Newton and Pepys

Idea. In each bet you need at least the expected number of sixes. With more dice, the count is more concentrated around its mean, and the chance of reaching at least the mean decreases towards $\frac{1}{2}$ instead of increasing.

Step 1: one six with 6 dice. $$1-{\left(\frac{5}{6}\right)}^6\approx \mathrm{0.665.}$$

Step 2: two sixes with 12 dice. $$1-{\left(\frac{5}{6}\right)}^{12}-12\cdot \frac{1}{6}{\left(\frac{5}{6}\right)}^{11}\approx \mathrm{0.619.}$$

Step 3: three sixes with 18 dice. Subtracting the chances of 0, 1 and 2 sixes gives about $0.597$.

Answer. The first bet is best, with probability $1-{\left(\frac{5}{6}\right)}^6\approx 0.665$.

History. Samuel Pepys asked Isaac Newton this question in 1693, expecting the 18-dice bet to win. Newton showed it was the other way round.