probabilitiy: roll die $12$ times get each side twice this is my 3rd question today ^^
Thanks a lot for any help :-)
I need to calculate how probable it is if i roll a die 12 rounds and i get each side twice.
Idea:
Because every side needs to be present twice and we roll the die 12 times:
$$ \left ( \frac{2}{6*2} \right )^{6} = \left ( \frac{2}{12} \right )^{6} = \left ( \frac{1}{6} \right )^{6} $$
Is this correct?
 A: We want $12$ letter words over the alphabet $\{1,2,3,\dotsc,6\}$ with exactly two of each letter. There are 
$$
\binom{12}{2,2,2,2,2,2}
$$
such words out of 
$$
6^{12}
$$
words without any restrictions. Hence the probability is 
$$
\frac{\binom{12}{2,2,2,2,2,2}}{6^{12}}.
$$
A: I'm not quite sure what you are trying to do with your idea but unfortunately, the answer is wrong.
There are probably better approaches but this is a possible one:
There are exactly $\binom{12}{2,2,2,2,2,2}$ possible results to satisfy the given conditions. (Example: One possible result is 112233445566. That is, we roll 1, then 1, then 2, then 2, ...)
That's the key idea! Those are all possibilities to hit each side exactly twice. We have to compute the probability of each of those results and add them.
Luckily, this is easy. The result of each single roll is already fixed. Hence, the probability for any of the aforementioned results is $\left(\frac16 \right)^{12}$, so our answer is
$ \left(\frac16 \right)^{12} \cdot \binom{12}{2,2,2,2,2,2}$.
EDIT: I guess I was sniped, heh.
