# How can probability and expected value be same in the limit?

I have seen similar arguments at other places but couldn't convince myself so far about it. I am reading some literature related to graph theory but let me post the analogous problem which avoids the graph theory terminology:

There are $M$ boxes and $m$ balls which we want to randomly uniformly distribute among those boxes.

Now author essentially says that the expected number of balls in each box $i$ is $p_{i} = \frac{m}{M}$. However, then he says that if $m << M$, this can also be considered as the probability of filling that box with one of the $m$ balls.

I don't understand this. I tried deriving expression for the expected value but that itself involves the probability that a ball goes into the box. Can somebody please clarify this? Thanks

The expected value of a random variable $X$ whose values are natural numbers is $$0\cdot P(X=0) + 1\cdot P(X=1) + 2\cdot P(X=2) + 3\cdot P(X=3) + \cdots$$
In your case, if $m\ll M$, then the probability that there's more than one ball in the box will be vanishingly small, so the terms with $P(X=2)$ and so forth can be ignored. So what is left is $$0\cdot P(X=0) + 1\cdot P(X=1)$$ which is of course just $P(X=1)$.
• @SnehalShekatkar: The expected value of the sum of the number of balls in any box is of course $m$ (because that is always the value). Because expectation is additive, this is also the sum of the expectations of the number of balls in each of the $M$ boxes, and since those must be equal (by symmetry), the expectations for each box can only be $m/M$, or they wouldn't sum to $m$. – Henning Makholm Jun 15 '16 at 9:45
• The probability of having exactly $1$ ball in box $i$ must be $m\frac1M(1-\frac1M)^{m-1}$, so the $m/M$ estimate is good as long as $(1-\frac1M)^{m-1}$ is close to $1$. For large $m$ we have $(1-\frac1M)^{m-1} \approx e^{-m/M}$, so when $m/M$ is small it is also a good approximation. – Henning Makholm Jun 15 '16 at 9:50