Three Out of Nine: A Carnival Probability Claim
The student council is designing a carnival game for the spring fundraiser. A player pays $1 to spin two fair spinners once each. Spinner A has four equal sectors numbered 1, 2, 3, 4; spinner B has six equal sectors numbered 1 through 6. The player wins a $3 prize if the sum of the two spins is 8 or more. Before ordering prizes, the council needs the probability of winning and the expected profit per play, and it hopes to clear $150 over an estimated 400 plays. During the meeting, one member argues the probability is 1/3 "because the sums run from 2 to 10, and 8, 9, and 10 are three of those nine sums." A second member wants to raise the price to $2 rather than change the prize. The order form is due in a week, and there is no budget to cover a loss.
1. Find the correct probability of winning and explain precisely what is wrong with the "three out of nine sums" reasoning.
Read the worked answer
There are 4 × 6 = 24 equally likely ordered outcomes (a, b). Count the winning ones with a ≥ 8:
a = 2: b = 6 gives 8 - 1 outcome. a = 3: b = 5, 6 give 8, 9 - 2 outcomes. a = 4: b = 4, 5, 6 give 8, 9, 10 - 3 outcomes. a = 1: none.
Total 6 outcomes, so P(win) = 6/24 = 1/4.
The flaw is treating the nine possible sums as equally likely. They are not. Sum 5 occurs as (1,4), (2,3), (3,2), (4,1) - four ways - while sum 2 occurs only as (1,1), one way, and sum 10 only as (4,6). Sum 8 occurs 3 ways, 9 occurs 2, and 10 occurs 1, so the winning sums carry 6 of the 24 outcomes, not 3 of 9. Listing outcomes rather than sums fixes the sample space so that each entry has probability 1/24.
A quick structural check supports the count: the average sum is 2.5 + 3.5 = 6, and the distribution is symmetric about 6, so a threshold of 8 or more should sit clearly below half - 1/4 is plausible, 1/3 is too high.
2. Compute the expected value of the game for the player and for the council. Is the game fair, and how much should the council expect to clear over 400 plays?
Read the worked answer
Let X be the player's net gain in dollars. With probability 3/4 the player loses the $1; with probability 1/4 the player receives $3 but has paid $1, netting $2.
E[X] = (3/4)(−1) + (1/4)(2) = −0.75 + 0.50 = −$0.25 per play.
Equivalently, compute the council's side: revenue $1, expected payout (1/4)($3) = $0.75, so expected profit is $0.25 per play. The council's gain is exactly the player's loss, which is the check that the arithmetic is consistent.
The game is not fair. A fair game has expected value zero for both parties, which here would require the expected payout to equal the $1 price, i.e., a prize of $4 at a win probability of 1/4.
Over 400 plays, expected profit is 400 × $0.25 = $100, short of the $150 goal but well above zero. That $100 is an average, not a guarantee. Per play the payout has standard deviation √(2.25 − 0.5625) ≈ $1.30, so over 400 plays the total varies with standard deviation ≈ $26. A loss would be about 3.8 standard deviations below the mean - unlikely but not impossible, and the council should know the downside before ordering 400 plays' worth of prizes.
3. The council needs $150 from 400 plays. Evaluate the two proposals - raising the price to $2, or changing the prize - and recommend one with justification.
Read the worked answer
Raise the price to $2, keep the $3 prize. Expected payout stays $0.75, so profit per play becomes 2 − 0.75 = $1.25, and 400 plays yield $500. That overshoots the goal fivefold and would likely suppress attendance: a $2 entry for a 1-in-4 chance at $3 is a poor deal that players may recognize. The council would be trading participation for margin, and the $150 target would be met with far fewer than 400 plays only if people still play.
Adjust the prize. Required profit per play = 150 ÷ 400 = $0.375. With $1 revenue, the expected payout must be $0.625, so the prize P satisfies (1/4)P = 0.625, giving P = $2.50. A $2.50 prize is a natural choice and keeps the entry at the psychologically easier $1.
Recommendation: lower the prize to $2.50, or keep $3 and accept the $100 expectation while planning a second revenue stream. The key reasoning is that expected value is linear in both price and prize, so any (price, prize) pair satisfying price − 0.25 × prize = 0.375 achieves the goal; the choice among them should be made on player appeal and risk, not on arithmetic alone. Whatever is chosen, the council should state the assumption that 400 plays actually occur, since the profit target scales directly with attendance.