An urn contains n+m balls, of which n are red and m are black. On each step, one ball is taken from the urn. If it is red, a black ball is instead returned to the urn. How would one find the expected value of the number of attempts before grabbing a black ball?
Put a different way: If the probability of success is 40% and increased by 10% for each failure (e.g. 40% fail, 50% fail, 60% succeed) what is the expected amount of attempts needed?
Edit: I think that I made some headway. I think it would be a modified geometric formula.
Again, using the percentage example above:
$Pr(X=1) = p_1 = 0.4$
$Pr(X=2) = (1-p_1)*p_2 = (1-0.4)*0.5 = 0.3$
$Pr(X=3) = (1-p_1)*(1-p_2)*p_3 = (1-0.4)*(1-0.5)*0.6 = 0.18$
Expected value would then be:
$ \mathbf{E}X=\sum_{1}^{k}\mathbf{Pr(X=k)*k} = 2.06$
Does that sound correct?
Ah, yes, this is for the case $m=4, n=6$.
You can easily generalise your solution for any integers $m, n$: $$\begin{align}\mathsf P(X=1) &= 0.4 &&=\frac{m}{(m+n)}\\[1ex] \mathsf P(X=2)&= (1-0.4)0.5 &&=\frac{n~(m+1)}{(m+n)^2}\\[0ex] &=0.6\cdot0.5 \\[1ex] \mathsf P(X=3)&= 0.6\cdot0.5\cdot0.6 &&=\frac{n~(n-1)~(m+2)}{(m+n)^3}\\[1ex] &~~\vdots\\[1ex] \mathsf P(X=7)&=0.6\cdot0.5\cdot0.4\cdot0.3\cdot0.2\cdot0.1\cdot1.0\\[1ex]\mathsf P(X=k) &=\frac {3+k}{10^{k}}\frac{6!}{(7-k)!}\mathbf 1_{k\in[[1..7]]}&&=\frac{n!~(m-1+k)}{(n+1-k)!~(m+n)^k}\mathbf 1_{k\in[1..n+1]]}\\[2ex]\mathsf E~X &= \sum_{k=1}^7 k\,\mathsf P(X=k)&&=\sum_{k=1}^{n+1}k~\mathsf P(X=k)\\&=6!\sum_{k=1}^7\frac{k~(3+k)}{(7-k)!~10^k}&&=n!\sum_{k=1}^{n+1}\frac{k~(m-1+k)}{(n+1-k)!~(m+n)^k}\\&=\dfrac{25799}{12500}\\&\approx 2.0639\end{align}$$
But there does not appear to be a closed form for this equation.