Carnival Token Booth: Expected Payout and a Rule Change
A carnival booth charges $2 per play. A player draws one token at random from a bag holding 6 red, 4 blue, and 10 green tokens, then returns the token to the bag. A red token pays $4, a blue token pays $1, and a green token pays nothing. The operator wants to know whether the game is worth offering, whether one night's results are evidence that something is wrong, and how a proposed rule change would affect the margin. On Saturday the booth took in $1,000 from 500 plays and paid out $640 in prizes. The operator considers a change: replace 5 green tokens with 5 gold tokens that pay $10, keeping the bag at 20 tokens and the price at $2.
1. Compute the expected payout per play and the expected profit per play at the $2 price. What total payout should the operator expect from 500 plays?
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Probabilities: P(red) = 6/20 = 0.30, P(blue) = 4/20 = 0.20, P(green) = 10/20 = 0.50.
Expected payout: E = 0.30(4) + 0.20(1) + 0.50(0) = 1.20 + 0.20 + 0 = $1.40 per play.
Expected profit per play: 2.00 − 1.40 = $0.60.
Expected payout over 500 plays: 500 × 1.40 = $700.
The booth is a favorable game for the operator: $0.60 of every $2 taken in is retained on average. Note that the green tokens must be counted in the probability denominator (20), even though they pay nothing.
2. Saturday's payout was $640 on 500 plays, not the expected $700. Is that a reason to suspect the bag was set up wrong? Support your answer with numbers.
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Realized result: 500 × $2 = $1,000 taken in, $640 paid out, profit $360, or $0.72 per play. That is better for the operator than the $0.60 expected, not worse.
Is a $60 shortfall unusual? The per-play payout has variance: E[X²] = 0.30(16) + 0.20(1) + 0.50(0) = 4.80 + 0.20 = 5.00, so Var = 5.00 − 1.40² = 5.00 − 1.96 = 3.04, and the standard deviation is √3.04 ≈ $1.74 per play. Over 500 plays the standard deviation of the total is about 1.74 × √500 ≈ 1.74 × 22.4 ≈ $39.
A $60 gap is roughly 1.5 standard deviations - an ordinary fluctuation for one night.
Concrete check: $640 could come from 140 red and 80 blue (140 × 4 + 80 = 640), 10 fewer reds and 20 fewer blues than the 150 and 100 expected. That is well within normal sampling variability, so the bag is not implicated.
3. If 5 green tokens are replaced by 5 gold tokens paying $10, what is the new expected payout, and what minimum price should the operator charge to keep at least the current $0.60 profit margin? What happens if the price stays at $2?
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New bag (still 20 tokens): 6 red, 4 blue, 5 green, 5 gold. Probabilities: red 0.30, blue 0.20, green 0.25, gold 0.25.
New expected payout: E = 0.30(4) + 0.20(1) + 0.25(10) + 0.25(0) = 1.20 + 0.20 + 2.50 + 0 = $3.90 per play.
Price needed for a $0.60 margin: 3.90 + 0.60 = $4.50 per play.
At $2: the booth would lose 2.00 − 3.90 = $1.90 on every play - roughly $950 lost over 500 plays.
The judgment: the gold tokens raise expected payout by $2.50 per play while the price is unchanged, so the change is only viable with a price more than doubled. The operator's instinct that "a bigger prize brings more players" may be right about traffic, but at $2 the added traffic multiplies a loss, not a profit.