$3$ indistinguishable dice are rolled twice. What is the probability that both throws have the same configuration? 
$3$ indistinguishable dice are rolled twice, independently. If each face of each die has the equal probability to appear uppermost, what is the probability that both the throws have the same configuration? 

What is the sample space of this experiment? For distinguishable I know that the sample space has size $6^6$. What will happen for indistinguishable case?
Please help me in this regard.
Thank you very much.
 A: The first throw has $6^3$ outcomes (some are equivalent as to configuration).
There are $\binom{6}{1}=6$ outcomes where all dices are the same, so in the next throw chances are only $1$ in $6^3$ you'll get the same configuration.
There are $\binom{6}{3}3!=120$ outcomes where all dices are distinct, and the next throw has $3!$ outcomes with the same configuration (differently permuted which is ignored/indistinguishable). 
The remaining $6^3-6-120=90$ outcomes have the form $(a,a,b)$ or $(a,b,b)$. In each such case the next throw has $3!/2!=3$ possible outcomes with the same configuration.
The weighted average of the probabilities in each case is
$$\frac{6\dfrac{1}{6^3}+120\dfrac{3!}{6^3}+90\dfrac{3}{6^3}}{6^3}=\frac{83}{3888}\approx 2.135\%$$
A: In each of the two rolls, there will be $\binom 63$ distinct outcomes of three distinct faces, $\binom 62\binom 21$ outcomes of a pair and singleton, and $\binom 61$ outcomes of a tripple.  
That is $56^2$ in total.
But... are these outcomes equally probable?
A: Could anyone please give help me for clarification?
If three indistinguishable dice are rolled for the first time, probability that any configuration will be formed is 1 (for in the question no specific conf is asked). Since the dice are indistinguishable the possible number of different configurations is 56. So that in the next throw, only 1 of the 56 configurations can be formed. The probability should then be 1/56.
If the dice are distinguishable, possible number of different confs are 216. First throw gives probability 1, and second throw gives 1/216, like the same principle above.
What is the error in this solution?
