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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.

+123456
1234567
2345678
3456789
45678910
567891011
6789101112

Now we construct a probability table listing the probabilities of the total being 1, 2, 3, ..., 12:

Total23456789101112
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 approximately 11.3%

P(John OR James pays) = 575/648, therefore P(John pays) = P(James pays) = 575/1296 approximately 44.4%. Hence the game of chance is definitely not fair!

Problem ID: 158 (Mar 2004)     Difficulty: 3 Star

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