Three dice are rolled twice. What is the probability that they show the same numbers (a) If the dice are distinguishable, (b) If they are not.
The answer is (a)$\frac{1}{6^3}$ (b)$\frac{(6×1+90×3+120×6)}{6^6}$
I don't understand why (a) is $\frac{1}{6^3}.$ I think that since the three dice are rolled twice, there are 6 attempts. It should be $\frac{1}{6^6}.$
I think your question is a bit unclear. Do you mean: You throw three dice, what is the probability that on the second throw you see the same numbers as the first throw?
In that case for a) you want to see the exact same number on the same die as the first throw. The result of each die is independent of the others. As each die has $\frac{1}{6}$ chance to have the same number, we find for three dice your answer $\frac{1}{6^3}$.