Probability of at least 3 sixes with 18 dice I am trying to logically think about this, not to just apply formulas that are readily available for similar problems. Can anyone help explain the math behind finding the probability of getting at least 3 sixes when rolling 18 (six sided) dice?

The probability of not rolling any six is given by $\left(\frac{5}{6}\right)^{18}$, the probability of getting exactly one six is given by $18\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{17}$ and the probability of getting exactly two sixes is given by $\binom{18}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{16}$, so the probability of getting three or more sixes is given by: $$ 1 - \left(\frac{5}{6}\right)^{18} - 18\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{17} - \binom{18}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{16} \approx \color{red}{59,73\%}.$$