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Highest Roll Wins
Problem
In deciding who should pay for lunch, Jane challenges John and James to a game of chance, "I shall take two ordinary 6-sided dice, roll them, and add their scores together. Then one of you shall do the same. If the second total is higher, John pays for lunch, if it is lower, James pays, or if it is same, I will pay. As there are three equally likely outcomes, the game is fair."
Is the game fair?
Solution
We begin by constructing a sample space to list the possible totals.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Now we construct a probability table listing the probabilities of the total being 1, 2, 3, ..., 12:
| Total | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Finally we calculate the probability that the second score is the same as the first score:
| P(same) | = | P(2 AND 2) + P(3 AND 3) + ... P(12 AND 12) |
| = | (1/36)(1/36) + (2/36)(2/36) + ... + (1/36)(1/36) | |
| = | (1/1296)(1 + 4 ... + 1) | |
| = | 146/1296 = 73/648  11.3% |
P(John OR James pays) = 575/648, therefore P(John pays) = P(James pays) = 575/1296 44.4%. Hence the game of chance is definitely not fair!
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