Fete game: expected takings before the day
The Year 6 student council is running a dice game at the school fete. Each play costs $2.00. The player rolls one standard six-sided die: a 5 or a 6 wins a $3.00 voucher, and any other roll earns a sticker that costs the council $0.20. Based on last year's gate, the council expects about 280 plays across the day. The council needs to clear at least $250 to pay for new library cushions, and the principal will not allow the price to be above $3.00 per play or the prize to be withdrawn once the fete opens.
1. Calculate the expected profit per play and the expected total profit from 280 plays, and state whether the $250 target is met.
Read the worked answer
Prizes per play
- P(win) = 2/6 = 1/3; P(lose) = 4/6 = 2/3
- Expected voucher cost: 1/3 × $3.00 = $1.00
- Expected sticker cost: 2/3 × $0.20 = $0.133... ≈ $0.13
- Expected payout: $1.133... ≈ $1.13
Profit
- Per play: $2.00 − $1.133... = $0.866... ≈ $0.87
- 280 plays: 280 × $0.866... = $242.67
This is about $7 short of the $250 target, so the plan as written does not guarantee the cushions.
2. Suggest a single change that keeps the game within the principal's rules and reliably clears $250, showing the arithmetic.
Read the worked answer
Two workable single changes:
1. Raise the price to $2.50 (allowed, since the cap is $3.00)
- Profit per play: $2.50 − $1.133... = $1.366...
- 280 plays: 280 × $1.366... = $382.67 - clears the target with a large margin.
2. Keep the price at $2.00 but pay the voucher only on a 6
- Payout: 1/6 × $3.00 + 5/6 × $0.20 = $0.50 + $0.166... = $0.666...
- Profit per play: $2.00 − $0.666... = $1.333...
- 280 plays: $373.33 - also clears the target.
Either works; the price rise is simpler to explain and keeps the game generous. Note the expected profit is a long-run average: 280 plays is enough for the result to land close to these figures, but not exact.
3. In the first hour, 30 students play and 9 win. A parent says the die must be loaded because fewer than a third won. Evaluate this claim.
Read the worked answer
- Expected wins in 30 plays: 30 × 1/3 = 10
- Observed: 9, one below expectation.
That is an entirely ordinary result. For 30 plays with p = 1/3, the typical spread around the mean is about √(30 × 1/3 × 2/3) ≈ 2.6 wins, so results from roughly 7 to 13 wins are unsurprising. A single win fewer than expected is well inside normal chance variation and gives no evidence that the die is biased.
Judgement: the parent is confusing 'less than expected' with 'unfair'. Note also that a bias claim would need far more than 30 rolls - several hundred - and each face should be counted, not just win/lose outcomes, since 5 and 6 are combined here.