Two Dice and a Fair Game
Two fair six-sided dice are rolled, and the sum S is recorded. A game charges 1 € to play: the player wins 5 € if S = 7, wins 2 € if S = 11, and wins nothing otherwise. A player argues that because 7 is the most likely sum, the game must be in the player's favour. The organiser wants to check the expected value per game and decide whether the entry fee should change. Separately, the organiser wonders how many distinct ordered outcomes (a, b) give a sum of 7 versus a sum of 8.
1. List the number of ordered outcomes giving S = 7 and S = 8, and state each probability.
Read the worked answer
Total ordered outcomes = 36. S = 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes, P = 6/36 = 1/6. S = 8: (2,6),(3,5),(4,4),(5,3),(6,2) → 5 outcomes, P = 5/36. So 7 is more likely than 8, but only 6 of 36 outcomes win anything on 7.
2. Compute the expected net gain per game (winnings minus the 1 € fee).
Read the worked answer
P(S=7) = 6/36 = 1/6; P(S=11) = 2/36 = 1/18 (outcomes (5,6),(6,5)). Expected winnings = 5·(1/6) + 2·(1/18) = 5/6 + 2/18 = 5/6 + 1/9. Common denominator 18: 15/18 + 2/18 = 17/18 ≈ 0.944 €. Net gain = 17/18 − 1 = −1/18 ≈ −0.056 € per game. The game favours the organiser, not the player.
3. The player's argument is that 7 is the most likely sum, so the game must favour the player. Refute this using the expected value.
Read the worked answer
Being the most likely single sum does not mean the payoff is favourable. Even when 7 appears, the player wins 5 € but paid 1 €, and 7 occurs only 1/6 of the time. Most rolls (36 − 6 − 2 = 28 outcomes) win nothing. The expected net is −1/18 € per game, a loss for the player. The argument confuses 'most likely outcome' with 'positive expected value'.