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.
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.
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 $$P(X=1620)=\sum_{i=120}^{500}\frac{1}{500}\sum_{j=620-i}^{500}\frac{1}{500}\sum_{k=1120-i-j}^{500}\frac{1}{500}\cdot\frac{1}{500}$$ $$=\frac{1}{(500)^4}\sum_{i=120}^{500}\sum_{j=620-i}^{500}\sum_{k=1120-i-j}^{500}1$$ $$=\frac{9,290,431}{(500)^4}\approx 0.000148646896$$
