22. Alice and Bob are playing a game with dice. They each roll a die six times, and take the sums of the outcomes of their own rolls. The player with the higher sum wins. If both players have the same sum, then nobody wins. Alice's first three rolls are 6,5 , and 6 , while Bob's first three rolls are 2,1 , and 3 . The probability that Bob wins can be written as a fraction $a / b$ in lowest terms. What is $a+b$ ?
22. 3895
Solution. Let $a_{i}$ denote the value of Alice's $i$ th roll, and $b_{i}$ denote the value of Bob's $i$ th roll. For Bob to win, the following inequality must hold:
$$
6+5+6+a_{4}+a_{5}+a_{6}<2+1+3+b_{4}+b_{5}+b_{6}
$$

Rearranging yields
$$
\left(a_{4}-1\right)+\left(a_{5}-1\right)+\left(a_{6}-1\right)+\left(6-b_{4}\right)+\left(6-b_{5}\right)+\left(6-b_{6}\right)<4 .
$$

Let $x_{i}=a_{i+3}-1$ and $x_{i+3}=6-b_{i=3}$ for $i=1,2,3$. The inequality then simplifies to
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}<4 \text {. }
$$

We claim that each nonnegative integer solutions to this inequality correspond to a valid situation of dice rolls. Note that the previous inequality implies that $0 \leq x_{i}<4$, so $1 \leq x_{i}+1<5$ and $2<6-x_{i} \leq 6$. Thus, the corresponding dice rolls for both Alice and Bob are within the bounds.
The expression on the left can have a value of either $0,1,2$, or 3 . Thus, the number of nonnegative integer solutions of the aforementioned equation is $\binom{5}{5}+\binom{6}{5}+\binom{7}{5}+\binom{8}{5}=1+6+21+56=84$. Thus, the probability that Bob wins is $\frac{84}{6^{6}}=\frac{7}{3888}$, and so $a+b=7+3888=3895$.