The Chevalier's double sixes

Idea. “At least one” is easiest through its complement: the probability of no double six in $n$ rolls is ${\left(\frac{35}{36}\right)}^n$, and we need it below $\frac{1}{2}$.

Step 1. We need $$n>\frac{\ln 2}{-\ln (35/36)}\approx \frac{0.6931}{0.02817}\approx \mathrm{24.6.}$$

Step 2: check. $$\begin{array}{rl}1-{\left(\frac{35}{36}\right)}^{24}& \approx 0.491,\\ 1-{\left(\frac{35}{36}\right)}^{25}& \approx \mathrm{0.506.}\end{array}$$ So $n=25$.

The history. The Chevalier de Méré knew that betting on at least one six in 4 rolls of one die was favourable ($1-(5/6{)}^4\approx 0.518$). He reasoned that a double six is 6 times rarer, so 24 rolls should be just as good, and lost money. The chance does not scale in proportion: the question led Pascal and Fermat to start probability theory.