Here is how I would go about calculating the optimal choice. We know that if the number of rolls were very high, tending to infinity, the proportion of the number of rolls for each sum will converge to the probability distribution of the sums over the numbers 2 - 12 in a single roll.
Now what I would do is calculate the probability of each number occurring
$$P(2) = \frac{1}{36};\ \ \ \ \ P(3) = \frac{2}{36};\ \ \ \ \ P(4) = \frac{3}{36};\ \ \ \ \ P(5) = \frac{4}{36}\\
P(6) = \frac{5}{36};\ \ \ \ \ P(7) = \frac{6}{36};\ \ \ \ \ ...\ \ \ \ \ P(11) = \frac{2}{36};\ \ \ \ \ P(12) = \frac{1}{36}$$
Now, it is easy to see that if the two dice were rolled 36 times, I would have picked the numbers so that 2 occurs once, 3 occurs twice .. 7 occurs 6 times etc...
Since they are rolled 11 times, however, you will have to do some rounding to get the distribution that closest approximates this. My guess: 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10.
The chance that $1$ comes up exactly $33$ times in $100$ comes from the binomial distribution. The chance of success is $\frac 16$ and failure is $\frac 56$ so it is ${100 \choose 33}(\frac 16)^{33}(\frac 56)^{67}\approx 0.00003$ If we sum from $33$ to $100$ we get the chance of at least $33\ 1$s, which is about $0.0005$ per Alpha. You can multiply these by $6$ to get the chance for any number, as it is very unlikely we doublecount by having at least $33$ of two different numbers. So the chance of a "lucky number" happening by chance is about $0.0003$ or one in $3300$. Pretty unlikely, but rarer things happen all the time.
Best Answer
I find it easier to consider the problem of not rolling a 6 with any of the three dice (the same problem with a different definition).
The chance of not rolling a 6 is 5/6. The chance of not rolling a six on any of your dice is therefor 5/6 * 5/6 * 5/6.
The chance of rolling at least one six is therefore the opposite of this:
1 - (5/6)^3 = 0.421