Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \begin{align*} \Pr[\text{bin } i \text{ has at least } k \text{ balls}] &\leqslant \left( \frac{e}{k} \right)^k = \left( \frac{e \ln \ln n}{3 \ln n} \right)^{\frac{3 \ln n}{\ln \ln n}} \\ &\leqslant \exp \left( \frac{3 \ln n}{\ln \ln n} (\ln \ln \ln n - \ln \ln n) \right) \\ &= \exp \left( - 3 \ln n + \frac{3 \ln n \cdot \ln \ln \ln n}{\ln \ln n} \right) \end{align*} $$

When $n$ is large enough, $$ \Pr[\text{bin } i \text{ has at least } k \text{ balls}] \leqslant \exp ( - 2 \ln n ) = \frac{1}{n^2}. $$

This was found in this set of lecture notes.

Can anyone explain why the last step ("When n is large enough..") is true?

share|cite|improve this question
up vote 2 down vote accepted

All you need is that for large enough $n$, we have $3\ln (\ln(\ln(n))) < \ln(\ln(n))$ which is true since for large enough $m$, $3 \ln(m) < m$ and take $m = \ln(\ln(n))$. For instance, take $n \geq e^{e^5}$.

Since $3\ln (\ln(\ln(n))) < \ln(\ln(n))$, we have $$-3 \ln(n) + \frac{3 \ln (n) \ln (\ln(\ln(n)))}{\ln(\ln(n))} < -3 \ln(n) + \ln(n) = - 2 \ln(n).$$

Hope it is clear now.

share|cite|improve this answer
What about that ln(n) multiplying 3ln(ln(ln(n)))/ln(ln(n))? – Leo Dec 9 '11 at 23:12
@Leo: That gets added to $-3\ln(n)$ to give $-2\ln(n)$. – user17762 Dec 9 '11 at 23:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.