Find the probability that if we roll a die 6 times, we find exactly two numbers repeated twice(e.g. 121234,335422) The probability that we  have 1 number twice is : 2/6
The probability that we have two numbers twice is 2/6*2/6
And the other 2 must be random: 2/6*2/6*1/6*1/6
However the answer is 0.347
How should I proceed to get such answer?
 A: Taking $R$ for repeated numbers and $S$ for single numbers, there are $\binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $\binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15\cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45\cdot 360 = 16200$ options. 
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $\dfrac{16200}{46656}=\dfrac{25}{72} \approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
A: The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled, 
and then $\binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $\frac{6!}{2!}$ ways to arrange those numbers
So the probability one exactly one repeat is 
$$P(1)= \frac{6\binom 54\times \frac {6!}{2!}}{6^6} \approx 0.23$$
For two repeats ...
there are $\binom 62 = 15$ ways to choose the numbers to be doubled, 
and then $\binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $\frac{6!}{2!2! }$ ways to arrange those numbers
SO ...
$$P(2)= \frac{\binom 62 \binom 42\times \frac {6!}{2!2!}}{6^6} \approx 0.347$$
