Fair or Fundraiser? Expected Value at the Dice Booth
The student council is running a dice game at the spring fair. A player pays $3 to roll two fair six-sided dice. The game pays $10 if the sum is 7, pays $5 if the sum is 11, and pays nothing for any other sum. The principal will allow the booth only if the expected net loss to a player is no more than $1.25 per play; otherwise she considers the game exploitative. The council needs to raise at least $0.75 per play on average to fund a field trip. One member suggests raising the payout for a sum of 7 to $16, arguing that 7 is the most likely sum and therefore deserves a bigger prize. Another member worries that changing the prize will make the game either too generous or not profitable enough.
1. Compute the expected payout and expected net gain per play. Does the game satisfy the principal's limit of no more than $1.25 expected loss per player?
Read the worked answer
There are 36 equally likely outcomes. A sum of 7 occurs in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). A sum of 11 occurs in 2 ways: (5,6) and (6,5). The expected payout is
(6/36)(10) + (2/36)(5) = 60/36 + 10/36 = 70/36 ≈ 1.944 dollars.
The player pays $3, so the expected net gain is 1.944 − 3 ≈ −1.056, an expected loss of about $1.06 per play. Since $1.06 < $1.25, the game satisfies the principal's limit, though not by a large margin.
2. The council wants to raise at least $0.75 per play. If the sum-of-11 prize stays at $5, find the whole-dollar prize for a sum of 7 that meets this goal while keeping the expected loss at or below $1.25. Show the inequality.
Read the worked answer
Let x be the prize for a sum of 7. Expected payout = (6x + 2·5)/36 = (6x + 10)/36. The council's average gain per play is 3 − (6x + 10)/36. For the council to raise at least $0.75, we need
3 − (6x + 10)/36 ≥ 0.75, so (6x + 10)/36 ≤ 2.25, 6x + 10 ≤ 81, x ≤ 71/6 ≈ 11.83.
For the principal's limit, the player's expected loss must be at most $1.25: 3 − payout ≤ 1.25, so payout ≥ 1.75. Thus (6x + 10)/36 ≥ 1.75, 6x + 10 ≥ 63, x ≥ 53/6 ≈ 8.83. Whole-dollar prizes from $9 to $11 satisfy both conditions. The largest such prize, $11, gives payout (66 + 10)/36 = 76/36 ≈ 2.111 and council gain 3 − 2.111 ≈ 0.889, about $0.89 per play. The proposed $16 is too high: payout = (96 + 10)/36 ≈ 2.944 and council gain is only about $0.056 per play.
3. Given that a player's roll includes at least one 3, what is the probability that the sum is 7? Explain why this is not the same as the unconditional probability of a sum of 7.
Read the worked answer
List the 11 outcomes with at least one 3: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3). Of these, exactly two have sum 7: (3,4) and (4,3). So the conditional probability is 2/11 ≈ 0.182. The unconditional probability is 6/36 = 1/6 ≈ 0.167. Conditioning on 'at least one 3' changes the sample space from all 36 rolls to those 11 rolls; it is not simply the number of 7s divided by 36, because the given information rules out most outcomes and slightly raises the chance of a 7 relative to the reduced set.