The question was asked couple of times here. First this question and second this question.
But unfortunately, I couldn't really understand that logic.
Since the dice are indistinguishable, there are only 41 ways of getting different results in the first throw of three dice. These are
All distinct (20)
(1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (1,3,6), (1,4,5), (1,4,6), (1,5,6), (2,3,4), (2,3,5), (2,3,6), (2,4,5), (2,4,6), (2,5,6), (3,4,5), (3,4,6), (3,5,6), (4,5,6)
Two repetition (30)
(1,2,2), (1,3,3), (1,4,4), (1,5,5), (1,6,6), (2,1,1), (2,3,3), (2,4,4), (2,5,5), (2,6,6), (3,1,1), (3,2,2), (3,4,4), (3,5,5), (3,6,6), (4,1,1), (4,2,2), (4,3,3), (4,5,5), (4,6,6), (5,1,1), (5,2,2), (5,3,3), (5,4,4), (5,6,6), (6,1,1), (6,2,2), (6,3,3), (6,4,4), (6,5,5)
Three repetition (6)
(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)
Thus we got total of 56 distinct ways of arranging the first throw of the dice. Now the second throw should exactly match the first throw. For each of the first throw, this can happen only way. So the probability has to be $\frac{56}{6^6}$.
What am I doing wrong ?
Best Answer
The $20$ all distinct "configurations" correspond to $20 \times 3! = 120$ actually different throws.
The $30$ from your second group correspond to $30 \times 3 = 90$ actually different throws.
(check $120 + 90 + 6 = 216 = 6^3$)
Call the event that their the same configuration in the end $S$, and call the groups I,II,III. If the first throw is from I we have $\frac{6}{216}$ chance to repeat the configuration. If from II, that is $\frac{3}{216}$, and from III, only $\frac{1}{216}$ so that (conditioning on the event that the first throw is in the respective group):
$$P(S) = P(S|I)P(I)+ P(S|II)P(II) + P(S|III)P(III)= \frac{6}{216}\frac{120}{216} + \frac{3}{216}\frac{90}{216} + \frac{1}{216}\frac{6}{216} \simeq .0213$$