The fairground spinner: expected value and redesigning the prizes
A school fair stall has a spinner with 8 equal sectors. One sector is marked 'P10', two sectors are marked 'P2', and the remaining five sectors are marked 'No prize'. A player pays P3 to spin once. The stall expects about 400 plays during the fair. After 60 plays the stall has paid out P116 in prizes and the organiser begins to worry that the prizes are 'too generous'. A helper says the game is definitely fair to the stall because 'one in eight is P10, so most people lose'. The organisers want to keep the P3 charge, keep the single P10 sector, and keep the game looking attractive, but they want the stall to make exactly P0.50 profit per play on average.
1. Calculate the expected payout per spin, the stall's expected profit per spin, and the expected total profit over 400 plays. Comment on the helper's reasoning.
Read the worked answer
Expected payout = sum of (probability x prize):
- P10 sector: (1/8) x 10 = 1.25
- Two P2 sectors: (2/8) x 2 = 0.50
- Five no-prize sectors: 0
Expected payout = P1.75 per spin.
Expected profit per spin = 3.00 - 1.75 = P1.25.
Over 400 plays: 400 x 1.25 = P500 expected profit, with about 400 x 1.75 = P700 paid out in prizes.
The helper has only counted the chance of winning, not the size of the prizes. It is true that five of eight sectors lose, but a player still gets back P1.75 for every P3 spent - a return of about 58%. 'Most people lose' and 'the stall is safe' are not the same statement; the arithmetic of expected value is what decides it.
2. After 60 plays the stall has paid out P116. Is this good evidence that the prizes are too generous or the spinner unfair? Support your answer with a calculation.
Read the worked answer
Expected payout over 60 plays = 60 x 1.75 = P105.
The stall actually paid P116, which is P11 more - about 10.5% above expectation.
A payout of exactly P116 arises, for example, from 8 spins landing on P10 (P80) and 18 landing on P2 (P36), with the other 34 spins winning nothing. The expected counts are 60 x 1/8 = 7.5 for P10 and 60 x 2/8 = 15 for P2, so observed counts of 8 and 18 are ordinary small-sample variation.
Conclusion: no evidence of a biased spinner or over-generous prizes. A run of 60 spins is far too short to detect anything except a very large bias; the long-run figure of P1.25 profit per spin is what matters over the 400 plays expected. Reacting to P11 on a sample of 60 would be over-reacting to noise.
3. Keeping the P3 charge and the single P10 sector, what should each of the two small-prize sectors pay so that the stall's expected profit is exactly P0.50 per spin? Check your answer.
Read the worked answer
Let each of the two small sectors pay x pula. For a P0.50 profit the expected payout must be 3.00 - 0.50 = P2.50.
(1/8)(10) + (2/8)(x) + (5/8)(0) = 2.50
1.25 + 0.25x = 2.50
0.25x = 1.25
x = 5
So each of the two sectors should pay P5.
Check: (1/8)(10) + (2/8)(5) = 1.25 + 1.25 = P2.50 expected payout, leaving P0.50 profit per spin; over 400 plays that is P200 profit.
Comment: the number of winning sectors is unchanged (still three of eight), so the game looks just as attractive, and a P5 prize is now larger than the P3 stake, which players will notice. Whether the stall makes money is decided by expected value, not by how generous a single prize looks.