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Suppose I want to hold lottery competition. I just want two people to win the lottery and there are 100 people buying my lottery. What should the maximum probability (of each player wining the lottery) be to ensure that exactly two people win the lottery?

I obviously think it's $2/100=1/50$. But I think it should not be so trivial. I would appreciate any comments.

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closed as unclear what you're asking by Saad, José Carlos Santos, Leucippus, Shailesh, JonMark Perry Mar 15 '18 at 5:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't see how this question makes sense, unless you have just 50 numbers to draw and assign each of the numbers to 2 tickets. If the people buying tickets are free to choose their numbers, how are you going to ensure anything? $\endgroup$ – Ron Gordon May 14 '13 at 14:03
  • $\begingroup$ @RonGordon This is very badly worded indeed! Imagine such a lottery competition where lottery is bought randomly, independently and suppose we just know deal with probability that any person wins the lottery, and not even care or know the system of anouncing winner. Better still wording would have been "there is a quiz competition where there are 100 participants and the probability that they give correct answer is $x$, what optimum $x$ gives exactly two winners?' $\endgroup$ – 007resu May 14 '13 at 14:26
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For the new question, you have a binomial distribution. If the probability that each player wins is $x$, the probability of exactly $2$ winning is ${100 \choose 2}x^2(1-x)^{98}$. The first term is the number of ways to choose the winners, the $x^2$ is the probability they both win, and the $(1-x)^{98}$ is the probability that all the others lose. Now take the derivative and set to zero. Yes, you should get $1/50$

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  • $\begingroup$ please see comment above. Sorry for bad wording. Is the answer still 1/50? $\endgroup$ – 007resu May 14 '13 at 14:28
  • $\begingroup$ You can't guarantee two winners, so all you can do is choose $x$ so the chance of two winners is maximized. Yes, the answer should be $1/50$ $\endgroup$ – Ross Millikan Aug 15 '13 at 3:38

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