A special six-sided die has its faces numbered 1 through 6 and has the property that rolling each number $x$ is $x$ times as likely as rolling a 1. For example, the probability of rolling a 5 is 5 times the probability of rolling a 1 , while the probability of rolling a 2 is 2 times the probability of rolling a 1. Robbie and Francine play a game where they each roll this die three times, and the total of their three rolls is their score. The winner is the player with the highest score; if the two players are tied, neither player wins. After two rolls each, Robbie has a score of 8 and Francine has a score of 10. The probability that Robbie will win can be written in lowest terms as $\frac{r}{400+s}$, where $r$ and $s$ are positive integers. What is value of $r+s$ ?
(A) 96
(B) 86
(C) 76
(D) 66
(E) 56
Robbie has a score of 8 and Francine has a score of 10 after two rolls each. Thus, in order for Robbie to win (that is, to have a higher total score), his third roll must be at least 3 larger than that of Francine.

If Robbie rolls 1, 2 or 3, his roll cannot be 3 larger than that of Francine.

If Robbie rolls a 4 and wins, then Francine rolls a 1.

If Robbie rolls a 5 and wins, then Francine rolls a 1 or a 2.

If Robbie rolls a 6 and wins, then Francine rolls a 1 or a 2 or a 3.

We now know the possible combinations of rolls that lead to Robbie winning, and so need to calculate the probabilities.

We recall that Robbie and Francine are rolling a special six-sided dice.

Suppose that the probability of rolling a 1 is $p$.

From the given information, the probability of rolling a 2 is $2 p$, of rolling a 3 is $3 p$, and of rolling a 4,5 and 6 is $4 p, 5 p$ and $6 p$, respectively.

Since the combined probability of rolling a 1, 2, 3, 4, 5, or 6 equals 1 , we get the equation $p+2 p+3 p+4 p+5 p+6 p=1$ which gives $21 p=1$ or $p=\frac{1}{21}$.

Thus, the probability that Robbie rolls a 4 and Francine rolls a 1 equals the product of the probabilities of each of these events, which equals $\frac{4}{21} \cdot \frac{1}{21}$.

Also, the probability that Robbie rolls a 5 and Francine rolls a 1 or 2 equals $\frac{5}{21} \cdot \frac{1}{21}+\frac{5}{21} \cdot \frac{2}{21}$.

Lastly, the probability that Robbie rolls a 6 and Francine rolls a 1, a 2, or a 3 equals

$$
\frac{6}{21} \cdot \frac{1}{21}+\frac{6}{21} \cdot \frac{2}{21}+\frac{6}{21} \cdot \frac{3}{21}
$$

Therefore, the probability that Robbie wins is

$$
\frac{4}{21} \cdot \frac{1}{21}+\frac{5}{21} \cdot \frac{1}{21}+\frac{5}{21} \cdot \frac{2}{21}+\frac{6}{21} \cdot \frac{1}{21}+\frac{6}{21} \cdot \frac{2}{21}+\frac{6}{21} \cdot \frac{3}{21}=\frac{4+5+10+6+12+18}{21 \cdot 21}=\frac{55}{441}
$$

which is in lowest terms since $55=5 \cdot 11$ and $441=3^{2} \cdot 7^{2}$.

Converting to the desired form, we see that $r=55$ and $s=41$ which gives $r+s=96$.