How can I win on Lottery?

Simply, you have $\dbinom{56}{6}$ different possibilities for the $\color{#3D99F6}{\textbf{Sample Space}}$ and out of them, only 1 is the correct choice.

Thus the probability of having it right is $\dfrac{1}{\binom{56}{6}}$ .

Hence $S^{-1} = \dbinom{56}{6} = \boxed{32468436}$

Let there be $n$ ways for a "good" selection and $m$ ways for a "bad" selection out of a total of $n+m$ possibilities. Take $N$ samples and let $x_i$ equal $1$ if selection $i$ is successful and $0$ if it is not. Let $x$ be the total number of successful selections,

$x\equiv\sum_{i=1}^Nx_i.$

The probability of $i$ successful selections is then

\eqalign{P(x=i) &= \dfrac{[\#\text{ ways for }i\text{ successes}][\#\text{ ways for }N-i\text{ failures}]}{[\text{total number of ways to select}]} \\ &= \dfrac{\dbinom ni\dbinom m{N-i}}{\dbinom{m+n} N}\\ &= \dfrac{m!n!N!(m+n-N)!}{i!(n-i)!(m+i-N)!(N-i)!(m+n)!}.}

---

In this problem we recognize that we have for "good" $n=6$ , for "bad" $m=50$ , for "drawn" $N=6$ , for successful selections $i=6$ . Therefore according to the formula, the probability is:

\eqalign{P(x=6)&= \dfrac{\dbinom 66\dbinom {50}{6-6}}{\dbinom{50+6} 6}\\ &= \dfrac{50!6!6!(50+6-6)!}{6!(6-6)!(50+6-6)!(6-6)!(50+6)!} \\ &=\dfrac{6!50!}{56!}. }

Which is equal to $1/32\,468\,436$ according to WolframAlpha , therefore the number $\cal S$ we specified is equal to $32\,468\,436$ .