Question

In your homework, you were introduced to a game, called Under-or-Over-Seven. It has many rules. Here is the particular rule that we are going to use for this question: a pair of fair dice is rolled once, and the resulting sum determines whether the player wins or loses his/her bet; the player wins $9 if the result of the roll is exactly equal to 7 and loses $3 otherwise. Compute the expected value of this game (i.e., the expected gain or loss).

Answer 1

Answer:

EX=-$1

Step-by-step explanation:

Expected Value of Probability Distribution

Assume a discrete probability distribution is

$P=\{p1,p2,p3,...pn\}$

For

$X=\{x1,x2,x3,...xn\}$

The expected value is

$EX=\sum x_i.p_i$

We have two possible outcomes from our random experience: The sum of the die is 7 or different from 7. If it sums 7, the player wins $9, otherwise, they lose $3. Thus

$x=\{9,-3\}$

We must find the probability of having a 7. Each dice can be a 1, 2, 3 ,4 , 5, or 6. The combinations to sum 7 are 1+6, 2+5, 3+4, 4+4, 5+2, and 6+1. That is 6 possibilities out of 36 in total. The probability of having a 7 is

$\displaystyle p_1=\frac{6}{36}=\frac{1}{6}$

The probability of not getting 7 is the negation of the previous event

$\displaystyle p_2=1-\frac{1}{6}=\frac{5}{6}$

The probability set is:

$\displaystyle P=\left\{\frac{1}{6},\frac{5}{6}\right\}$

The expected value is:

$\displaystyle EX=9\cdot \frac{1}{6}-3\cdot \frac{5}{6}$

$\displaystyle EX=\frac{3}{2}-\frac{5}{2}=-1$

Therefore, the player can expect to lose $1