Calculation of probability with arithmetic mean of random variables There are 4 people, each of whom has one deck of cards with 500 cards that are numbered from 1 to 500 with no duplicates.
Each person draws a card from his deck and I would like to calculate the probability of the event that "the arithmetic mean of the number on the 4 cards is 405".
How to make that?

Some explanation is welcome.
 A: You have a correct answer already, but I would like to add that there is a standard combinatorics method for getting this answer.
You have four numbers $a, b, c, d$ which are the numbers drawn by
(respectively) the first, second, third, and fourth player.
Each of those four numbers is a positive integer and in order for the arithmetic mean to be $405,$ we are looking for an event where the sum is
$$ a + b + c + d = 1620. $$
There is a well-known method for finding the number of ways to reach any given sum with four positive numbers, but in this case the usual method would count sums such as $1 + 1 + 1 + 1617,$ which you have ruled out by stating that the highest number on any card is $500.$
There is, however, another almost-as-well-known way of dealing with the
maximum of $500$ per card, and that is to "count from the top".
Let's look at the numbers
$a' = 500 - a,$ $b' = 500 - b,$ $c' = 500 - c,$ and $d' = 500 - d.$
The four drawn cards give us the four numbers $a,b,c,d$ but also give us the "complementary" number $a',b',c',d'.$
Notice that if (and only if) $ a + b + c + d = 1620,$ then 
\begin{align}
 a' + b' + c' + d' &= (500 - a) + (500 - b) + (500 - c) + (500 - c) \\
&= 2000 - (a + b + c + d)\\
&= 2000 - 1620\\
&= 380.
\end{align}
So rather than look for four positive integers that add to $1620$ with the restriction that none can be greater than $500,$
we can look for four non-negative integers (not necessarily positive, because
$a' = 0$ when $a = 500$) whose sum is $380.$
The integers each have to be less than $500,$ but in fact none can be greater than $380$ anyway so the "less than $500$" restriction actually has no effect and can be ignored.
This gives us a standard problem with a standard solution.
The solution (explained in the answers to How to use stars and bars?) is that the number of ways to add four non-negative integers to a sum of $380$ is
$$
\binom{380 + 4 - 1}{4 - 1} = \binom{383}{3} = \frac{383\cdot382\cdot381}{6}
 = 9290431.
$$
So that's the number of ways the four cards can add up to $1620,$
the same number obtained by nested sums in Ben W's answer,
confirming that the sums were correctly computed.
A: This is equivalent to asking if the sum is 1620.  The individual variables are i.i.d. discrete uniform, so there is probably some well-developed theory on this.  However, we can do it elementary style ; )
To get a total of 1620, the first player must have at least a 120.  So we have $\sum_{i=120}^{500}$ to consider.  Now the second player must have at least $620-i$, so we take $\sum_{j=620-i}^{500}$.  The third player must have at least $1120-i-j$, so we take $\sum_{k=1120-i-j}^{500}$.  The fourth player must now draw exactly $500-i-j-k$.  Each draw has a probability of $1/500$.  So we obtain
$$P(X=1620)=\sum_{i=120}^{500}\frac{1}{500}\sum_{j=620-i}^{500}\frac{1}{500}\sum_{k=1120-i-j}^{500}\frac{1}{500}\cdot\frac{1}{500}$$
$$=\frac{1}{(500)^4}\sum_{i=120}^{500}\sum_{j=620-i}^{500}\sum_{k=1120-i-j}^{500}1$$
$$=\frac{9,290,431}{(500)^4}\approx 0.000148646896$$
