A pair of fair dice is rolled once. Suppose that you lose $8 if the dice sum to 7 and win $12 if the dice sum to 10 or 6. How much should you win or lose if any other number turns up in order for the game to be fair?

Question

A pair of fair dice is rolled once. Suppose that you lose ​$8 if the dice sum to 7 and win ​$12 if the dice sum to 10 or 6. How much should you win or lose if any other number turns up in order for the game to be​ fair?

oobleck · Answer

figure the probabilities of the win/lose events. Then if P1 = P(7) P2 = P(6 or 10) P3 = 1 - P1 - P2 Then the win/loss of $x needed to make the game fair (Expected value = 0) is -8P1 + 12P2 + P3*x = 0 come on back if you have trouble with P1 or P2

Explain Bot · Answer

To determine how much you should win or lose if any other number turns up in order for the game to be fair, we need to calculate the probability of each possible outcome and then determine the expected value. Let's start by finding the probabilities of each outcome: 1. Sum of 7: There are 6 ways to obtain a sum of 7 with two fair dice (1+6, 6+1, 2+5, 5+2, 3+4, 4+3). Since there are 36 total possible outcomes with two dice, the probability of rolling a sum of 7 is 6/36. 2. Sum of 10 or 6: There are 3 ways to obtain a sum of 10 (4+6, 5+5, 6+4) and 5 ways to obtain a sum of 6 (1+5, 5+1, 2+4, 4+2, 3+3). So, the probability of rolling a sum of 10 or 6 is (3+5)/36 = 8/36. Now, let's calculate the expected value: 1. If the dice sum to 7, you lose ​$8. 2. If the dice sum to 10 or 6, you win ​$12. 3. If any other number turns up, let's denote the amount you win or lose as "x". The expected value is calculated by multiplying each outcome by its probability and adding them together: Expected value = (Probability of summing to 7) × (Amount you lose) + (Probability of summing to 10 or 6) × (Amount you win) + (Probability of any other number) × (Amount you win or lose) Since the game is fair, the expected value should be zero. Therefore, we can solve for "x" using the equation: 0 = (6/36) × (-8) + (8/36) × (12) + (x) × (1 - (6/36) - (8/36)) Simplifying the equation: 0 = -(48/36) + (96/36) + (13/36)x Combining like terms: 0 = (48/36) + (13/36)x Solving for "x": -(48/36) = (13/36)x -(4/3) = (13/36)x x = -(4/3) × (36/13) x = -12 Therefore, in order for the game to be fair, you should win ​$12 or lose ​$12 if any other number turns up.

A pair of fair dice is rolled once. Suppose that you lose ​$8 if the dice sum to 7 and win ​$12 if the dice sum to 10 or 6. How much should you win or lose if any other number turns up in order for the game to be​ fair?

figure the probabilities of the win/lose events. Then if

To determine how much you should win or lose if any other number turns up in order for the game to be fair, we need to calculate the probability of each possible outcome and then determine the expected value.

A pair of fair dice is rolled once. Suppose that you lose $8 if the dice sum to 7 and win $12 if the dice sum to 10 or 6. How much should you win or lose if any other number turns up in order for the game to be fair?