Tylenol problem For quite some time I whenever I am taking a Tylenol, often I take half a caplet, and put the other half back in the bottle. Next time again I take a caplet, break it in half and put the other half in the bottle. Now many times when I pick another caplet, it is a whole caplet I am getting, putting half back in the bottle. So the number of "wholes" are getting less and the number of "halves" are getting more. I have been wondering: How many caplets should I pick this way, until I pick a half caplet? In other words, what is the expected number of whole caplets I have to take before I pick my first half caplet? I realized that with 100 caplets, this computation is getting out of control, so I tried to look at this problem with only 20 whole caplets to begin with. So on the first draw, you don't get a half caplet: $1(0/20)$. Then if the first is a whole ($20/20$) and the second is the (only) half caplet: $(20/20)(1/20)$ times $2$ because you did $2$ draws. If the third draw is a half caplet: $(20/20)(19/20)(2/20)$ times $3$. (There are now $2$ half caplets in the bottle). So expectation: $$E(X)=1(0/20)+2(20/20)(1/20)+3(20/20)(19/20)(2/20)+4(20/20)(19/20)(18/20)(3/20)+5(20/20)(19/20)(18/20)(17/20)(4/20)+...$$ I did this work in my TI and arrived at 6.29 (rounded) and so according to the expectation, you can expect the 6th draw (or so) to be a half caplet. My question is now: Using sequences/series (or some other method), how can the expectation be converted into some algebra problem, instead of having to type it up in the TI? 
 A: Let $e(w,h)$ be the expected number of draws needed to get a half caplet when the bottle contains $w$ whole and $h$ half caplets.  We want to compute $e(w,0)$.  By conditioning on the first draw, we obtain
$$
e(w,h)=
\begin{cases}
1 &\text{if $w=0$}\\
1 + \frac{w}{w+h}e(w-1,h+1) + \frac{h}{w+h}\cdot 0 &\text{otherwise}
\end{cases}
$$
Hence 
\begin{align}
e(w,0)
&=1+\frac{w}{w}e(w-1,1)\\
&=1+\left(1+\frac{w-1}{w}e(w-2,2)\right)\\
&=1+1+\frac{w-1}{w}e(w-2,2)\\
&=1+1+\frac{w-1}{w}\left(1+\frac{w-2}{w}e(w-3,3)\right)\\
&=1+1+\frac{w-1}{w}+\frac{w-1}{w}\cdot\frac{w-2}{w}e(w-3,3)\\
&\dots\\
&=1+1+\frac{w-1}{w}+\frac{w-1}{w}\cdot\frac{w-2}{w}+\dots+\prod_{j=0}^{w-1}\frac{w-j}{w}e(0,w)\\
&=1+\sum_{k=0}^{w-1}\prod_{j=0}^k\frac{w-j}{w}\\
&=1+\sum_{k=0}^{w-1}\frac{1}{w^{k+1}}\prod_{j=0}^k (w-j)\\
&=1+\sum_{k=0}^{w-1}\frac{1}{w^{k+1}}\cdot\frac{w!}{(w-k-1)!}\\
&=1+\sum_{j=0}^{w-1}\frac{w!}{w^{w-j}j!}\\
&=1+\frac{w!}{w^w}\sum_{j=0}^{w-1}\frac{w^j}{j!}\\
&=\frac{w!}{w^w}\sum_{j=0}^w\frac{w^j}{j!}
\end{align}
