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. Suppose that the probability of rolling a 1 is \( p \). From the given information, the probability of rolling a 2 is \( 2p \), of rolling a 3 is \( 3p \), and of rolling a 4, 5, and 6 is \( 4p, 5p \) and \( 6p \), respectively. Since the combined probability of rolling a 1, 2, 3, 4, 5, or 6 equals 1, we get the equation \( p + 2p + 3p + 4p + 5p + 6p = 1 \) which gives \( 21p = 1 \) 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 \( r + s = 96 \).
What is the probability that Robbie will win if he and Francine each roll a special six-sided die three times, and after two rolls each, Robbie has a score of 8 and Francine has a score of 10?