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Sagot :
To determine whether Jessica should play odds or evens in the game of rolling two dice, let's analyze the expected values for the sums, considering each possible outcome and its probability. This analysis will guide her in making the best choice.
1. Understand the Probabilities:
From rolling two dice, the probabilities for each sum (from 2 to 12) are given below:
- P(2) = [tex]\(\frac{1}{36}\)[/tex]
- P(3) = [tex]\(\frac{2}{36}\)[/tex]
- P(4) = [tex]\(\frac{3}{36}\)[/tex]
- P(5) = [tex]\(\frac{4}{36}\)[/tex]
- P(6) = [tex]\(\frac{5}{36}\)[/tex]
- P(7) = [tex]\(\frac{6}{36}\)[/tex]
- P(8) = [tex]\(\frac{5}{36}\)[/tex]
- P(9) = [tex]\(\frac{4}{36}\)[/tex]
- P(10) = [tex]\(\frac{3}{36}\)[/tex]
- P(11) = [tex]\(\frac{2}{36}\)[/tex]
- P(12) = [tex]\(\frac{1}{36}\)[/tex]
2. Calculate the Expected Value for Odd Sum (E(odds)):
The expected value for odd sums is calculated by multiplying each odd sum by its probability and then summing these products.
[tex]\[ E(odds) = 3 \cdot \frac{2}{36} + 5 \cdot \frac{4}{36} + 7 \cdot \frac{6}{36} + 9 \cdot \frac{4}{36} + 11 \cdot \frac{2}{36} \][/tex]
Simplifying, we get:
[tex]\[ E(odds) = \frac{2 \cdot 3}{36} + \frac{4 \cdot 5}{36} + \frac{6 \cdot 7}{36} + \frac{4 \cdot 9}{36} + \frac{2 \cdot 11}{36} \][/tex]
[tex]\[ E(odds) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} \][/tex]
[tex]\[ E(odds) = \frac{6 + 20 + 42 + 36 + 22}{36} = \frac{126}{36} = 3.5 \][/tex]
3. Calculate the Expected Value for Even Sum (E(evens)):
The expected value for even sums is calculated by multiplying each even sum by its probability and then summing these products.
[tex]\[ E(evens) = 2 \cdot \frac{1}{36} + 4 \cdot \frac{3}{36} + 6 \cdot \frac{5}{36} + 8 \cdot \frac{5}{36} + 10 \cdot \frac{3}{36} + 12 \cdot \frac{1}{36} \][/tex]
Simplifying, we get:
[tex]\[ E(evens) = \frac{1 \cdot 2}{36} + \frac{3 \cdot 4}{36} + \frac{5 \cdot 6}{36} + \frac{5 \cdot 8}{36} + \frac{3 \cdot 10}{36} + \frac{1 \cdot 12}{36} \][/tex]
[tex]\[ E(evens) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} \][/tex]
[tex]\[ E(evens) = \frac{2 + 12 + 30 + 40 + 30 + 12}{36} = \frac{126}{36} = 3.5 \][/tex]
From the calculations, we see that both the expected values for the sums of odd and even rolls are the same—3.5. Therefore, neither odds nor evens has an advantage based on this calculation.
Thus, neither odds nor evens will have a higher expected value for the score per roll. Jessica or anyone playing the game can choose either odds or evens, and it won't make a difference in terms of expected points based on the sums of the rolls.
1. Understand the Probabilities:
From rolling two dice, the probabilities for each sum (from 2 to 12) are given below:
- P(2) = [tex]\(\frac{1}{36}\)[/tex]
- P(3) = [tex]\(\frac{2}{36}\)[/tex]
- P(4) = [tex]\(\frac{3}{36}\)[/tex]
- P(5) = [tex]\(\frac{4}{36}\)[/tex]
- P(6) = [tex]\(\frac{5}{36}\)[/tex]
- P(7) = [tex]\(\frac{6}{36}\)[/tex]
- P(8) = [tex]\(\frac{5}{36}\)[/tex]
- P(9) = [tex]\(\frac{4}{36}\)[/tex]
- P(10) = [tex]\(\frac{3}{36}\)[/tex]
- P(11) = [tex]\(\frac{2}{36}\)[/tex]
- P(12) = [tex]\(\frac{1}{36}\)[/tex]
2. Calculate the Expected Value for Odd Sum (E(odds)):
The expected value for odd sums is calculated by multiplying each odd sum by its probability and then summing these products.
[tex]\[ E(odds) = 3 \cdot \frac{2}{36} + 5 \cdot \frac{4}{36} + 7 \cdot \frac{6}{36} + 9 \cdot \frac{4}{36} + 11 \cdot \frac{2}{36} \][/tex]
Simplifying, we get:
[tex]\[ E(odds) = \frac{2 \cdot 3}{36} + \frac{4 \cdot 5}{36} + \frac{6 \cdot 7}{36} + \frac{4 \cdot 9}{36} + \frac{2 \cdot 11}{36} \][/tex]
[tex]\[ E(odds) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} \][/tex]
[tex]\[ E(odds) = \frac{6 + 20 + 42 + 36 + 22}{36} = \frac{126}{36} = 3.5 \][/tex]
3. Calculate the Expected Value for Even Sum (E(evens)):
The expected value for even sums is calculated by multiplying each even sum by its probability and then summing these products.
[tex]\[ E(evens) = 2 \cdot \frac{1}{36} + 4 \cdot \frac{3}{36} + 6 \cdot \frac{5}{36} + 8 \cdot \frac{5}{36} + 10 \cdot \frac{3}{36} + 12 \cdot \frac{1}{36} \][/tex]
Simplifying, we get:
[tex]\[ E(evens) = \frac{1 \cdot 2}{36} + \frac{3 \cdot 4}{36} + \frac{5 \cdot 6}{36} + \frac{5 \cdot 8}{36} + \frac{3 \cdot 10}{36} + \frac{1 \cdot 12}{36} \][/tex]
[tex]\[ E(evens) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} \][/tex]
[tex]\[ E(evens) = \frac{2 + 12 + 30 + 40 + 30 + 12}{36} = \frac{126}{36} = 3.5 \][/tex]
From the calculations, we see that both the expected values for the sums of odd and even rolls are the same—3.5. Therefore, neither odds nor evens has an advantage based on this calculation.
Thus, neither odds nor evens will have a higher expected value for the score per roll. Jessica or anyone playing the game can choose either odds or evens, and it won't make a difference in terms of expected points based on the sums of the rolls.
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