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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?

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  • $\begingroup$ 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? $\endgroup$ – lulu Mar 2 at 14:42
  • $\begingroup$ 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? $\endgroup$ – lulu Mar 2 at 15:00
  • $\begingroup$ @lulu Since the OP says "two such dice are rolled," I take it that they are both biassed in the same way. $\endgroup$ – saulspatz Mar 2 at 15:19
  • $\begingroup$ 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. $\endgroup$ – Aneesh Ghosh Mar 2 at 15:26
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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.

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