In a single cast with four dice what is the chance of throwing two doublets? In a single cast with four dice what is the chance of throwing two doublets ?
My working:
Either we will get aabb kind or we will get aaaa kind favourable cases 
Hence required probability is $\frac{{6\choose 2}\frac{4!}{2!2!}+6}{6^4}=\frac{2}{27}$
But answer given is $\frac{1}{12}$
 A: As mentioned in the comments, the $1 \over 12$ answer is wrong by both interpretations of "two doublets".  What leads to this "answer" is the flawed reasoning that:
$$P_{\,{\rm 2-doublets}} = \eqalign{
{\rm (*\,aabb\,*)\;} &(1) (1/6) (1) (1/6)\; + \cr
{\rm (*\,abab\,*)\;} &(1) (1) (1/6) (1/6)\; + \cr
{\rm (*\,abba\,*)\;} &(1) (1) (1/6) (1/6)\;   \cr
} = {1 \over 12}$$
Using a probability of 1 for the first occurrence of b means that b can be the same as a. But in that case, this formula overcounts the occurrence of aaaa as it matches all three cases.
If you want to exclude the case aaaa, you need to use $5 \over 6$ as the probability for the first occurrence of b to make sure it is distinct from a : 
$$P_{\,{\rm 2-doublets}} = \eqalign{
{\rm (*\,aabb\,*)\;} &(1) (1/6) (5/6) (1/6)\; + \cr
{\rm (*\,abab\,*)\;} &(1) (5/6) (1/6) (1/6)\; + \cr
{\rm (*\,abba\,*)\;} &(1) (5/6) (1/6) (1/6)\;   \cr
} = {5 \over 72}$$
If you wanted to include the case aaaa, you would just add the probability of that occurrence to the above formula:
$$P_{\,{\rm 2-doublets(non-distinct)}} = \eqalign{
{\rm (*\,aabb\,*)\;} &(1) (1/6) (5/6) (1/6)\; + \cr
{\rm (*\,abab\,*)\;} &(1) (5/6) (1/6) (1/6)\; + \cr
{\rm (*\,abba\,*)\;} &(1) (5/6) (1/6) (1/6)\; + \cr
{\rm (*\,aaaa\,*)\;} &(1) (1/6) (1/6) (1/6)\;   \cr
} = {2 \over 27}$$
A: Using the counting principle, I get the same answer as the question contributor:
$6+5+4+3+2+1=21$ different pairings of doubles which includes $6$ identical pairs.
There are $\frac{4!}{2!2!}= 6$ ways to get each pairing of different doubles
Hence $P = \frac{15\cdot \frac{4!}{2!2!}+6}{6^4} = \frac{2}{27}$
