Final answer:
To make the dice game fair, we solve for the win/lose amount for other sums by setting the expected value to zero and accounting for the probabilities and payouts of rolling sums 9, 4, and 8.
Step-by-step explanation:
To establish a game as fair, we need to calculate the expected value for a player and make sure it is zero. Here, we have a pair of fair dice and different outcomes lead to different monetary results. The student is asking to determine the win/lose amount for other sums to maintain a fair game.
First, we should calculate the probabilities of winning or losing. For a dice sum of 9, there are 4 outcomes: (3,6), (4,5), (5,4), and (6,3). There are 36 possible outcomes when two dice are rolled, so the probability of getting a sum of 9 is 4/36 or 1/9.
Similarly, the sums of 4 or 8 each have three possible outcomes: (1,3), (2,2), (3,1) for a sum of 4 and (2,6), (3,5), (5,3), (6,2) for a sum of 8. Therefore, the probability of getting either 4 or 8 is 3/36 + 3/36 which simplifies to 1/6.
To calculate the expected value (EV), we multiply each outcome's probability by its respective monetary value and sum those products for all outcomes. Using EV for expectation, p for probability, and v for value:
EV = p(sum to 9) * v(lose $10) + p(sum to 4 or 8) * v(win $15) + p(other sums) * v(win/lose amount).
To find the win/lose amount for the other sums, we set the EV to zero since it's a fair game:
0 = (1/9)*(-$10) + (1/6)*$15 + p(other sums) * v(win/lose amount)
We know that p(other sums) is the remainder of the probabilities, which is 1 - (1/9) - (1/6), and v(win/lose amount) is the amount we're solving for.
Let's solve for v(win/lose amount):
v(win/lose amount) = -[ (1/9)*(-$10) + (1/6)*$15 ] / [1 - (1/9) - (1/6)]
Plugging in the numbers, we can calculate the fair win/lose amount for other sums to ensure the game's fairness. Remember that a positive value for v(win/lose amount) indicates a win, while a negative value indicates a loss for the player.