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 $50\$ 50. If a sum less than six is rolled, the player receives $20\$ 20. If a player rolls two odd numbers, then they receive $8\$ 8. A player can only receive one prize. Therefore, if a roll meets the description of fnore 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? yes  \checkmark \checkmark ~ ০^{\infty}

Studdy Solution

STEP 1

1. A standard six-sided die is used, with faces numbered from 1 to 6.
2. The outcomes of the dice rolls are independent.
3. The player receives only the highest prize if multiple conditions are met.
4. The cost to play the game is \$10.

STEP 2

1. Identify all possible outcomes of rolling two dice.
2. Calculate the probability of each prize condition.
3. Calculate the expected value of the game.
4. Determine if the game favors the player.

STEP 3

Identify all possible outcomes of rolling two dice. There are 6×6=366 \times 6 = 36 possible outcomes.

STEP 4

Calculate the probability of each prize condition.
- **Sum greater than 10**: Possible outcomes are (5,6), (6,5), (6,6). There are 3 outcomes. $ P(\text{Sum} > 10) = \frac{3}{36} = \frac{1}{12} \]
- **Sum less than 6**: Possible outcomes are (1,1), (1,2), (2,1), (1,3), (3,1), (2,2). There are 6 outcomes. $ P(\text{Sum} < 6) = \frac{6}{36} = \frac{1}{6} \]
- **Two odd numbers**: Possible outcomes are (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5). There are 9 outcomes. $ P(\text{Two odd numbers}) = \frac{9}{36} = \frac{1}{4} \]

STEP 5

Calculate the expected value of the game.
- **Prize for sum > 10**: \$50 - **Prize for sum < 6**: \$20 - **Prize for two odd numbers**: \$8 - **No prize**: \$0
Calculate the expected payout:
E(payout)=(112×50)+(16×20)+(14×8)+(12×0)E(\text{payout}) = \left(\frac{1}{12} \times 50\right) + \left(\frac{1}{6} \times 20\right) + \left(\frac{1}{4} \times 8\right) + \left(\frac{1}{2} \times 0\right)
E(payout)=(5012)+(206)+(84)E(\text{payout}) = \left(\frac{50}{12}\right) + \left(\frac{20}{6}\right) + \left(\frac{8}{4}\right)
E(payout)=4.17+3.33+2.00=9.50E(\text{payout}) = 4.17 + 3.33 + 2.00 = 9.50
Calculate the expected value of the game:
E(game)=E(payout)Cost to play=9.5010=0.50E(\text{game}) = E(\text{payout}) - \text{Cost to play} = 9.50 - 10 = -0.50

STEP 6

Determine if the game favors the player.
Since the expected value is negative (\$-0.50), the game does not favor the player.
The expected value of the game is:
0.50\boxed{-0.50}
In the long run, the game does not favor the player.

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