Q: If some of your boomerangs don't come back, how many throws will you get? Let's say you're practicing throwing boomerangs. You're not an expert, and only 50% of the time does a boomerang return to you. 
So you stand out in a field with 16 boomerangs and start throwing them. After the first pass, 8 have come back, and you can throw them again. Out of those, 4 will come back, etc.  When you throw the last one, you will have made 31 throws altogether.
If N is your starting number of boomerangs and P is the percent that come back, what is the formula for T, the total number of throws you can make?
EDIT: Apparently I needed to add "more details", but since this is a problem that I kind of made up, I'm not sure what I could have left out.
This situation was inspired by an online game I'm playing. My archer buys arrows by the stack. Normally each arrow in the stack can only be fired once, but there are also arrows which have a 75% chance of returning. I thought "A stack of 100 of these arrows must be the equivalent of some-big-number of regular arrows."  
I can get that equivalent number by using a spreadsheet. Rather than roll a die for each arrow, I assume that 75% of the stack would return the first time, then 75% of the remainder the next time, and so on.
Although the spreadsheet works, I believed that there was some formula that would apply to any number of arrows with any specific return rate. Probably involving integrals, since (as was pointed out here) this was the sum of a bunch of related steps.  
In fact, I suspected that my hypothetical formula was already well-known, like all those in the textbooks involving coin-flipping and dice-throwing. But other than googling "Boomerangs" I couldn't be sure where to start.  
It seemed to make more sense to ask people who post about math for fun. 

A: So, if I understand it correctly, every time I throw the boomerang there is a 50% chance that I get it back?
In that case, let $E(n)$ be the expectation value for the number of throws you can make when you have $n$ boomerangs left.
After throwing one, there is a chance of $\frac 1 2$ to be left with the same amount, and a chance of $\frac 1 2$ to have one boomerang less:
$$ E(n) = 1 + \frac 1 2 E(n) + \frac 1 2 E(n-1) $$
We solve for $E(n) = 2 + E(n-1)$, this leads to $E(n) = 2n$ (by induction).
I'm a bit confused about your "31", though.
Why do you assume that you always miss the last throw?

A: You get your first $N$ throws no matter what. Of those you will get $NP$ back. So throw those. Of those you get $NP^2$ back. And so on... 
So $T = N + NP + NP^2 + ... $
However, there is also the issue of rounding that Hagen brought up above. I hope this helps as a bit of a start to the question (which is hopefully less ambiguous).  
