# Balls and bins problem - expected number of balls needed to throw [closed]

Suppose we have n boxes and we start randomly and independently throwing balls into the boxes.

(a) For a given box, what is the expected number of balls we need to throw before one of the balls lands in that box?

(b) What is the expected number of balls we need to throw before every bin contains at least one ball?

## closed as off-topic by TMM, Michael Hoppe, Magdiragdag, user99914, Andrew D. HwangMar 11 '14 at 19:25

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• $\ln(n)$ and $n\ln(n)$ Check this book – Guy Mar 11 '14 at 11:22
• @Sabyasachi: The answer to the first is definitely $n$, not $\ln n$. The second is indeed $n \ln n$, by the coupon-collector problem. – ShreevatsaR Mar 11 '14 at 15:52
• @ShreevatsaR $n$ is you consider worst case. $\ln(n)$ in average case. – Guy Mar 11 '14 at 15:53
• Although I guess we're supposed to consider worst case anyway. – Guy Mar 11 '14 at 15:54
• @Sabyasachi: No, the question is precise (there is no such thing as average case or worst case): it wants the expected number of balls thrown until one lands in that box, which is $n$. See answer by Stefanos below. – ShreevatsaR Mar 11 '14 at 15:54

For (a). This is the geometric distribution with parameter $1/n$ since we count trials until the first success (with success we denote the event that a ball lands in the given box) and the probability of success in each trial is equal to $1/n$, assuming that each ball has equal probability to land in each box. So the expected number of trials is the expectation of the Geometric random variable $X$ with parameter $p=1/n$ which is equal to $$E[X]=\frac{1}{p}=\frac{1}{\frac{1}{n}}=n$$ (Note however that this is the expected number of trials until one ball lands in the given box. If you mean "before one lands" then you should take the geometric starting from $0$ which has expectation $(1-p)/p$ which is equal to $n-1$ for $p=1/n$. Same way of thinking, slight difference, it is only that I do not understand properly which of the two is required.)