This is problem 11 in "A Collection of Dice Problems" by Matthew Conroy. Let state i be the state that i different sides have appeared, then 6 is the state that all sides have appeared, make it the absorbing state and we can solve the problem. My problem is that the transition probabilities I calculated are different from the sollution he gives: this.
For Example, according to the solution, P(1,3)=20/36, but my calculation is P(1,3) = 5/6*5/6=25/36, because the chance of rolling something not yet appeared is 5/6 for both dice and by independence we can multiply them. Is the solution wrong or I'm missing something?
$P(1,3)$ is the probability that both dice come up with a face other than the one that you've already seen and that they do not come up with the same face. If they do come up with the same face, then you transition to state 2 rather than state 3.
To correct your derivation, note that chance of the two dice coming up with the same face, and that it's different than the one you've already seen, is 5/36 (i.e., there are five outcomes out of 36 in which this happens.) And so $$ P(1,3) = \left( \frac{5}{6} \right)^2 - \frac{5}{36} = \frac{20}{36}. $$