Auditing a Raffle's Expected Payout and Fairness Claim
A student council runs a raffle to fund a trip. Each ticket costs $5. There are 400 tickets sold. The prizes are: one grand prize of $300, two second prizes of $100 each, and ten third prizes of $20 each. A council member claims the raffle is 'fair' because the total prize value ($300 + $200 + $200 = $700) is close to the total revenue ($2,000) divided by three. Another member argues fairness should be judged per ticket, not in total. Separately, the council wants to know the probability that a single buyer who purchases 3 tickets wins at least one prize. Tickets are drawn without replacement, and each ticket can win at most one prize. A parent complains that because 13 prizes exist among 400 tickets, the chance of winning is 'about 13 in 400 for each ticket, so buying 3 tickets triples it to about 39 in 400.'
1. Compute the expected net gain per ticket and assess the 'fair' claim.
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Total prize value is $300 + 2($100) + 10($20) = $300 + $200 + $200 = $700. With 400 tickets at $5, revenue is $2,000. Expected prize value per ticket is $700 / 400 = $1.75. Expected net gain per ticket is $1.75 - $5.00 = -$3.25. So a ticket holder loses $3.25 on average; the raffle is strongly favourable to the organiser, not fair. The 'total prize close to revenue divided by three' reasoning is not a valid fairness criterion: fairness in a lottery is defined by expected value per ticket, comparing expected winnings to ticket price. Here the expected return is 35% of the price. The per-ticket view is the correct one, and it shows the raffle is far from fair to buyers, though appropriate as a fundraiser if disclosed.
2. Evaluate the parent's 'triples it' argument and compute the correct probability that a 3-ticket buyer wins at least one prize.
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The parent's reasoning is flawed because the three tickets are not independent: they are drawn without replacement, and the events 'ticket 1 wins' and 'ticket 2 wins' are negatively associated. The correct approach is the complement. The probability a specific ticket wins nothing is 1 - 13/400 = 387/400. For 3 distinct tickets, the probability all three win nothing is (387/400)(386/399)(385/398). Compute: 387/400 = 0.9675; 386/399 ≈ 0.96742; 385/398 ≈ 0.96734. Product ≈ 0.9675 × 0.96742 × 0.96734 ≈ 0.9055. So P(at least one prize) ≈ 1 - 0.9055 = 0.0945, about 9.4%. The naive 'triple' estimate gives 39/400 = 9.75%, so the parent overstates the chance slightly. The exact value is lower because winning with one ticket removes a prize and a ticket from the pool, reducing the chance for the others.