Probability of biased die.
Let 0 < x < 1/6 be a real number. When a certain biased dice is rolled, a particular face F occurs with probability 1/6 − x and and its opposite face occurs with probability 1/6 + x; the other four faces occur with probability 1/6. Recall that opposite faces sum to 7 in any dice. Assume that the probability of obtaining the sum 7 when two such dice are rolled is 13/96. What is the value of x?

Are we assured that the two unfair dice are biased in the same way?  That is...can we assume that the same pair of opposed faces is biased on each die?

Can you clarify your question?  If, say, the $5$ on the first die comes up with probability $\frac 16 +x$, can we assume that this is also true for the second die?  Assuming this to be the case...what's the problem?  Just write out the ways to get a $7$ and compute the probability in terms of $x$, then solve for $x$.  If you meant to assume something else, could you be specific?

@lulu Since the OP says "two such dice are rolled," I take it that they are both biassed in the same way.

Yes they are biased in the same way...the point of interest is that corresponding to a particular probability of appearance of a number on one dice, will there be three{(1/6+x),(1/6-x),1/6} different probability cases for the other dice or not. The computation is getting difficult.

There are $6$ ways to roll $7$ originally and each way involves a pair of opposite faces.  therefore, the probabilities of $4$ of the rolls are unchanged.  The probability of the loaded pair becomes $$2\left(\frac16+x\right)\left(\frac16-x\right)$$ so we have$$\frac19+2\left(\frac16+x\right)\left(\frac16-x\right)={13\over96}$$ 
Not complicated at all, actually.
