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Formula (using binomial combination) for X-win raffle probability with no replacement?

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  • Formula (using binomial combination) for X-win raffle probability with no replacement?


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Answer #1 | 02/01 2014 12:51
Most of what you said is true and it sounds like you're getting a good grasp so far. But you said: ... + Chance of winning 4 + ...Chance of Winning 40 Sure but 11 to 40 all have 0 chance since you can't win 11 drawings with only 10 tickets. So just terminate the sum at 10. Also, you said, "the greater the number of wins, the smaller the chances for that incremental win." That's not necessarily true. The chance might increase at first and then decrease (so it has a peak somewhere). For instance if you purchased 100 tickets the peak is at exactly 4, you can verify that Pr(4) > Pr(3) > Pr(5) (Edit -- I changed the example above to a less extreme one and one with an actual peak instead of a constant increase.) Edit 2 -- there's an easy way to know in advance where the "peak" I'm talking about will be. For the case of 100 purchases, just multiply 40 * 100 / 1000 = 4, that's the mean of a hypergeometric distribution (4 is the average or expected number of wins). The maximum exact probability corresponds to the number closest to the mean. Now to answer your question. To win exactly 5 prizes, you need exactly 5 of the drawn tickets to be yours and exactly 35 to be from the 990 that aren't yours. Pr(exactly 5) = (10 C 5)*(990 C 35) / (1000 C 40) And the name of this type of distribution is "hypergeometric". It's the non-replacement version of binomial distribution.

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