what is probability of throwing $n$ dice $m$ times to get at least one 6 for each dice We have $n$, 6-face fair dice. At each time $n$ dice are thrown independently. I want to calculate the probability of number of times we should throw dices before having at least one 6 from each of the dice.
We know that for one die to have one 6, the probability of $k$ failures obeys a Geometric Distribution with $p=1/6$ and expectation of 6. 
How should I extend this to $n$ dices?

Label the dice $1$ to $n$. Let $X_1$ be the number of throws of the first die until we get a $6$. Define random variables $X_2, X_3,\dots, X_n$ analogously.
Let $Y=\max(X_1,X_2,\dots,X_n)$. Your question asks for an expression for $\Pr(Y=y)$.
It is easier to find $\Pr(Y\le y)$. That will do it, because 
$$\Pr(Y=y)=\Pr(Y\le y)-\Pr(Y\le y-1).\tag{$1$}$$
Now we go after $\Pr(Y\le y)$. We have $Y\le y$ iff we have at least one $6$ on all the dice by time $y$.
The probability that there was at least one $6$ on die $1$ by time $y$ is $1$ minus the probability of no $6$, so it is
$$1-\left(\frac{5}{6}\right)^y.$$
For $n$ dice, take the $n$-th power. We get
$$\Pr(Y\le y)= \left(1-\left(\frac{5}{6}\right)^y\right)^n.$$ 
Now use $(1)$.  
