2006 AIME II 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 $F$ 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 $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

For now, assume that face $F$ has a 6, so the opposite face has a 1. 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. 7 can be obtained by rolling a $A(n)=2$ and $B(n)=5$ , 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$ , totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$ . Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ 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 $B$ :

$A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}$

Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$

$A(1)\cdot A(6)=\frac{5}{192}$

Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$ , so:

$A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}$

$A(1)+A(6)=\frac{1}{3}$

Combining the equations:

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$

$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$

$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$

$192 A(6)^2 - 64 A(6) + 5 = 0$

$A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}$

$A(6)=\frac{64\pm16}{384}$

$A(6)=\frac{5}{24}, \frac{1}{8}$

We know that $A(6)>\frac{1}{6}$ , so it can't be $\frac{1}{8}$ . Therefore, it has to be $\frac{5}{24}$ and the answer is $5+24=29$ .

Note also that the initial assumption that face $F$ 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.

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2006 AIME II Problems/Problem 5

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