Matt's expected value

$\begin{aligned} E &= \frac{\sum\limits_{a=1}^{50} {\sum\limits_{b=1}^{50} {a b}} - \sum\limits_{k=1}^{50} k^2}{50 \cdot 49} \\[1em] &= \frac{\left( \sum\limits_{k=1}^{50} k \right) ^2 - \sum\limits_{k=1}^{50} k^2}{50 \cdot 49} \\[1em] &= \frac{\left( \frac{50 \cdot 51}{2} \right) ^2 - \frac{50 \cdot 51 \cdot 101}{6}}{50 \cdot 49} \\[1em] &= \frac{3^2 \cdot 5^4 \cdot 17^2 - 5^2\cdot 17 \cdot 101}{50 \cdot 49} \\[1em] &= \frac{5^2 \cdot 17 ( 3^2 \cdot 5^2 \cdot 17 - 101)}{50 \cdot 49} \\[1em] &= \frac{2^2 \cdot 5^2 \cdot 7^2 \cdot 17 \cdot 19}{2 \cdot 5^2 \cdot 7^2} \\[1em] &= 2 \cdot 17 \cdot 19 \\[1em] &= \boxed{646} \end{aligned}$

Probability of choosing the first integer and second integer of values $I_1$ and $I_2$ is $\frac {1}{50}$ and $\frac {1}{49}$ respectively.

The expected value of their product is

$E[XY] = \displaystyle \sum_{i=1}^{50} \sum_{j=1}^{50} \space i j \frac {1}{50} \space \frac {1}{49}$ , with $i \ne j$

$\Rightarrow E[XY] = \frac {1}{50} \space \frac {1}{49} \displaystyle \sum_{i=1}^{50} \sum_{j=1}^{50} ij - \frac {1}{50} \space \frac {1}{49} \displaystyle \sum_{i=1}^{50} i^2$

With the first summation allowing $i=j$ to occur, and then we remove that occurence in the second summation

$\Rightarrow E[XY] = \frac {1}{50} \space \frac {1}{49} ( \displaystyle \sum_{i=1}^{50} i )^2 ( \sum_{j=1}^{50} j )^2 - \frac {1}{50} \space \frac {1}{49} \displaystyle \sum_{i=1}^{50} i^2$

Apply the case for $\sum i = \frac {n}{2} (n+1)$ , $\sum i^2 = \frac {n}{6} (n+1)(2n+1)$ , we get $E[XY] = \boxed{646}$

The solution is the total amount that can obtained from getting every possible integer multiplication and dividing that total to the number of possible integer pairs.

Sum of all possible pair multiplication thus is:

$(1 \times 1) + (1 \times 2) + ... + (1 \times 50) +$ $(2 \times 1) + (2 \times 2) + ... + (2 \times 50) +$

$...$

$(50 \times 1) + (50 \times 2) + ... + (50 \times 50)$

Which is equal to

$(1 \times \sum\limits_{x=1}^{50} x) + (2 \times \sum\limits_{x=1}^{50} x) + ... + (50 \times \sum\limits_{x=1}^{50} x)$

which in turn is equal to:

$(\sum\limits_{x=1}^{50} x) \times (\sum\limits_{y=1}^{50} y)$ $= \frac{50 \times 51}{2} \times \frac{50 \times 51}{2}$

This the total amount without regarding the distinctness of the pairs of integers. Therefore we must subtract every value where the pairs are the same:

$\frac{50 \times 51}{2} \times \frac{50 \times 51}{2} - (1\times 1 + 2\times 2 + ... + 50 \times 50)$ $=\frac{50 \times 51}{2} \times \frac{50 \times 51}{2} - (\sum\limits_{x=1}^{50} x^2)$ $= \frac{50 \times 51}{2} \times \frac{50 \times 51}{2} - \frac{50 \times 51 \times 101}{6}$

The total number of distinct integer pairs is given by: $50 \times 49$ Thus the expected value of their product would be:

$\big(\frac{50 \times 51}{2} \times \frac{50 \times 51}{2} - \frac{50 \times 51 \times 101}{6}\big) \div (50 \times 49)$ $=646$

For the sake of generality, expand the range of integers to $n$

The answer is obviously

$\frac { \sum_{1\leq i<j\leq n}ij}{\sum_{k=1}^{n-1} k} = \frac{\sum_{j=2}^n\sum_{i=1}^{j-1} ij} {\sum_{k=1}^{n-1} k } = \frac {3n^{2}+5n+2}{12}$

Plug $n=50$ in and the result is 646

We first compute the sum of the product of the integers in every possible ordered pair.

We denote the ordered pairs as $(a,b)$ . If we have a given $a$ , then the sum of all possible products is $(1)(1+2+...+50)=(a)\frac{(50)(51)}{2}$ . Hence the sum of the product of the integers in every possible ordered pair is $(1+2+...+50)\frac{(50)(51)}{2}$ =(\frac{(50)(51}{2})^2 )=1,625,625.

Now we compute the sum of the product of the integers in the ordered pairs where the 2 number are distinct. From the previous sum, we deduct the sum of the product of the integers of the ordered pairs where the 2 numbers are the same. That sum if $1^2+2^2+...+50^2=\frac{(50)(51)(101)}{6}=42,925$ . Hence, the remaining sum is 1,582,700. But each unordered pairs counts in $2$ ordered pairs, so the sum of the product of the integers in every ordered pair with $2$ distinct numbers is $\frac{1,582,700}{2}=791,350$ , (which is different from the first), so there are a total of $2,450$ ordered pairs, so a total of $\frac{2,450}{2}=1,275$ unordered pairs.

Hence the expected value of their product is $\frac{791,350}{1,275}=\boxed{646}$

While trying to manually calculate the summations, I noticed that each time int a (the multipliers (1 and 2) in 1(2 + 3 + .... + 50) = 1 52 49 (by arithmetic series formula n (a1 + an) -> 2(3 + 4 .... 50) = 2 53*48 ... ) goes up by one each increment, int b, the sum of lowest value and highest value (2+50, 3+50) also goes up by one each increment. Int c, the count of how many terms goes down by 1 each time.

I disregard the divide by 2 in arithmetic series since the denominator of 50c2 is 50*49/2 so i disregard the div by 2 in both numerator and denominator.

int a = 1;
    int b = 52;
    int c = 49;

    int n = 0;

    while(c>=1)
    {
        n = n+a*b*c;
        a++;
        b++;
        c--;
    }

    System.out.println(n/50/49);

We begin by choosing two (not necessarily distinct) integers from 1 to 50. The sum of all such possible pairs is $\left(\frac{50\left(50 + 1\right)}{2}\right)^2 = 1275^2.$ We now consider the criterion of distinct integers by removing the cases where the numbers of the same, $1275^2 - \frac{50\left(50 + 1\right)\left(2 \cdot 50 + 1\right)}{6}.$ To find the expected value, we divide by the number of cases, $\frac{1275^2 - 42925}{50 \cdot 49} = 646.$

Since we are getting the expected value of a finite number of data points, we simply average it. To get the sum of ALL the possible values, we make use of the idea of a method similar to "foil": sum = (1+2+3+....+50)(1+2+3+....+50) but take note that 1^2, 2^2, 3^2, .., 50^2 are included in the sum, so we'll have to subtract them

real sum = (1+2+3+....+50)^2 - (1^2 + 2^2 + 3^2 + .. +50^2)

We can solve for these easily by using their respective formula:

real sum = (51 50/2)^2 - (50 51*101)/6 real sum = 1 582 700

The number of terms are 50 49 because they are distinct, and the way we computed for the sum considers the orders of the terms (e.g. 1 50 AND 50*1 were included in the sum).

Therefore, E(x) = 1 582 700/(49*50) = 646

Let $A$ be the sum of the product of distinct integers from 1 to 50. The desired expected value is easily seen to be $\frac{2A}{50\times49}$ since we pick one number of 50 then another of 49, and we can do it in either order. This quantity can be quickly calculated by noting that the cross terms of $(1+2+\cdots+50)^2$ are precisely $2A$ . Thus, our answer is

$\frac{(1+2+\cdots+50)^2 - (1^2+2^2+\cdots+50^2)}{2450} = \frac{(\frac{50 \times 51}{2})^2 - \frac{50\times 51 \times 101}{6}}{2450} = 646$ .

The EV is going to have a denominator of 50 choose 2 = 25*49, and a numerator of $\sum_{i,j} i*j = \sum_{i=1}^{50} i*\sum_{j>i}^{50} = \sum_{i=1}^{50} i * (\frac{50(51)}{2} - \frac{i(i+1)}{2} )$

$= (25*51)^2 - (\frac{1}{2} )*\sum_{i=1}^{50} i^3 + i^2$

$= (25*51)^2 - (\frac{1}{2})(\frac{50*51}{2})^2 + \frac{50*51*101}{6}) = 791350$

So the EV is: $= (791350)/(25*49) = 646$ as desired.