Calculation of probability with arithmetic mean of random variables

Question

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?

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Some explanation is welcome.

Answer 1

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

Answer 2

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.