Question

A certain game consist of rolling a single fair die and based off as a following numbers listed in the picture

Answer 1

Given:

A fair die is rolled.

It pays off $10 for 6, $7 for a 5, $4  for a 4 and no payoff otherwise.

To find:

The expected winning for this game.

Solution:

If a die is rolled then the possible outcomes are 1, 2, 3, 4, 5, 6.

The probability of getting a 6 is:

$P(6)=\dfrac{1}{6}$

The probability of getting a 5 is:

$P(5)=\dfrac{1}{6}$

The probability of getting a 4 is:

$P(4)=\dfrac{1}{6}$

The probability of getting other numbers (1,2,3) is:

$P(\text{Otherwise})=\dfrac{3}{6}$

$P(\text{Otherwise})=\dfrac{1}{2}$

We need to find the sum of product of payoff and their corresponding probabilities to find the expected winning for this game.

$E(x)=10\times P(6)+7\times P(5)+4\times P(4)+0\times P(\text{Otherwise})$

$E(x)=10\times \dfrac{1}{6}+7\times \dfrac{1}{6}+4\times \dfrac{1}{6}+0\times \dfrac{1}{2}$

$E(x)=\dfrac{10}{6}+\dfrac{7}{6}+\dfrac{4}{6}+0$

$E(x)=\dfrac{10+7+4}{6}$

$E(x)=\dfrac{21}{6}$

$E(x)=3.5$

Therefore, the expected winnings for this game are $3.50.