There are $\binom{55}6$ total draws, and we need to count how many draws have two balls with the same number. This is best done with inclusion-exclusion.
There are $\binom{53}4$ ways to draw both $1$s (we have to choose the two $1$s and $4$ of the other $53$ balls), and the same number of ways to draw both $2$s, etc. Simply adding these up would count all the draws which have (say) both $1$s **and** both $2$s twice, so we need to subtract those.
There are $\binom{51}2$ draws which have both $1$s and both $2$s, and the same number for every other pair of choices from $\\{1,2,3,4,5,6\\}$. Once we subtract all those off, we need to add the draws which have three pairs; there are $\binom63=20$ of these.
So the final answer is $$\frac{\binom61\binom{53}4-\binom62\binom{51}2+\binom63}{\binom{55}6}\approx 0.06.$$