Two students have devised a dice game named "Sums" for their statistics class. The game consists of choosing to play odds or evens.

Probabilities for "Sums":
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Roll & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline[tex]$P$[/tex] (roll) & [tex]$\frac{1}{36}$[/tex] & [tex]$\frac{2}{36}$[/tex] & [tex]$\frac{3}{36}$[/tex] & [tex]$\frac{4}{36}$[/tex] & [tex]$\frac{5}{36}$[/tex] & [tex]$\frac{6}{36}$[/tex] & [tex]$\frac{5}{36}$[/tex] & [tex]$\frac{4}{36}$[/tex] & [tex]$\frac{3}{36}$[/tex] & [tex]$\frac{2}{36}$[/tex] & [tex]$\frac{1}{36}$[/tex] \\
\hline
\end{tabular}

Each person takes turns rolling two dice. If the sum is odd, the person playing odds gets points equal to the sum of the roll. If the sum is even, the person playing evens gets points equal to the sum of the roll. Note that the points earned are independent of who is rolling the dice.

If Jessica is challenged to a game of "Sums," which statement below is accurate in guiding her to the correct choice of choosing to play odds or evens?

A. [tex]$E$[/tex] (evens) will be more because there are more even numbers that result from rolling two dice. Therefore, Jessica should play evens.

B. [tex]$O$[/tex] (odds) will be more because there are more odd numbers that result from rolling two dice. Therefore, Jessica should play odds.



Answer :

Let's analyze Jessica's situation in the dice game "Sums" to determine whether she should choose to play odds or evens.

We are given the probabilities associated with each of the possible sums from rolling two dice, as shown in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Roll} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(\text{Roll}) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \\ \hline \end{array} \][/tex]

We will calculate the expected value of points for both the even and odd sums.

### Even Sums and Their Probabilities
The even sums are [tex]$2, 4, 6, 8, 10, 12$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(2) &= \frac{1}{36},\\ P(4) &= \frac{3}{36},\\ P(6) &= \frac{5}{36},\\ P(8) &= \frac{5}{36},\\ P(10) &= \frac{3}{36},\\ P(12) &= \frac{1}{36}. \end{aligned} \][/tex]

Next, we calculate the expected value of points for even sums, \(E(\text{evens})\):
[tex]\[ E(\text{evens}) = 2 \cdot \frac{1}{36} + 4 \cdot \frac{3}{36} + 6 \cdot \frac{5}{36} + 8 \cdot \frac{5}{36} + 10 \cdot \frac{3}{36} + 12 \cdot \frac{1}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} = \frac{126}{36} = 3.5 \][/tex]

### Odd Sums and Their Probabilities
The odd sums are [tex]$3, 5, 7, 9, 11$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(3) &= \frac{2}{36},\\ P(5) &= \frac{4}{36},\\ P(7) &= \frac{6}{36},\\ P(9) &= \frac{4}{36},\\ P(11) &= \frac{2}{36}. \end{aligned} \][/tex]

Next, we calculate the expected value of points for odd sums, \(E(\text{odds})\):
[tex]\[ E(\text{odds}) = 3 \cdot \frac{2}{36} + 5 \cdot \frac{4}{36} + 7 \cdot \frac{6}{36} + 9 \cdot \frac{4}{36} + 11 \cdot \frac{2}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} = \frac{126}{36} = 3.5 \][/tex]

### Conclusion
The expected value of points for both even and odd sums is the same, \(3.5\).

So, the accurate statement in guiding Jessica is:
Both evens and odds have the same expected value of points, [tex]\(3.5\)[/tex]. Therefore, it makes no statistical difference whether Jessica chooses to play evens or odds, as her expected score will be the same in either case.