A box contains $3$ large donuts and $7$ small donuts. Each hour, you randomly choose a donut from the box. If it's a small donut, it's eaten; If it's a large donut, it's cut in half, where one half is eaten and the other half is returned into the box as a small donut. After how many hours is it expected for the box to only have small donuts left?
The only way I can think of approaching this problem is calculating the probability of finishing the small donuts in n minutes and multiplying that by n for all possible n. However, this seems quite tedious. Is there a more efficient approach to tackling a problem like this?
The answer is $E(3,7) = 113/12$, where $E(m,n)$ is the expected hours for the box to only have small donuts left if there are $m$ large donuts and $n$ small donuts initially.
We have formulas $$ \begin{aligned} E(1,n) &=1+\frac{n}{2}\\ E(2,n) & = \frac{5}{2} + \frac{2n}{3}\\ E(3,n)& = \frac{25}{6} + \frac{3n}{4}\,. \end{aligned} $$ The proof is using induction.