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Each of 1000 balls is thrown and lands randomly into one of 1000 pots. What is the probability that at least one pot will have 34 or more balls?

David G. Stork
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user1526836
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  • What have you tried? – David G. Stork Feb 20 '22 at 22:05
  • See this generic duplicate, ignore the first answer as it is wrong. The second answer is conceptually useful but, as it mentions, the computations are so unwieldy that people tend to either simulate the thing or fall back on asymptotic analysis. – lulu Feb 20 '22 at 22:06
  • @lulu - given there are three answers there (currently two with the same numbers of votes) it might be worth identifying which one you are pointing at – Henry Feb 20 '22 at 22:16
  • @Henry Thanks for pointing that out. I meant the answer from Ian. The incorrect one is the one from Mostafa Ayaz. And the links provided in the comments might be helpful as well. – lulu Feb 20 '22 at 22:17
  • The answer is going to be very small in your case: you can work out the probability the first pot has $34$ or more, and thus the expected number of pots with $34$ or more. That will be an upper bound on the probability any are (and in this case, very close to the probability) – Henry Feb 20 '22 at 22:26
  • @Henry How would you go about calculating the probability for the first pot only, and what makes you sure the result will be an upper bound for the total probability? – user1526836 Feb 20 '22 at 22:42
  • @user1526836 For the first pot, this is a binomial distribution (each ball is either in the first pot or not). The expectation single counts when there is one pot 34+, double counts when there are two pots 34+, triple counts when there are three etc. when you want to single count in each case for the probability of any 34+. So it is an upper bound. But here having a single pot 34+ is very unlikely and having more than one 34+ is extremely unlikely so it will be extremely close to the correct answer – Henry Feb 21 '22 at 00:15

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