Maximum load of a bin in the $n$ balls with weights into $m$ bins problem $n$ balls, each with a weight $p_i$, are thrown into $m$ bins. Each bin is chosen with uniform probability.
Prove or disprove that the expected value of the maximum load among the loads of bins is $\frac1m\sum_{j=1}^n p_j$, where with "load" means the sum of the weights of the balls in that bin.
Now, I was able to model the problem on the expected value of each bin and this is:
$E[X_i]=\frac1m\sum_{j=1}^n p_j$, where $X_i$ is the load of the bin $i$.
Should I calculate something like this:
$$E[\max_{1 \leq i \leq n} {X_i}]$$
Do you have any idea? Or is the equation to disprove? But, I have no idea how to have to find a counterexample to disprove with expected values.
 A: I want to disprove that $s = \frac1m\sum_{j=1}^n p_j$  is the expected value of the maximum for $m>1$.
(When $m=1$, this is trivially true - all the balls are in the same unique bin.)
But, when $m>1$, this expression is the expected value of the load in any of the bins (use linearity of expectation to see this - each ball has probability $\frac{1}{m}$ to be thrown in a specific bin). So if we let $X_i = \text{load of bin }i$, we have 
$$\forall i: EX_i = s$$
$$\sum_{i} X_i = \sum_{j} p_j = ms$$
Since $\sum_{i} X_i = ms$, unless $X_i=s$ for each $i$, we have $\max_{i} X_i > s$. So 
$$E \max_{i} X_i = Pr(X_1 = X_2 = \cdots = X_n=s) s + $$
$$Pr(\text{Not all }X_i \text{ are equal})E(\max_{i} X_i | \text{Not all }X_i\text{ are  equal})$$
When $Pr(\text{Not all }X_i \text{ are equal}) > 0$ (true whenever $m>1,n>0$ - for example, you can have the first bin empty and the last full, and vice versa), we have:
$$E \max_{i} X_i > ( Pr(X_1 = X_2 = \cdots = X_n=s)  + Pr(\text{Not all }X_i \text{ are equal})) s = s$$
since $E(\max_{i} X_i | \text{Not all }X_i\text{ are  equal}) > s$.
