I don't know whether you're interested in approximate and asymptotic answers – there's a straightforward estimate for large $n$. The distribution for the sum tends to a normal distribution. The variance for one die is
$$
\langle x^2\rangle-\langle x\rangle^2=\frac{1+4+9+16+25+36}6-\left(\frac{1+2+3+4+5+6}6\right)^2=\frac{35}{12}\;,
$$
so the variance for $n$ dice is $n$ times that. The probability of a tie is the sum over the squares of the probabilities, which we can approximate by the integral over the square of the density, so this is
$$
\int_{-\infty}^\infty\left(\frac{\exp\left(-x^2/\left(2\cdot\frac{35}{12}n\right)\right)}{\sqrt{2\pi\frac{35}{12}n}}\right)^2\,\mathrm dx=\int_{-\infty}^\infty\frac{\exp\left(-x^2/\left(\frac{35}{12}n\right)\right)}{2\pi\frac{35}{12}n}\,\mathrm dx=\sqrt{\frac3{35\pi n}}\approx\frac{0.1652}{\sqrt n}\;.
$$
The approximation is already quite good for $n=3$, where it yields about $0.095$ whereas your exact answer is about $0.093$.
I am not sure of this answer but here it is any way.
Green dice have a 1 in 6 chance of rolling a shotgun. You roll 3 dice per turn so on average you will roll one half a shotgun per turn. It takes 3 shotguns to eliminate a player with no points for the round so on average that will take 6 turns.
Green dice have a 1 in 2 chance of rolling a brain. You roll 3 dice per turn so on average you will roll one and one half brains per turn. In the 6 turns that it takes to eliminate a player on average green dice will roll 9 brains.
Yellow dice have a 1 in 3 chance of rolling a shotgun. You roll 3 dice per turn so on average you will roll 1 shotgun per turn. It takes 3 shotguns to eliminata a player with no points for the round so on average that will take 3 turns.
Yellow dice also have a 1 in 3 chance of rolling a brain. So in the 3 turns it takes to roll roll 3 shotguns on average you will also roll 3 brains.
Red dice have a 1 in 2 chance of rolling a shotgun. You roll 3 dice per turn so on average you will roll one and one half shotguns per turn. It takes 3 shot guns to eliminate a player with no points for the round so on average that will take 2 turns.
Red dice have a 1 in 6 chance of rolling a brain. You roll 3 dice per turn so on average you will roll one half a brain on each turn. In the two turns on average that it takes to eliminate you, you would roll 1 brain.
If we give one point for each green dice. 3 are rolled each turn, in 6 turns that would be 18 points plus the 9 points from the brains rolled for a total of 27 points.
For yellow to get 27 points in the 3 turns they would have to add 24 to their brains. It would take 8 points a turn. With 3 dice per turn that is 2 and 2/3 points per dice.
For red to get 27 points in 2 turns they would have to add 26 points to their one brain. It would take 13 points per turn. With 3 dice per turn that would be 4 and 1/3 points per dice.
Thus the green:yellow:red ratio should be 1:2:3.
EDIT ---
I may have miss understood part of this question. I thought points were awarded for each dice rolled and added to the total obtained from the brains rolled. If we are only changing the number of points obtained by each brain on the given color of dice it changes things.
With the ratios being points per brain rolled:
Green on average will roll 9 brains, with 1 point each.
Yellow on average will roll 3 brains, requiring 3 points per brain to equal green.
Red on average will roll 1 brain requiring 9 points per brain to equal green.
Making the green:yellow:red ration 1:3:9
Best Answer
Roll the three dice. Record the result as $(a,b,c)$ where $a$ is the number on the blue die, $b$ the number on the white, and $c$ the number on the red. All $6^3$ such sequences are equally likely.
There are $5^3$ sequences consisting of non-$6$, and therefore $216-125$ sequences with at least one $6$. Thus the probability of at least one $6$ is $\frac{216-125}{216}$.
Note that $216-125$ is $91$, not your $93$. You counted everything correctly, except there is only one sequence $(6,6,6)$, not three.
For this problem, I prefer to work directly with probabilities. The probability of a non-$6$ is $5/6$, so the probability of three of them in a row is $(5/6)^3$, so the probability of at least one $6$ is $1-(5/6)^3$.