Fair Booth: Expected Value and Redesigning the Bag
A school fair booth uses a bag of 10 tiles: 5 red, 3 blue, and 2 green. A player pays $2.50, draws one tile without looking, and receives a prize based on color: $1 for red, $3 for blue, and $6 for green. The tile is returned to the bag after every play, and the booth expects roughly 200 plays during the fair. The organizer wants the booth to raise money for the club rather than lose it, but also does not want the game to look obviously ungenerous, so the prize amounts must stay the same and the bag must keep exactly 10 tiles. Any change is to the number of tiles of each color.
1. Compute the expected prize payout per play and the booth's expected profit or loss per play and over 200 plays.
Read the worked answer
Probabilities: red = 5/10 = 0.50, blue = 3/10 = 0.30, green = 2/10 = 0.20.
Expected payout = 0.50($1) + 0.30($3) + 0.20($6) = $0.50 + $0.90 + $1.20 = $2.60 per play.
Net per play for the booth = $2.50 − $2.60 = −$0.10.
Over 200 plays: 200 × (−$0.10) = −$20 expected.
The booth is expected to lose money. Judgement point: the $2.50 price is above the $1 minimum prize and above the average of the three prize values ($10 ÷ 3 ≈ $3.33 is not the right comparison either) - only the probability-weighted payout answers the question. Also note −$0.10 per play is small enough that a single lucky run of green tiles could make the booth look profitable on the day.
2. Keeping the prizes and the $2.50 price, what is the lowest price in whole cents that gives the booth an expected profit of at least $0.40 per play?
Read the worked answer
Required: price − $2.60 ≥ $0.40 → price ≥ $3.00.
So $3.00 is the lowest whole-cent price meeting the target, giving exactly $0.40 expected profit per play and 200 × $0.40 = $80 expected over the fair.
Check that the price is not being confused with a target based on worst cases: a $3.00 price still pays out $6 on a green tile, so individual plays can lose money - the booth only profits in expectation. If the organizer instead set the price at $2.90, expected profit would be $0.30 per play ($60 over 200 plays), which misses the target even though the booth would look busy.
3. The organizer wants an expected payout of exactly $2.00 with 10 tiles, the same prizes, and at least some tiles of every color (red, blue, and green all present). Is that possible? Justify.
Read the worked answer
Let r, b, g be the counts. Constraints: r + b + g = 10, and (1r + 3b + 6g)/10 = 2.00 → r + 3b + 6g = 20. Each color present means r ≥ 1, b ≥ 1, g ≥ 1.
Substitute r = 10 − b − g: (10 − b − g) + 3b + 6g = 20 → 2b + 5g = 10.
Now test whole numbers with b ≥ 1, g ≥ 1:
- g = 1 → 2b = 5 → b = 2.5 ✘
- g = 2 → 2b = 0 → b = 0 ✘ (no blue)
- g ≥ 3 → 2b becomes negative ✘
No valid solution. The closest options are g = 2, b = 0, r = 8 (payout exactly $2.00 but no blue tiles) or g = 0, b = 5, r = 5 (payout $2.00 but no green tiles). So the organizer must relax one requirement - either allow a missing color or accept a payout near, not equal to, $2.00 (for example g = 1, b = 3, r = 6 gives (6 + 9 + 6)/10 = $2.10, a $0.40 profit at a $2.50 price).