conditional expectation of discrete random variable given sum constraint Given $k$ discrete iid random variables $X_1,\ldots,X_k$, the conditional expectation
$E[X_1\,|\,X_1+\cdots+X_k=n]$
is intuitively given by 
$\frac{n}{k}$
since the variables are iid and, thus, given that their sum is $n$, each one is expected to contribute $\frac{n}{k}$. This leads to a nice (and well-known) recurrence formula for sums of random variables, namely, 
$P[X_1+\cdots+X_k=n] = \frac{k}{n}\sum_{j} jP[X_1=j]P[X_2+\cdots+X_k=n-j].$
It would seem that one could extend this argumentation by claiming that, intuitively,
$E[X_1^2\,|\, X_1+\cdots+X_k=n]=(\frac{n}{k})^2$
but this appears to be wrong. Why so? What can be said about $E[X_1^2\,|\, X_1+\cdots+X_k=n]?$
 A: Ok, I answer this question myself, given the help provided by Anthony Quas. 
In order to prevent my question from being classified as "on hold", I'll also give a nice application of the derivations, showing that they imply the (non-obvious) relationship
(AA) $\binom{k}{n}=\frac{k}{n^2}\left(\binom{k-1}{n-1}+(k-1)\binom{k-2}{n-2}\right)$ 
holds for binomial coefficients. 
As Anthony has outlined
$E[X_1^2|S_k=n]=\frac{1}{k}E[X_1^2+\cdots+X_k^2|S_k=n] = \frac{1}{k} E[S_k^2-2\sum_{i<j}X_iX_j|S_k=n] = \frac{1}{k}\left(n^2-k(k-1)E[X_1X_2|S_k=n]\right)$.
Now, by definition,
$E[X_1X_2|S_k=n] = \sum_{x,y}xyP[X_1=x,X_2=y|S_k=n] = \sum_{x,y}f(x)f(y)\frac{P[S_{k-2}=n-x-y]}{P[S_k=n]}$.
So, a closed-form formula is established for $E[X_1^2|S_k=n]$. Computing alternatively, we have $E[X_1^2|S_k=n] = \sum_{x}x^2\frac{f(x)P[S_{k-1}=n-x]}{P[S_k=n]}$, where $f(x)=P[X_1=x]$. 
Now, equate both formulas for $E[X_1^2|S_k=n]$, rearrange, let $X_i$ be Bernoulli distributed with $p=1/2$ and obtain the curious relationship (AA)
