1. Set up the probabilities

Let the particular face be $i$, so its opposite face is $7-i$.

Given:

• $P(i)=\dfrac{1}{6}-x$
• $P(7-i)=\dfrac{1}{6}+x$
• Each of the remaining $4$ faces has probability $\dfrac{1}{6}$

We roll this same biased die twice.

2. Find the probability that the sum is $7$

For a standard die labeling, the pairs summing to $7$ are: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

So for two independent rolls, $P(\text{sum}=7)=\sum_{k=1}^{6} P(k)P(7-k).$

Now among these 6 ordered pairs:

• Two pairs involve the special opposite faces: $(i,7-i)$ and $(7-i,i)$
• The other 4 pairs involve ordinary faces

Thus, $P(\text{sum}=7)=2\left(\frac16-x\right)\left(\frac16+x\right)+4\left(\frac16\cdot\frac16\right).$

3. Simplify

First, $\left(\frac16-x\right)\left(\frac16+x\right)=\frac{1}{36}-x^2.$

Therefore, $P(\text{sum}=7)=2\left(\frac{1}{36}-x^2\right)+4\cdot\frac{1}{36}.$

$=\frac{2}{36}-2x^2+\frac{4}{36}$

$=\frac{6}{36}-2x^2=\frac16-2x^2.$

Given this equals $\dfrac{13}{96}$, so $\frac16-2x^2=\frac{13}{96}.$

4. Solve for $x$

Since $\frac16=\frac{16}{96},$ we get $\frac{16}{96}-2x^2=\frac{13}{96}$

$2x^2=\frac{3}{96}=\frac{1}{32}$

$x^2=\frac{1}{64}$

$x=\frac18$

Since $0<x<\frac16$, this is valid.

5. Check options

• A: $\frac{1}{16}$
• B: $\frac{1}{8}$ ✅
• C: $\frac{1}{9}$
• D: $\frac{1}{12}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.