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A newspaper has a weekly crossword puzzle, for which a winner gets a prize. The paper receives ~4000 solutions a week of which ~25% are correct solutions. The newspaper reads the solutions until they find the first correct one, which then receives a prize. What is the probability that a reader has to send more than 2000 correct solutions before he gets his second prize?

This reminded me of the negative binomial probability distribution, so I found what I think is a correct exact answer:

$$P(Y > 2000) = 1 - P(Y \leq 2000) = \displaystyle 1- \sum_{y=1}^{2000} {y-1 \choose 1} (\dfrac{1}{1000}) (\dfrac{999}{1000})^{y-2})$$

However, I want to numerically approximate using a probability distribution. I don't know how though. Normal doesn't work I think because $np$ is too small, so Poisson remains but I have no idea how to tackle the problem with Poisson.

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  • $\begingroup$ I can't parse "a reader has two send more than $2000$ correct solutions before he gets his second prize". $\endgroup$ – Brian Tung Oct 3 '16 at 20:59
  • $\begingroup$ @BrianTung I forgot some details. $\endgroup$ – NAbag Oct 3 '16 at 21:00
  • $\begingroup$ Ahh! I just had to read it out loud. $\endgroup$ – Brian Tung Oct 3 '16 at 21:01
  • $\begingroup$ Not sure I get your formula. You are asking "what is the probability that someone fails to get two prizes in $2000$ trials", yes? Well...the only ways to fail are to get exactly $0$ or $1$ prizes. $P(0)=\left(1-\frac 1{1000}\right)^{2000}=0.135199925$ and $P(1)=2000\times \frac 1{1000}\times \left(1-\frac 1{1000}\right)^{1999}=0.270670521$ and your answer is just $P(0)+P(1)=0.405870447$, no? $\endgroup$ – lulu Oct 3 '16 at 21:19
  • $\begingroup$ @lulu That's a lot better indeed $\endgroup$ – NAbag Oct 3 '16 at 21:21

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