Fair Game? Expected Value and a 40-Play Sample
Ms. Alvarez's seventh-grade class designed a carnival game for family night. A spinner has 8 equal sectors: 1 sector reads 'Win $3,' 2 sectors read 'Win $2,' and 5 sectors read 'Try again.' A play costs $1. After 40 plays at the booth, the class treasurer reports a net gain of $6 for the players and argues the game is losing money for the school. During discussion, Devon says, 'It can't be unfair - I played six times and won twice.' Priya answers, 'With eight sectors, you win only three times out of eight, so the game is rigged.' Ms. Alvarez wants the class to settle the question with mathematics rather than impressions, and she has one class period before the booth runs again at the spring fair. She also needs to decide whether to change the prize structure or the price.
1. Compute the expected net value of one play to a player, showing the full distribution. State whether the game favors the school or the player, and identify the most likely computational slip.
Read the worked answer
Work with net payoffs (prize minus the $1 fee), because the fee is paid on every play:
| Outcome | Net to player | Probability |
|---|---|---|
| Win $3 | +2 | 1/8 |
| Win $2 | +1 | 2/8 |
| Try again | -1 | 5/8 |
E = 2(1/8) + 1(2/8) + (-1)(5/8) = 0.25 + 0.25 - 0.625 = -$0.125 per play.
The player loses 12.5 cents per play on average, so the school gains about $0.125 per play, or $5 per 40 plays. The most common slip is weighting the gross prizes ($3 and $2) and forgetting that the $1 fee applies even to losing plays; that error gives 3(1/8) + 2(2/8) = 0.875 and makes the game look far friendlier than it is. A second slip is treating the five 'Try again' sectors as costing nothing.
2. Evaluate Devon's and Priya's arguments, then judge whether the treasurer's 40-play result contradicts the expected value. Include a rough measure of how surprising the result is.
Read the worked answer
Devon argues from 6 plays: 2 wins in 6 is a frequency of 0.33 against a true win probability of 3/8 = 0.375. That is entirely ordinary sampling variation, and 6 trials cannot detect a 12.5-cent edge. His claim also ignores which prize he won.
Priya confuses win probability with expected value. A game can be won 3/8 of the time and still favor the house, or be won only 1/8 of the time and still favor the player, because value depends on payout sizes, not just on how often you win.
Treasurer. Expected player net over 40 plays = 40(-0.125) = -$5. Observed player net = +$6, a deviation of +$11. Per-play spread: E[X²] = 4(1/8) + 1(2/8) + 1(5/8) = 1.375, so Var = 1.375 - 0.015625 ≈ 1.359 and SD ≈ 1.17. The SD of a 40-play sum is about 1.17√40 ≈ $7.40. A deviation of $11 is about 1.5 standard deviations - roughly a 1-in-7 event, not evidence the game is unfair. Detecting a 12.5-cent edge reliably would need many hundreds of plays.
3. Recommend a specific change that makes the game exactly fair, justifying it with expected value. Give one reason to prefer that change over raising the price.
Read the worked answer
Fairness requires expected net value 0, i.e. total prizes paid out must average the $1 fee per play. Over 8 plays the booth collects $8, so the weighted prize total must be $8:
1(g₁) + 2(g₂) = 8, where g₁ and g₂ are gross prizes.
Currently 3 + 2(2) = 7, which is $1 short - exactly the 12.5-cent-per-play edge (8 plays × $0.125 = $1).
Recommendation: raise the top prize from $3 to $4. Then 4 + 2(2) = 8, so E = (1/8)(3) + (2/8)(1) + (5/8)(-1) = 0.375 + 0.25 - 0.625 = 0. A fair game.
Alternatives: lower the $2 prize to $2.50 (awkward to pay), or set the price at $0.875 (unusable coins). Changing the top prize keeps the price at a clean $1 and leaves the spinner and payout routine untouched, so students can run the booth without making change in odd amounts. Raising the price would also work but makes the arithmetic messy and obscures the conservation argument students just built.