2006 AIME II Problems/Problem 5
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face is greater than , the probability of obtaining the face opposite is less than , the probability of obtaining any one of the other four faces is , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is . Given that the probability of obtaining face is where and are relatively prime positive integers, find
Without loss of generality, assume that face has a 6, so the opposite face has a 1. Let be the probability of rolling a number on one die and let be the probability of rolling a number on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of , totaling . Subtracting all these probabilities from leaves chance of getting a 1 on die and a 6 on die or a 6 on die and a 1 on die :
Since the two dice are identical, and so
Also, we know that and that the total probability must be , so:
Combining the equations:
We know that , so it can't be . Therefore, the probability is and the answer is .
Note also that the initial assumption that face was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was. We could have labelled as , for example, and replaced the others with variables too, but the notation would have been harder to follow.
We have that the cube probabilities to land on its faces are , , , , , we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: multiplying by 288 we get: dividing by 16 and rearranging we get: so the probability F which is greater than is equal
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