Obtain your best roll!

Firstly, we roll the die 6 times and count success if 1 or more sixes appear. For each roll the probability of a number that is not a six is

$\frac{5}{6}$ . Since the outcomes are all independent, we can multiply the individual outcomes to find the overall probability. Hence, the chance of no sixes in 6 rolls is

$\frac{5}{6}^{6}=\frac{15625}{46656}$ .

Therefore P(1 or more sixes)= $1-\frac{15625}{46656}=0.66510202332$ .

Second, we now need to subtract the probabilities of zero sixes and of 1 six to find the probability of 2 or more sixes in 12 rolls.

P(0 sixes in 12 rolls)= $\frac{5}{6}^{12}=0.112156654785$ .

P(a six on the 1 roll and non-six on all other rolls)= $\frac{1}{6}\times\frac{5}{6}^{11}=0.022431330957$ .

The six can be in any of the 12 rolls, so we need to multiply by 12 (= 0.269175971483).

And thus, P(2 or more sixes in 12 rolls)= $1-0.112156654785-0.269175971483=0.618667373732$

Finally, the probability of 3 or more sixes in 18 rolls, we have to again subtract the probabilities of 0, 1, or 2 sixes in 18 rolls. The first two are similar to our previous calculations:

P(0 in 18)= $\frac{5}{6}^{18}=0.03756$ and P(1 in 18)= $\frac{1}{6}\times\frac{5}{6}^{17}\times 18=0.13522$ .

For 3 sixes out of 18 rolls, we can calculate $\frac{1}{6}$ (chance of a six on the first roll) $\times\frac{1}{6}$ (chance of another six) $\times\frac{5}{6}^{16}$ (chance of no sixes in the other 16 rolls).

But since the first six can occur in any of the 18 rolls and the second in any of the remaining 17, we need to multiply the previous calculation by the number of ways to pick the 2 numbers out of 18 rolls, which is

18 (for the 1st six) x 17 (for the 2nd six) divided by 2 (since order of the two sixes does not matter).

Hence, P(3 or more sixes in 18 rolls)= $1-0.03756-0.13522-0.22987=0.59735$ .

All in all:

A=0.66510202332,

B=0.618667373732,

C=0.59735

Therefore, $6A+12B+18C=22.166920...=22.2$ (1d.p.)