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I'm looking for different ways to calculate payouts after I deposit some units.

Idea: There is a raffle for total of 1 000 000 000(TP). Everyone can deposit their Tickets(OT) (everyone has different amount of owned tickets) and at the end, I'll get the percentage of total tickets(TT) to deposited tickets(DT)

Example, someone deposited 1000 tickets, and I'm going to deposit 234:

$$currentPricePerTicket = \frac{TP}{TT} = 1000000$$ $$TTafterMyDeposit = TT + DT = 1000 + 234 = 1234$$ $$pricePerTicketAfterDeposit = \frac{TP}{TTafterMyDeposit} = \frac{1000000000}{1234} = 810372.77$$ $$myPercentOfDeposited = DT \div \frac{TTafterMyDeposit}{100} = 234 \div \frac{1234}{100} = 18.96\%$$ $$iWillGet = TP \cdot myPercentOfDeposited = 1000000000 \cdot 18.96\% = 189 600 000$$

This is what I'll get if nobody after me deposited their tickets.

This is pretty straight way to calculate. My main question, is there more "short" version to calculate the answer (189 600 000)?

Also, my second question: how do I invert this this? For example, if I want to get 249 365 000, how many tickets i should deposit?

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up vote 2 down vote accepted

So that I will not have to type so many zeros, let the prize be $P$. If the current total is $t$, and I deposit $d$, then my fractional share of the prize is $\dfrac{d}{t+d}$. So the amount I get is $$\frac{Pd}{t+d}.$$ That is precisely what you did. There does not seem to be anything simpler.

The question now is, if we want our share to be the amount $S$, what should $d$ be? So we have the equation $$S=\frac{Pd}{t+d},$$ and want to solve for $d$. Multiply both sides of the above equation by $t+d$. We get $$S(t+d)=Pd.$$ So $St+Sd=Pd$, and therefore $St=(P-S)d$. It follows that $$d=\frac{St}{P-S}.$$

In your question, we have $P=1000000000$, $t=1000$, and $S=249365000$. Then the above equation gives $d\approx 332.2$.

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I think this is a simplification, as it avoids going through the percentages. –  Ross Millikan Oct 7 '12 at 16:34
    
Thank you very much, that's awesome :) –  Bubonic Pestilence Oct 7 '12 at 16:43
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