Carnival Spinners: Expected Winners and a Suspicious Result
The student council at Jefferson Middle School is designing a carnival booth. Each player pays $2.00 to spin two spinners. Spinner A has four equal sections labeled 1, 2, 3, and 4. Spinner B has three equal sections, two of them red and one blue. A player wins a prize only when Spinner A lands on an even number and Spinner B lands on red. Each prize costs the council $1.25, and last year's booth drew about 300 plays, which the council expects again. On the first afternoon the booth ran 60 plays and 28 players won, which led one council member to say that the game must be rigged because far too many players were winning. The council meets Monday to decide whether to keep the game, and the faculty advisor wants an expected profit figure for 300 plays plus a judgment on the rigged claim.
1. What is the probability that a single player wins? Support the answer with a count of equally likely outcomes.
Read the worked answer
The spinners are independent, so multiply the probabilities. On Spinner A, even results are 2 and 4 out of 4 equal sections: 1/2. On Spinner B, red is 2 of 3 equal sections: 2/3.
P(win) = 1/2 × 2/3 = 2/6 = 1/3.
Confirming by listing outcomes: Spinner A has 4 results and Spinner B has 3, so there are 12 equally likely pairs. The winning pairs are (2, red), (2, red), (4, red), (4, red) - 4 of the 12, which is 1/3, matching the multiplication.
2. Compute the council's expected profit for 300 plays, and state the expected profit per play.
Read the worked answer
Expected wins = 1/3 × 300 = 100.
Prize cost = 100 × $1.25 = $125. Revenue = 300 × $2.00 = $600.
Expected profit = 600 − 125 = $475, which is $475 ÷ 300 ≈ $1.58 per play.
This is an expected value, not a guarantee. Each additional winner beyond 100 costs $1.25, so profit falls as wins rise, but the council would need roughly 480 winners out of 300 plays to break even - nowhere near the likely range - so the booth is a safe fundraiser as designed.
3. In the first 60 plays, 28 players won. Decide whether this is strong evidence that the game is rigged, and describe the record-keeping that would settle the question.
Read the worked answer
Expected wins in 60 plays = 1/3 × 60 = 20. The booth saw 28, which is 8 more than expected.
With only 60 plays, random variation is large: a short run of reds and even numbers can easily push the total several wins above or below 20, so 28 is not strong evidence that the spinners are weighted. A sample that small cannot distinguish an unusual streak from a biased spinner.
To settle it, record every play, including both spinner results, and compare a much larger total against the model. Over 600 plays, about 200 wins are expected; a total near 280 would be genuinely suspicious, while a total near 200 would support the 1/3 model. Keep the current game running while the data accumulate.