Question

Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win ​$ if you succeed and you lose ​$ if you fail. What is the expected value of this game to​ you? Should you expect to win or lose the expected value in the first​ game? What can you expect if you play ​times? Explain.

Answer 1

Complete question :

Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 if you fail. What is the expected value of this game to​ you? What can you expect if you play 100 times.

Answer:

$0.5 ; win $50 with 100 rolls

Explanation:

From a roll of two fair dice; probability of obtaining an even number :

Even numbers = (2, 4, 6) = 3

P = 3 /6 = 1 /2

For 2 fair dice ; probability of rolling two even numbers : independent event.

1/2 * 1/2 = 1/4

Hence, p(success) = 1/4 ; P(failure) = 1 - 1/4 = 3/4

Probability table

Winning = $8 or loss = - $2

X : ____ 8 ______ - 2

P(x) __ 1/4 ______ 3/4

Expected value : E(x) = ΣX*P(x)

E(x) = (8 * 1/4) + (-2 * 3/4)

E(x) = 2 - 1.5

E(x) = $0.5

Since expected value is positive, the expect to win

If played 100 times;

Expected value = 100 * $0.5 = $50