We may pick $\binom{50}{6}$ different sets of tickets. Any set of tickets without consecutive tickets can be mapped bijectively into a septuple $(a,b,c,d,e,f,g)$ such that $a,g\in\mathbb{N}$, $b,c,d,e,f\in\mathbb{N}^+$ and $a+b+c+d+e+f+g = 44$. By stars and bars, the number of sets of tickets without consecutive tickets equals the number of ways of writing $46$ as the sum of $7$ positive integers, i.e. $\binom{45}{6}$. It follows that the wanted probability is:
$$ \frac{\binom{45}{6}}{\binom{50}{6}}=\frac{45\cdot 44\cdot 43\cdot 42\cdot 41\cdot 40}{50\cdot 49\cdot 48\cdot 47\cdot 46\cdot 45}=\frac{19393}{37835}\approx \color{red}{51,257\%}.$$