A blue die and a red die are thrown in a game. If the sum of the two numbers is 7 or 11, the player wins $10. If the sum of the two numbers is 12, then the player wins $20. In all other cases, the player loses a dollar. What are the expected winnings of the player in one game

Given - A blue die and a red die are thrown in a game. If the sum of the two numbers is 7 or 11, the player wins $10. If the sum of the two numbers is 12, then the player wins $20. In all other cases, the player loses a dollar.

To find - What are the expected winnings of the player in one game ?

Solution -

Given that,

A blue die and a red die are thrown in a game.

The Sample Space becomes -

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (4, 4), (6, 5), (6, 6) }

So,

n(S) = 36

Now,

Let A be the outcome that, the sum of the two numbers is 7 or 11

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

(6, 5), (5, 6) }

So,

n(A) = 8

So,

P(A) = n(A)/ n(S)

= 8/ 36

= 2/9

⇒P(A) = 2/9

Now,

Let B be the outcome that, the sum of the two numbers is 12

B = {(6, 6)}

So,

n(B) = 1

So,

P(B) = n(B)/ n(S)

= 1/ 36

⇒P(B) = 1/36

Now,

Expected value, E(x) = ∑ x P(x)

= 10 × P(A) + 12 × P(B)

= 10 × 2/9 + 12 × 1/36

= 20/9 + 12/36

= (20×4 + 12)/36

= (80 + 12)/36

= 92/36

= 2.55

⇒Expected value, E(x) = 2.55

∴ we get

The expected winnings of the player in one game = 2.55