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I have 200 tickets, and I'll sort 20 awards, 1st,2nd,3rd...20th.

I ask you to help me creating a probably calculator who should say:

Probability of winning 1st award
Probability of winning one of the first tree awards (1st,2nd or 3rd)
Probability of winning one of the awards.
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Hm .. $\frac 1{200}$, $\frac 3{200}$, $\frac 1{10}$? – martini Sep 8 '12 at 10:48
That's not as simple as this.. Following your reasoning, if I bought 100 tickets, I would have this probabilities of winning: 100/200, 300/200, 100/10 – William Santos Sep 8 '12 at 12:02
And I need a formula.. Please help me. – William Santos Sep 8 '12 at 12:04
You didn't write in your question about bying more than one ticket. I'll write an answer. – martini Sep 8 '12 at 12:33

Assume you buy $k$ tickets. There are $\binom{200}k$ possibilities of choosing $k$ tickets out of the 200. We now just have to find out the winning combinations for each of the three cases.

If we want to calculate in how many cases we win one of the first $m$ awards ($m \in \{1,3,20\}$ in your cases), we argue as follows: It is easier to calculate the number of combinations where we miss these awards: There are $200-m$ corresponding tickets, and $\binom{200-m}k$ possibilities. So with probiability \[ p = 1 - \frac{\binom{200-m}k}{\binom{200}k} = 1 - \frac{(200-m)!k!(200-k)!}{200!k!(200-m-k)!} = 1 - \frac{(200-m)\cdots (200-m-k+1)}{200\cdots(200-k+1)} \] we'll win one of the first $m$ awards.

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Can you help me converting this to PHP? – William Santos Sep 9 '12 at 11:57
It was very helpfull, thanks. – William Santos Sep 9 '12 at 15:50

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