Probability of being matched against a pair of people $\textbf{Question:}$ Suppose you are playing a game in which two teams of five people, call them Team A and Team B, compete. Each of the ten people is randomly assigned a unique role (no two people share the same role) from a pool of 100 different roles, call them Role 1, Role 2, $\dots$ , Role 100. Assume you are on Team A. What is the probability that Role 1 and Role 2 will be assigned to people on Team B?
$\textbf{Full Disclosure:}$ For those of you familiar with the game DotA 2, I'm wondering what is the probability of facing a specific pair of heroes, such as Keeper of the Light and Phantom Lancer, while playing All Random.

$\textbf{Attempted Solution:}$ The number of ways to choose ten roles from the pool of 100 is $\binom{100}{10}$. The total number of ways to choose ten roles that include Role 1 and Role 2 is $\binom{98}{8}$, so the probability that Role 1 and Role 2 are assigned to two of the ten players is $\frac{\binom{98}{8}}{\binom{100}{10}}$. The total number of ways to form two teams of five from ten people is $\binom{10}{5}$. If two of the ten people are assigned Role 1 and Role 2, the total number of ways to form teams of 5 such that they are on opposite teams is $\binom{8}{4}$, so the probability that the two people assigned Role 1 and Role 2 are on the same team is $1-\frac{\binom{8}{4}}{\binom{10}{5}}$. The probability that you are placed on either team is $\frac 1 2$. Thus, the probability that Role 1 and Role 2 are assigned to two of the ten people AND that those two people are on the same team AND that you are on the team facing them is $$\frac{\binom{98}{8}}{\binom{100}{10}}\left(1-\frac{\binom{8}{4}}{\binom{10}{5}}\right)\frac{1}{2}=\frac{13}{3960}$$

Is my solution correct? Also, let me know if anything can/should be clarified. Thanks!

As you wrote, the probability that Role 1 and Role 2 are both chosen is
$$\frac{\binom{98}{8}}{\binom{100}{10}}.\tag{1}$$
Now line up the $10$ chosen roles, with say Roles 1 and 2 first. The probability Role 1 will be assigned to somebody on Team B is $\frac{5}{10}$. Given this has happened, the probability Role 2 is assigned to someone on Team B is $\frac{4}{9}$.  
For our probability, multiply the result of (1) by $\frac{5}{10}\cdot\frac{4}{9}$. 
Alternately, there are $\binom{10}{2}$ ways to choose the pair of people who will get Roles 1 and 2 (here we are just counting the number of pairs of people, not which one gets which role). And there are $\binom{5}{2}$ ways to choose them from Team B. Thus $\frac{5}{10}\cdot\frac{4}{9}$ can be replaced by $\dfrac{\binom{5}{2}}{\binom{10}{2}}$. 
Remark: An analysis somewhat like yours could be made. Suppose Team A is to wear blue, and Team B is to wear red. Given that we have chosen the people who will get the various roles, we find the probability that the two particular people who got Roles 1 and 2 end up wearing red. There are $\binom{10}{5}$ ways to decide who will wear red. And there are $\binom{8}{3}$ ways of choosing a team that will wear red if two of the members are to be a certain two fixed people. So the required multiplier is 
$$\frac{\binom{8}{3}}{\binom{10}{5}}.$$ 
Simplify. We get the right number. 
