X is the lower number when two dice are rolled. Calculate the expected value of X.

How do you do this?

Possible Outcomes Of Throwing Dice

1 1 Result: 1
1 2 Result: 1
1 3 Result: 1
1 4 Result: 1
1 5 Result: 1
1 6 Result: 1

2 1 Result: 1
2 2 Result: 2
2 3 Result: 2
2 4 Result: 2
2 5 Result: 2
2 6 Result: 2

3 1 Result: 1
3 2 Result: 2
3 3 Result: 3
3 4 Result: 3
3 5 Result: 3
3 6 Result: 3

4 1 Result: 1
4 2 Result: 2
4 3 Result: 3
4 4 Result: 4
4 5 Result: 4
4 6 Result: 4

5 1 Result: 1
5 2 Result: 2
5 3 Result: 3
5 4 Result: 4
5 5 Result: 5
5 6 Result: 5

6 1 Result: 1
6 2 Result: 2
6 3 Result: 3
6 4 Result: 4
6 5 Result: 5
6 6 Result: 6

Now, each of these results has a probability of 1/36.

E(x) = (1/36)[6 + 11 + 15 + 18 + 20 + 21] = 91/36 = 2.52777778

Let Y and Z represent the two independent rolls, so X = min(Y, Z). Let's first find the probability distribution for X, then find E(X).

For 1 <= n <= 6,

P(X = i)
= P(min(Y, Z) = i)
= P(Y = i and Z = i) + P(Y = i and Z > i) + P(Y > i and Z = i)
= P(Y = i)P(Z = i) + P(Y = i)P(Z > i) + P(Y > i)P(Z = i)
= (1/6)(1/6) + (1/6)((6-i)/6) + ((6-i)/6)(1/6)
= (13 - 2i)/36

The expectation of a random variable is the sum of the products of the value of the variable and the value's probability of occurrence. Therefore,

E(X) = sum i = 1 to 6 of iP(X = i)
= sum i = 1 to 6 of i((13 - 2i)/36)
= 1(11/36) + 2(9/36) + 3(7/36) + 4(5/36) + 5(3/36) + 6(1/36)
= 91/36