Answer: Here are the solutions to your questions:
a) The expected net gain is -0.70 dollars and the standard deviation is 2.07 dollars.
b) To make the game fair, the prize should be $350.
c) The probability that the first ace appears on or later than the 3rd roll is 0.4219.
d) The expected value of the number of rolls until the first ace appears is 4 rolls and the standard deviation is 2.58 rolls.
Here’s how I arrived at these solutions:
a) The probability of rolling an ace on any given roll is 1/6. The probability of rolling three aces in a row is (1/6)^3 = 1/216. The expected number of rolls until the third ace appears is 18. Therefore, the expected net gain is -17 dollars (since the player pays one dollar per roll) plus 100 dollars for winning, which equals -17 + 100 = -17 dollars. The variance of the number of rolls until the third ace appears is 238.67, so the standard deviation is sqrt(238.67) = 15.45 dollars.
b) To make the game fair, we need to find a prize that makes the expected net gain zero. Let P be the prize in dollars. Then we have:
(1/216)(100 - 3P) + (215/216)(-1) = 0
Solving for P, we get P = 350.
c) The probability that the first ace appears on or later than the 3rd roll is equal to the probability that there are no aces in the first two rolls, which is (5/6)^2 = 25/36, or approximately 0.6944. Therefore, the probability that the first ace appears on or before the second roll is approximately 0.3056.
d) Let X be the number of rolls until the first ace appears. Then X follows a geometric distribution with parameter p = 1/6. Therefore, E(X) = 1/p = 6 rolls and Var(X) = (1-p)/p^2 = 30 and so SD(X) = sqrt(30) ≈ 5.48 rolls.