Rolling two dices n-times. Probabillity of getting doubles

The chance you don't get double $1$ in $n$ rolls is $\left(\frac {35}{36}\right)^n$ so the chance you do get double $1$ is $1-\left(\frac {35}{36}\right)^n$. The naive thing is to say the chance you get all six doubles is then $\left(1-\left(\frac {35}{36}\right)^n\right)^6$ This ignores the correlation introduced because the fact that you got a double $1$ means you have fewer chances to get double $2$. As $n$ gets large this will not be a problem, but clearly for $n=5$ the chance of getting all the doubles is zero.

This is a job for inclusion-exclusion. There are $36^n$ total sets of rolls. Of that $35^n$ do not have double $1$s so subtract $6\cdot 35^n$ for the sets missing each double. You have subtracted the ones missing two pairs of doubles twice, so add them back in. This is ${6 \choose 2}34^n$. Now we have subtracted the ones missing three doubles three times and added them back three times, so subtract them once, which is ${6 \choose 3}33^n$. Now the ones missing four doubles...