Maybe related to [this question](https://math.stackexchange.com/questions/36308/expected-value-of-max-min-of-random-variables)

In the comments of this question they say that it gets easier if the variables are identically and independently distributed.
But i don't see how because in my case the variable is discrete

**Here is my problem :**
I toss 4 dice and keep the 3 best results. What is the expected value of the result ?

I think tossing 4 dice and keep the 3 best is like tossing 4 dice and removing the minimum.

* Let X be the result of a standard die.
* Let Y be tossing 4 dice and keeping the 3 best

Is that correct : $E(Y) = 4\*E(X) - E(min)$ ?

So how calculate E(min) ?
I know if the variable was uniform on [0,1] I could have started with $F\_Y = 1 - ( 1-F\_X )^p$ where p is the number of dice I toss, but here the variable is discrete so i don't know where to start.

**Generalization :**
How to calculate the expected value of k realizations of a discrete random variable in [0-n]?

It's been a while since i studied probability, so my basic calculation may be wrong. Also, 
English is not my mother tongue, so please forgive my mistakes.

edit : spelling mistakes
['https://math.stackexchange.com/questions/355820', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/71720/']
For clarity, suppose that the dice have ID numbers $1,2,3,4$. Let $X\_i$ be the result on die $i$. Let $Y$ be the sum of the three largest of the $X\_i$, and let $W$ be the minimum of the $X\_i$. 

Then $Y=X\_1+X\_2+X\_3+X\_4-W$. By the **linearity of expectation**, it follows that
$$E(Y)=E(X\_1)+E(X\_2)+E(X\_3)+E(X\_4)-E(W).$$
The linearity of expectation is a very useful result. Note that linearity *always* holds: independence is not required.

The expectation of the minimum can be calculated by first finding the distribution of the minimum $W$. 

The minimum is $1$ unless the dice all show a number $\ge 2$. The probability of this is $1-\left(\frac{5}{6}\right)^4$. We rewrite this as $\frac{6^4-5^4}{6^4}$.

The minimum is $2$ if all the dice are $\ge 2$ but not all are $\ge 3$. The probability of this is $\frac{5^4-4^4}{6^4}$/

The minimum is $3$ if all results are $\ge 3$ but not all are $\ge 4$. This has probability $\frac{4^4-3^4}{6^4}$.

And so on. Now use the ordinary formula for expectation. We get that the expectation of $W$ is
$$\frac{1}{6^4}\left(1(6^4-5^4)+ 2(5^4-4^4)+3(4^4-3^4)+4(3^4-2^4)+5(2^4-1^4)+6(1^4-0^4) \right).$$
We leave you the task of computing. *Before* computing, simplify!

**Generalization:** Suppose we toss $k$ "fair" $(n+1)$-sided dice, with the numbers $0$ to $n$ written on them. For $i=1$ to $k$, let $X\_i$ be the number showing on the $i$-th die. Let $S$ be the sum of the dice. Then $S=X\_1+\cdots+X\_k$. The expectation of $X\_i$ is $\frac{0+1+\cdots +n}{n+1}$. By the usual expression for the sum of consecutive integers, $E(X\_i)=\frac{n}{2}$ and therefore $E(S)=\frac{kn}{2}$.

The analysis of the minimum $W$ goes along the same lines as the earlier one. The probability that the minimum is $j$ is $\frac{(n+1-j)^k -(n-j)^k}{(n+1)^k}$. If we use the ordinary formula for expectation, and simplify, we find that 
$$E(W)=\frac{1^k+2^k+\cdots+n^k}{(n+1)^k}.$$ 

**A nice way to find $E(W)$:** The following is a useful general result. Let $X$ be a random variable that only takes non-negative integer values. Then
$$E(X)=\sum\_{i=1}^\infty \Pr(X\ge i).$$
We apply that to the case of the random variable $W$ which is the minimum of $X\_1,\dots,X\_4$. The probability that $W\ge i$ in that case is $\frac{(7-i)^k}{6^k}$.

The same procedure works for the more general situation you asked about.
Yes, your equation is correct, and this is a good and efficient way to perform this calculation. To find the expected value of the minimum, consider the hypercubes $[k,6]^4$ for $k=1,\dotsc,6$. The number of these hypercubes that a result is in is given by the minimum value of the dice. Thus we get the expected value of the minimum by adding up the probabilities of hitting these hypercubes:

$$\mathbb E[\min]=\frac{\displaystyle\sum\_{k=1}^6k^4}{6^4}=\left.\frac1{30}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{6^4}\right|\_{n=6}=\frac{2275}{1296}\approx1.7554\;.$$