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."
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."
