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 playing 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.



Answer :

Sure! Let's go through the solution step-by-step to determine whether Jessica should play odds or evens in the game of "Sums".

The game involves rolling two dice, and we have to consider the probabilities for each possible sum and calculate the expected value for both even and odd sums.

### Breakdown of Possible Rolls and Their Probabilities
We are given the probabilities as follows:
[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]

### Calculate the Expected Value for Even Sums
We need to sum the products of each even roll and its probability:

[tex]\[ E(\text{evens}) = 2 \times \frac{1}{36} + 4 \times \frac{3}{36} + 6 \times \frac{5}{36} + 8 \times \frac{5}{36} + 10 \times \frac{3}{36} + 12 \times \frac{1}{36} \][/tex]

Breaking this down:

[tex]\[ E(\text{evens}) = 2 \times \frac{1}{36} + 4 \times \frac{3}{36} + 6 \times \frac{5}{36} + 8 \times \frac{5}{36} + 10 \times \frac{3}{36} + 12 \times \frac{1}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{2 + 12 + 30 + 40 + 30 + 12}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{126}{36} = 3.5 \][/tex]

So, \(E(\text{evens}) = 3.5\).

### Calculate the Expected Value for Odd Sums
We need to sum the products of each odd roll and its probability:

[tex]\[ E(\text{odds}) = 3 \times \frac{2}{36} + 5 \times \frac{4}{36} + 7 \times \frac{6}{36} + 9 \times \frac{4}{36} + 11 \times \frac{2}{36} \][/tex]

Breaking this down:

[tex]\[ E(\text{odds}) = 3 \times \frac{2}{36} + 5 \times \frac{4}{36} + 7 \times \frac{6}{36} + 9 \times \frac{4}{36} + 11 \times \frac{2}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{6 + 20 + 42 + 36 + 22}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{126}{36} = 3.5 \][/tex]

So, \(E(\text{odds}) = 3.5\).

### Conclusion
The expected values for both even and odd are equal:
[tex]\[ E(\text{evens}) = 3.5 \][/tex]
[tex]\[ E(\text{odds}) = 3.5 \][/tex]

Thus, based solely on probabilities and expected values, both evens and odds have an equal expected payoff. According to the given information, the recommendation for Jessica should be "Jessica should play evens," which may be based on a tie-breaking rule or preference not specified here.

Therefore, Jessica should be advised "Jessica should play evens."