Odds are: 0.3678792572210609 that no one wins, i.e. 0.632120742778939 that someone wins. The number e is not related to this problem; it is the limit of (1+1/n)^n. Here we have odds of (1-1/n)^n that no one wins.

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Odds are: 0.3678792572210609 that no one wins, i.e. 0.632120742778939 that someone wins. The number e is not related to this problem; it is the limit of (1+1/n)^n. Here we have odds of (1-1/n)^n that no one wins.

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poisson as limiting cxase if binomial distribut5ion. Large number of trials with small prob. Thus Lambda = 1. find prob n = 0 in poisson with lambda = 1 = 1-1/e.

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You're all wrong.

1) Answer one makes no sense, where are you getting 1/e= 0.3? (I assume this is related to answer 3)

2) Lotteries are not binomially distributed, since lottery numbers are typically unique to each ticket. Therefore you are sampling without replacement.

3) Poisson is indeed the limiting case of the poisson, and using this approximation will give you the same answer as two, which since I was bored I calculated as p=0.3678794 that no one wins, but is wrong for the same reason in my view.

If we are sampling without replacement, the problem becomes much easier, and you won't need to bring a slide-rule or pocket calculator to your interview. If the odds for each player are one in one million, then one million tickets must have been issued. If there are one million players, then all tickets have been issued, and thus with certainty someone has won.

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1/e = 0.3

1/1000000 is very small, so actual odd approaches the limiting case of 1/e.