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On a certain day, $N$ lottery ticket are sold and $M$ win. To have a probability of at least $\alpha$ of winning on that day, approximately how many ticket should he purchased?
The answer is the least integer greater than $- \frac{M}{N} \log(1-\alpha)$ but I don't know why

Gamma Function
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user134927
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  • If I need $ \alpha = 1$, the answer should be N-M+1 but the model answer is infinity – user134927 Jun 02 '14 at 01:29
  • Also, as $M$ gets bigger - as more tickets win - this formula suggests that you need to buy more tickets to get the same chance of winning? – coolpapa Jun 02 '14 at 01:45

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Hm....how are you approximating this answer? If we assume that each ticket has an equal chance of winning (which is maybe not accurate - if $N$ and $M$ are fixed, we should be using a hypergeometric RV), then I get the following:

Each ticket that you buy has a probability of losing of $\frac{N-M}{N}$. So if you buy $x$ tickets, the probability of them all losing is $(\frac{N-M}{N})^{x}$. So your probability of winning is $1 - (\frac{N-M}{N})^{x}$. Then we want: $$1 - (\frac{N-M}{N})^{x} \geq \alpha $$ $$ 1-\alpha \geq (\frac{N-M}{N})^{x} $$ $$ \frac{\ln(1-\alpha)}{\ln(\frac{N-M}{N})} \leq x.$$ (Notice that the inequality flips in that last line since $\frac{N-M}{N} < 1$, so its natural log is negative.)

Now, using the power series $\ln(1+x) = x - \frac{x^{2}}{2} + \ldots $, we can say that $\ln(\frac{N-M}{N}) = \ln(1 + \frac{-M}{N})$, which is thus about $-M/N$. But then I get $x \geq \frac{-N}{M}\ ln(1-\alpha)$.

coolpapa
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