The Spinner Booth: Reading 80 Spins and Rescuing a Prize Budget
The fifth grade runs a coin-spinner booth at the spring fair. The spinner has four unequal sectors: red covers 1/2 of the circle, blue 1/4, green 1/8, and yellow 1/8. A prize is awarded whenever the spinner lands on red. The volunteer in charge expected red "about half the time" and brought only 30 prizes for the planned 80 spins, reasoning that some children would decline a prize. The booth recorded 80 spins: red 34, blue 22, green 15, yellow 9. The volunteer ran out of prizes and blamed bad luck. The class must decide whether the spinner is behaving as designed, whether the shortage was bad luck or bad planning, and then redesign the prize rule so that a $30 budget at $1.50 per prize survives a 200-spin schedule.
1. Compute the expected count for each color in 80 spins, compare with the recorded results, and judge whether the data give strong evidence that the spinner is unfair.
Read the worked answer
Expected counts are each fraction times 80:
- Red: 1/2 × 80 = 40 (observed 34, 6 below)
- Blue: 1/4 × 80 = 20 (observed 22, 2 above)
- Green: 1/8 × 80 = 10 (observed 15, 5 above)
- Yellow: 1/8 × 80 = 10 (observed 9, 1 below)
All four observed counts are within 6 spins of expectation, and 34 out of 80 is 0.425, close to 1/2. Green is the largest proportional gap (15 versus 10), but 5 extra hits in only 80 spins is ordinary variation, not proof of bias.
A stronger test would use far more spins - for example 800, where expected green would be 100 and a gap of 50 would be meaningful - or compare the booth's results with a second spinner of the same design. With 80 spins, the honest conclusion is "consistent with the design, not yet decisive."
2. The volunteer blames bad luck for running out of 30 prizes. Evaluate that claim using expected values, and state how many prizes should have been on hand for 80 spins.
Read the worked answer
For 80 spins, red is expected 1/2 × 80 = 40 times, so 40 prizes were the reasonable plan. Only 30 were brought, which is 10 below the expected need before any luck is considered.
The actual result, 34 reds, was below expectation, so luck actually reduced the demand by about 6 prizes. Even with that helpful luck, 34 > 30, so the booth still ran short.
Conclusion: this is a planning error, not bad luck. A defensible plan states the expected need (40) and adds a small cushion, since in 80 spins the red count will sometimes exceed 40. Blaming randomness cannot explain a shortage that persists even when the random outcome favored the booth.
3. Design a prize rule so that a $30 budget at $1.50 per prize covers a 200-spin schedule. Support your rule with expected-value computations and identify any trade-off.
Read the worked answer
Budget test: $30 ÷ $1.50 = 20 prizes for 200 spins, so the prize probability must be at most 20/200 = 0.1.
The smallest sector is 1/8 = 0.125, so awarding a prize on any single color is expected to give 200 × 1/8 = 25 prizes, costing 25 × $1.50 = $37.50 - $7.50 over budget, with no cushion at all.
Two workable redesigns:
- Keep the $1.50 prize but shorten the booth to 160 spins. Expected prizes = 160 × 1/8 = 20, exactly $30. Trade-off: no cushion, so a slightly high count of that color breaks the budget; 152 spins (19 expected prizes, $28.50) leaves a small margin.
- Keep 200 spins but drop the prize to $1.00. Expected prizes remain 25, costing $25, leaving a $5 cushion. Trade-off: smaller prizes, but the booth lasts the full schedule.
Either way the rule should name the specific color(s) that win, and the class should note that expected value is a planning target, not a guarantee.