Is the Dice Game Unfair?
Ms. Rivera's fifth graders play a two-cube game: Player A scores a point when the sum is 6, Player B scores when the sum is 7, and other sums score nothing. After 60 rolls the tally reads B 11, A 8. Player A complains the game is rigged, and a classmate adds, "There are 11 possible sums from 2 to 12, so every sum has a 1-in-11 chance - 6 and 7 should be equally likely." Ms. Rivera has two numbered cubes per pair, a results table on the board, and 30 minutes. She wants the class to reason from the full sample space and to decide whether the observed lead actually needs explaining.
1. Build the 36-outcome sample space for two number cubes, compute the probability of each target sum, and identify exactly what is wrong with the classmate's 1-in-11 claim.
Read the worked answer
Two cubes give 6 × 6 = 36 equally likely ordered pairs.
Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes → 5/36 ≈ 0.139. Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes → 6/36 = 1/6 ≈ 0.167.
So B has a genuine edge: 6/36 versus 5/36, a ratio of 6 to 5.
The classmate has correctly counted 11 possible sums but has assumed the 11 sums are equally likely. They are not, because each sum corresponds to a different number of ordered pairs - sum 2 has one (1,1), sum 7 has six, sum 12 has one. The counting must be done on outcomes, not on sums. Sums 6 and 7 are adjacent and both central, which is why the difference of a single outcome is easy to overlook: 7 is the only sum reachable in six ways, and every sum is mirrored around it.
2. Given 60 rolls and the observed tally of 8 for A and 11 for B, does the data support Player A's claim that the game is rigged? Support your conclusion with expected counts and a rough measure of typical variation.
Read the worked answer
Expected counts under the true probabilities:
- A: 60 × 5/36 ≈ 8.3 points
- B: 60 × 6/36 = 10 points
Observed 8 and 11 are essentially on top of those expectations - a difference of 0.3 and 1 point respectively.
Rough variation: for B's count, treat it as binomial with n = 60 and p = 1/6. Standard deviation ≈ √(60 × 1/6 × 5/6) = √8.33 ≈ 2.9. B's observed 11 is about one-third of a standard deviation above the mean - well within ordinary sampling noise. For A, sd ≈ √(60 × 5/36 × 31/36) ≈ 2.7, and 8 versus 8.3 is negligible.
Conclusion: the data are consistent with a fair pair of cubes and the small 6-versus-5 edge. The lead of 3 points in 60 rolls is not evidence of rigging. Two instructional points follow: (1) the game is unfair, but because of the probabilities, not because of the dice; (2) a small sample cannot detect a 5/36 versus 6/36 difference, so confirming the edge requires many more trials - which is itself a worthwhile discussion about why 60 rolls is not enough.
3. Redesign the game so each player has an equal chance of scoring, and describe how the class could test the redesign.
Read the worked answer
Redesign option 1 (matched probabilities): A scores on a sum of 6, B scores on a sum of 8. Both sums occur in exactly 5 of the 36 outcomes, so each player's probability of scoring on a roll is 5/36 and the game is perfectly balanced.
Redesign option 2 (guaranteed scoring, fair coin): A scores on even sums, B on odd sums. Even sums account for 18 outcomes and odd sums for 18, so each player scores with probability 1/2 on every roll - the cleanest possible design.
Redesign option 3 (reroll): keep 6 for A and 7 for B but declare 7 a reroll for both players; then each scores only on 6, making the game pointless - a good contrast showing that fairness requires both balance and comparable scoring rates.
Testing: have each pair roll 200 times, record counts, and compare with predicted values (option 2 predicts about 100 each). Better still, pool the class data - 30 pairs × 200 rolls gives 6,000 trials, where the predicted 50/50 split has a standard deviation of about 39, so a genuinely unbalanced rule would show up. Ask students to state in advance what result would make them doubt the design; this turns the simulation into a test rather than a demonstration.