup$

With replacement, the urn always has $B+W$ balls. Conditioned on the first ball being Black, the (conditional) expected waiting time to get one ball of each color is $1 + \frac{B+W}{W}$ where the fraction is the expected waiting time to get a White ball. Similarly, conditioned on the first ball being White, the (conditional) expected waiting time to get one ball of each color is $1 + \frac{B+W}{B}$. Now put these answers together to get the unconditional waiting time.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.