The dice game "Crazy 7" is about rolling the eye sum 7 with two dice. Since one of the players doesn't have much luck with the dice, he tries to help it along. To do this, he has faked two dice by using cavities and weights inside them. The dice both have the numbers 1 to 6, and the counterfeiting has increased the probability of rolling a 1 on the first die to 1/5. The probabilities for the numbers from 2 to 6 are the same. For the second dice, the probability of rolling a 6 has increased to 1/5. The probabilities for the remaining numbers from 1 to 5 are also equal.
How much higher is the probability for the cheater to achieve the eye sum 7 by rolling his two rigged dice together? Is the forgery worth it?
In a roll with two normal dice, 6 · 6 = 36 different pairs of eyes can be rolled. The six pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) result in the eye sum 7. Since the probability for each pair of eyes is the same for normal dice, the probability to roll the eye sum 7 is 6/36 = 1/6.
However, for the loaded dice, the probability of rolling a 1 with the first die or a 6 with the second die is 1/5. Therefore, for any other pair of eyes, it is (1 – 1/5)/5 = 4/25. The pair of eyes (1, 6) is consequently reached with both dice with a probability of 1/5 · 1/5 = 1/25, while each of the other five pairs of eyes that result in a 7 is rolled only with a probability of 4/25 · 4/25 = 16/625.
In total, therefore, the probability of rolling a 7 is 1/25 + 5 · 16/625 = 21/125. The probability of rolling the eye sum 7 is therefore increased by the forgery by only 21/125 – 1/6 = 1/750. It is therefore hardly worthwhile for the cheater.
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