Math  /  Data & Statistics

QuestionIt costs $10\$ 10 to play a dice game. For this game, two dice are rolled. If a sum greater than 10 is rolled, the player receives $45\$ 45. If a sum less than six is rolled, the player receives $30\$ 30. If a player rolls two odd numbers, then they receive $7\$ 7. A player can only receive one prize. Therefore, if a roll meets the description of more than one prize, the player only receives the higher prize value (not both). The expected value (to the nearest cent) of the game is $\$ \square In the long run, does the game favor the player? \square

Studdy Solution

STEP 1

1. The game involves rolling two six-sided dice.
2. The possible outcomes for the sum of two dice range from 2 to 12.
3. The player receives only the highest prize if multiple conditions are met.
4. The player pays \$10 to play the game.
5. We need to calculate the expected value to determine if the game favors the player.

STEP 2

1. Calculate the probability of each prize condition.
2. Calculate the expected payout for each condition.
3. Determine the expected value of the game.
4. Assess if the game favors the player.

STEP 3

Calculate the probability of rolling a sum greater than 10. The possible sums greater than 10 are 11 and 12.
- Sum of 11: (5,6), (6,5) — 2 outcomes - Sum of 12: (6,6) — 1 outcome
Total outcomes for sum > 10: 3
Total possible outcomes when rolling two dice: 6×6=366 \times 6 = 36
Probability of sum > 10:
P(sum>10)=336=112 P(\text{sum} > 10) = \frac{3}{36} = \frac{1}{12}

STEP 4

Calculate the probability of rolling a sum less than 6. The possible sums less than 6 are 2, 3, 4, and 5.
- Sum of 2: (1,1) — 1 outcome - Sum of 3: (1,2), (2,1) — 2 outcomes - Sum of 4: (1,3), (2,2), (3,1) — 3 outcomes - Sum of 5: (1,4), (2,3), (3,2), (4,1) — 4 outcomes
Total outcomes for sum < 6: 10
Probability of sum < 6:
P(sum<6)=1036=518 P(\text{sum} < 6) = \frac{10}{36} = \frac{5}{18}

STEP 5

Calculate the probability of rolling two odd numbers. The odd numbers on a die are 1, 3, and 5.
- Possible outcomes: (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5) — 9 outcomes
Probability of two odd numbers:
P(two odd numbers)=936=14 P(\text{two odd numbers}) = \frac{9}{36} = \frac{1}{4}

STEP 6

Calculate the expected payout for each condition:
- Expected payout for sum > 10: 112×45=4512=3.75 \frac{1}{12} \times 45 = \frac{45}{12} = 3.75 - Expected payout for sum < 6: 518×30=15018=8.33 \frac{5}{18} \times 30 = \frac{150}{18} = 8.33 - Expected payout for two odd numbers: 14×7=74=1.75 \frac{1}{4} \times 7 = \frac{7}{4} = 1.75

STEP 7

Calculate the total expected payout considering the highest prize only:
- The highest prize is \$45 for sum > 10, which overlaps with other conditions. - The next highest prize is \$30 for sum < 6, which does not overlap with sum > 10. - The lowest prize is \$7 for two odd numbers, which overlaps with other conditions.
The expected value (EV) of the game is:
EV=3.75+(8.333.75)+(1.753.75)=3.75+4.582=6.33 \text{EV} = 3.75 + (8.33 - 3.75) + (1.75 - 3.75) = 3.75 + 4.58 - 2 = 6.33
Subtract the cost of playing the game:
Net EV=6.3310=3.67 \text{Net EV} = 6.33 - 10 = -3.67

STEP 8

Assess if the game favors the player:
Since the expected value is negative (\$-3.67), the game does not favor the player.
The expected value of the game is:
3.67 \boxed{-3.67}
In the long run, the game does not favor the player.

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