Question

Consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total. What is the expected value of the points earned on a single roll in this game?

Answer 1

Answer:

The expected value of the points earned on a single roll in this game is $\dfrac{1}{6} = 0.1667$ .

Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

Note that the probability of getting any of the specific number on the dice is   $\dfrac{1}{6}$ .

      Numbers on the dice (X)                       P(X)

                      +1                                                 $\frac{1}{6}$

                      +2                                                $\frac{1}{6}$

                      +3                                                $\frac{1}{6}$

                      -4                                                 $\frac{1}{6}$

                      +5                                                $\frac{1}{6}$

                      -6                                                 $\frac{1}{6}$

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X)  =  $\sum X \times P(X)$

   =  $(+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}$

   =  $\frac{1}{6} + \frac{2}{6} + \frac{3}{6} - \frac{4}{6} + \frac{5}{6} - \frac{6}{6}$

   =  $\frac{1+2+3-4+5-6}{6}$

   =  $\frac{11-10}{6}= \frac{1}{6}$

Hence, the expected value of the points earned on a single roll in this game is  $\frac{1}{6} = 0.1667$ .