We have the box containing $m$ red and $n$blue balls.

Question

We have the box containing $m$ red and $n$ blue balls. The ball is sequentially removed and returned back until red balls appear twice in a row.
What is the average number of extracts will it take?
Note:-my answer is $\frac{{m}\choose{r}}{{m+n}\choose{r}}$. is it correct ?

Answer 1

There are two live states here, according to whether you drew red in the prior turn or not, call them $s_1,s_0$ respectively. Let $E_1$ be the expected number of turns remaining conditioned on having just drawn a red, and $E_0$ be the expected number conditioned on not having just drawn a red. The answer to the question is $E_0$.

Working from $s_1$. We consider the next draw and deduce $$E_1=\frac m{m+n}\times 1 +\frac n{m+n}\times (E_0+1)$$

Working from $s_0$. We consider the next draw and deduce $$E_0=\frac m{m+n}\times (E_1+1)+\frac n{m+n}\times (E_0+1)$$

This is easily solved to yield $$E_0=\boxed{\frac {(2m+n)(m+n)}{m^2}}$$

Sanity checks: if $n=0$ this becomes $\frac {2m^2}{m^2}=2$ as desired. If $m=0$ the denominator is $0$ so $E_0$ is not defined, as expected. If $m=n$ we get $E_0=6$ confirming the well known result about trying to toss $HH$ with a fair coin (see, e.g., this).