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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.

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    $\begingroup$ Is the percentage rounded up or down? As in: What is 50% of 11? 10% of 11? 10% of 19? $\endgroup$ – Hagen von Eitzen Apr 7 '15 at 21:02
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    $\begingroup$ Hi Shawn, I think the reason why your question was closed was because it was missing details necessary to solving the question. Please try to be as complete as possible in asking. It's also common here to include a sincere attempt at the problem (this is a convention in order to reduce the number of people asking for others to do their homework). $\endgroup$ – Newb Apr 8 '15 at 5:44
  • $\begingroup$ I think the best way to reword this question would be to rephrase it in terms of probabilities rather than exact percentages. I.e. rather than saying 50% of the boomerangs come back, say that a boomerang has a 50% chance of coming back. Judging by your edit, I think this is what you meant (and I assumed this to be the case before the edit). That way, the rounding at each step doesn't matter. What you're asking for can be expressed in terms of the sum of the first n terms of a geometric series. $\endgroup$ – Randy E Apr 9 '15 at 3:06
  • $\begingroup$ Sorry, I should have said "What you're asking for can be expressed in terms of the sum a geometric series" in the last sentence. $\endgroup$ – Randy E Apr 9 '15 at 13:22
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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?

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  • $\begingroup$ Because I don't know how to round off a boomerang. If I assume I always succeed with the last throw, then I'll be throwing that last boomerang forever and the total will be infinity. I'm not sure how to handle this issue. $\endgroup$ – Shawn V. Wilson Apr 8 '15 at 4:43
  • $\begingroup$ So if P is the chance (as a fraction) of the boomerang coming back, would the formula be: E(n) = (1+PE(n))/(1-P)? $\endgroup$ – Shawn V. Wilson Apr 10 '15 at 20:37
  • $\begingroup$ I don't understand your final step -- the "induction". Can you explain that briefly? $\endgroup$ – Shawn V. Wilson Apr 10 '15 at 20:43
  • $\begingroup$ If $E(n) = 2 + E(n-1)$ and $E(0) = 0$, then $E(n) = 2n$. This should be intuitively clear, just try it out: $E(1) = 2 + E(0) = 2$, $E(2) = 2 + E(1) = 4$ etc. If you want to formally prove this, it is an easy proof by induction. $\endgroup$ – Noiralef Apr 10 '15 at 21:38
  • $\begingroup$ For the other question (probability P instead of $\frac 1 2$): No, how did you get this? It would be $E(n) = 1 + P E(n) + (1-P) E(n-1)$. $\endgroup$ – Noiralef Apr 10 '15 at 21:39
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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).

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