Fun-Fair Game Expected Value
At the school fun fair, a game uses a bag of 10 marbles: 5 red, 3 blue, and 2 green. A player pays $2 to draw one marble, then replaces it. Prizes are $1 for red, $3 for blue, and $5 for green. The organizer expects 200 plays and wants the game to raise money for the library. After 40 plays, the volunteer reports 18 red, 14 blue, and 8 green, and concludes the game is not fair because red came up less often than expected. The organizer asks the class to analyse the prize structure and the early results before changing anything.
1. Calculate the expected payout per play and the organizer's expected profit or loss for 200 plays. Show the probability calculation.
Read the worked answer
Probabilities: red = 5/10 = 0.5; blue = 3/10 = 0.3; green = 2/10 = 0.2. Expected payout = 0.5($1) + 0.3($3) + 0.2($5) = $0.50 + $0.90 + $1.00 = $2.40 per play. Revenue is $2.00 per play, so expected profit per play = $2.00 − $2.40 = −$0.40. For 200 plays, expected profit = 200 × (−$0.40) = −$80. The game is expected to lose $80, not raise money.
2. The organizer wants an expected profit of $0.25 per play while keeping the same prizes. What price should the game charge? Round sensibly and check.
Read the worked answer
Needed revenue per play = expected payout + desired profit = $2.40 + $0.25 = $2.65. Charging $2.65 gives expected profit $2.65 − $2.40 = $0.25 per play, or 200 × $0.25 = $50 over 200 plays. Check: 0.5(1) + 0.3(3) + 0.2(5) = 2.40 payout; 2.65 − 2.40 = 0.25. A charge of $3.00 would give $0.60 profit per play ($120 over 200 plays) but might reduce participation.
3. The volunteer's 40-play results were 18 red, 14 blue, and 8 green. Compare these with expected counts and evaluate the conclusion that the game is unfair because red appeared less often than expected.
Read the worked answer
Expected counts in 40 plays: red = 0.5 × 40 = 20; blue = 0.3 × 40 = 12; green = 0.2 × 40 = 8. Observed: 18 red, 14 blue, 8 green. The differences are −2, +2, and 0, which are small and consistent with ordinary random variation in a sample of only 40 plays. The volunteer's conclusion is not justified. More importantly, the game's expected loss comes from the prize structure, not from a short-run shortage of red marbles, so changing prizes or the price is the reliable fix.