I have two bags of 6 dice each. In one bag (bag 1), all dice are fair. In the other (bag 2), each die is weighted so as to favour a different side by probability $\epsilon$, i.e., one of the dice rolls '1' with probability $\frac{1}{6}$ + $\epsilon$ and any other number with probability $\frac{1}{6} - \frac{\epsilon}{5}$; a second die rolls '2' with probability $\frac{1}{6}$ + $\epsilon$ and any other number with probability $\frac{1}{6} - \frac{\epsilon}{5}$, etc.
Say I choose a bag at random, take out a single die out of the bag and roll it six times over. I then report the sum S of the numbers obtained. Assuming I know $\epsilon$, how can one perform an informed guess of which bag I have chosen?
Current thoughts: I presume that to solve this problem, one needs to compute the probabilities $P(bag=1 \vert sum = S)$, $P(bag=2 \vert sum = S)$. I have determined that I can use Bayes rule, and that given bag 1 (fair dice), the probability of reporting $sum=S$, $p(sum=S \vert bag=1)$ is determined by the coefficient of $x^S$ in the polynomial ($x^1 + x^2 + x^3 +x^4 +x^5 + x^6)^6$ over $6^6$. However, I dont know how to approach the problem for bag 2. Isn't the expected total of the sum for this bag the same than for bag 1 (as we take into account the probabilities of choosing each loaded die in the first place)?