Fair game expected value and player target
A sixth-form class plans a fundraising game at the school fair. A player pays $5 and draws one token without looking from a bag containing 10 tokens: 6 tokens are blank, 3 tokens win a $4 snack voucher, and 1 token wins a $15 book voucher. The class must pay $60 to rent the game stall and $0.80 per player for insurance. Tokens are replaced after each draw. The organiser expects 120 players but is unsure; the class needs at least $250 profit to fund a trip. A student suggests changing the game to two draws for $8, but the second draw would be without replacement; another suggests keeping one draw but lowering the top prize to a $10 book voucher. The teacher wants a decision based on expected value and risk, not just on the most likely outcome.
1. For the original $5 game, calculate the expected prize per play, the expected net income per player after insurance, and the expected profit from 120 players after the stall rent. Determine how many players are needed to reach the $250 target.
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Expected prize per play = (6/10)(0)+(3/10)(4)+(1/10)(15)=0+1.20+1.50=$2.70. Net per player after insurance =5.00-2.70-0.80=$1.50. For 120 players: contribution=120×1.50=$180; after stall rent $60, profit=$120. This is short of $250 by $130. Required players: (250+60)/1.50=206.67, so at least 207 players are needed.
2. Evaluate the two-draw $8 version. Assume the player pays $8, insurance is still $0.80 per player, and the two tokens are drawn without replacement. Calculate the expected prize and expected profit at 120 players, and the player count needed for the target. Compare with the original.
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Two draws without replacement: by linearity, the expected total prize is 2×2.70=$5.40. Each draw has the same marginal expected prize, even though the draws are dependent. Net per player =8.00-5.40-0.80=$1.80. At 120 players: profit=120×1.80-60=$156, which is $36 better than the original but still below $250. Required players: 310/1.80=172.22, so 173 players. Comparison: the two-draw game needs 173 players for the target, versus 207 for the original, and gives a larger profit at the expected attendance of 120.
3. Evaluate the variant with one draw, $5 price, and a $10 book voucher instead of $15. Calculate the expected profit at 120 players and the break-even number of players. Recommend between this and the two-draw variant, considering both target and appeal.
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$10 book voucher variant: expected prize =(3/10)(4)+(1/10)(10)=1.20+1.00=$2.20. Net per player =5.00-2.20-0.80=$2.00. At 120 players: profit=120×2.00-60=$180. Required players:310/2.00=155.
This variant has the highest expected profit per player and the lowest player target. If the organiser is confident of at least 155 players, it is the best of the three on expected value. The two-draw version has a higher prize appeal but needs 173 players and yields only $156 at 120 players. The original is weakest, needing 207 players. Risk: lowering the top prize could reduce attendance; if expected attendance falls below 155, the target is not met. Recommendation: choose the $10-voucher single-draw game if attendance is expected to stay near 120; otherwise use the two-draw game only if at least 173 players are likely.