2006 AIME A Problems/Problem 5
Problem
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 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, 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 47/288. Given that the probability of obtaining face is 
where 
and 
are relatively prime positive integers, find 
Solution
For now, assume that face F has a 6 on it and that the face opposite F has a 1 on it. Let A(n) be the probability of rolling a number n on one die and let B(n) be the probability of rolling a number n on the other die. One way of getting a 7 is to get a 2 on die A and a 5 on die B. The probability of this happening is A(2)*B(5)=1/6*1/6=
. Conversely, one can get a 7 by getting a 2 on die B and a 5 on die A, the probability of which is also 
. Getting 7 with a 3 on die A and a 4 on die B also has a probability of , as does getting a 7 with a 4 on die A and a 3 on die B. Subtracting all these probabilities from 
leaves a 
chance of getting a 1 on die A and a 6 on die B or a 6 on die A and a 1 on die B:
A(6)*B(1)+B(6)*A(1)=
Since both die are the same, B(1)=A(1) and B(6)=A(6):
A(6)*A(1)+A(6)*A(1)=
2*A(6)*A(1)=
A(6)*A(1)=
But we know that A(2)=A(3)=A(4)=A(5)=
, so:
A(6)+A(1)=
Now, combine the two equations:
A(1)=-A(6)
A(6)*(-A(6))=
-A(6)^2=
A(6)^2-+=0
A(6)=5/24, 1/8
We know that A(6)>1/6, so it can't be 1/8. Therefore, it has to be 5/24 and the answer is 5+24=29.
