Expected value (and distribution) of sum of six balls labeled 1-49, no replacement. The problem stems from the Spanish lottery, in which 6 balls are drawn from an urn with 49 balls, labeled 1-49, without replacement. My goal is to figure out the expected value of their sum, and if it is possible, the distribution. I tried to work out on partitions of natural numbers, but unsuccessful.
It is obvious that the sum ranges between 21 (1+2+3+4+5+6) and 279 (49+48+47+46+45+44) but I can't find out the expected value or the probability of obtain a certain number, say a 60.
Could anybody give some clue or idea?
Thanks in advance.
 A: If we are to be completely formal in solving this problem, here's how we would do it:
Let $X_1, \cdots, X_6$ denote the draws, and let $S = \sum_{i=1}^{6}{X_6}$. So, we have
\begin{align*}
P(S = s) = \frac{1}{\binom{49}{6}}\cdot (\text{Number of distinct partition of s into 6}) = \frac{q(s, 6)}{\binom{49}{6}}
\end{align*}
where $q(n, k)$ denotes the number of ways of partitioning $n$ into $k$ parts. There is no closed-form for this, and the best we can do is its generating function.
Of course, expected value is easier to find using indicators, as André Nicolas suggested. We have
\begin{align*}
\mathbb{E}S = \sum_{i=1}^{6}{\mathbb{E}{X_i}} = \sum_{i=1}^{6}{25} = 150
\end{align*}
and while it may seem weird to use this method because a realization of, say, $X_1$ obviously affects the expected value of further draws, $X_1$ itself is equiprobable itself about 25, and the symmetries of the problem will indeed turn out an expected value of 25; the linearity of the expectation operator very subtly captures this information.
A: Many thanks for all your answers. Of course the expected value is 150. I don't know way but I thought that the linearity was not met here... when is always met with the expectancy. But as you pointed out Tom CHen, is  a kind of weird...
However, I think you make a mistake in the distribution of $S$. Notice that you write:
\begin{align*}
\text{Number of distinct partition of $s$ into 6}
\end{align*}
But what should be written, according to the problem is:
\begin{align*}
\text{Number of distinct partition of $s$ into 6 numbers ranging in 1-49.}
\end{align*}
For instance, in the first case, it is included 165=150+1+2+3+4+5 while in the second it is not.
Let explain in further details the problem, to motivate it. (I am a mathematician and I am trying to help and statistician to solve the following problem).
In this lottery, one has to guess the numbers that will appear in the balls. I have data of the combination appeared in the balls and of the number of guessers. What we want to see is if people tend to bet for small numbers rather than for big numbers. We have seen this by simply computing the correlation between the sum $S$ of the balls and the number of winners, which turns out to be negative.
But now, what I proposed is to divide the data set in two cases: 
\begin{align*}
\text{$S<\mathbb{E[S]}=150$ and $S>\mathbb{E[S]}=150$ }
\end{align*}
and compare the number of winners in each group. We already did, before know the theoretical expectancy, since the mean of the sum in our data set was 150.5.
Moreover, the histogram of $S$ shows symmetry. I need this symmetry because when I divide my dataset in these two parts, only make sense to directly compare the number of winners in each part if the probability mass in each of them is 0.5. As I said, the histogram shows symmetry, but I need to proof it directly (so I asked for the distribution, but it seems very difficult to deduce. As ADG said, I also thought to write a code, but I realized that the number of combinations was in the order of $10^{10}$)
I will think about this symmetry, but any idea about how to prove the symmetry or to approach the problem are welcomed.
