Go, Super Boomers!

First situation: They both get chosen. That would be described as: $\frac{1}{11} \times \frac{1}{10} \times 2 = \frac{2}{110} = \frac{1}{55}$ (Notice that you have to multiply it by 2, because it can be either Alfred and Bruce, or Bruce and Alfred).

Since the chance of the name being chosen is 100%, we just have to multiply the result by 1.

Now, second situation: Alfred gets chosen, but not Bruce. That would be described as: $\frac{1}{11} \times \frac{9}{10} \times 2 = \frac{18}{110} = \frac{9}{55}$ (In this situation, you need to multiply by 2 because Alfred can either be the first chosen player or the second one)

Since the chance of Alfred convincing the other players is 80%, we shall multiply the results by $\frac{4}{5}$ : $\frac{9}{55} \times \frac{4}{5} = \frac{36}{275}$ Now, last situation: Bruce gets chosen, but not Alfred. That would also be described as: $\frac{1}{11} \times \frac{9}{10} \times 2 = \frac{18}{110} = \frac{9}{55}$ (Multiply by 2 for the same reason)

Since the chance of Bruce convincing the other player is only 40%, we shall multiply the result by $\frac{2}{5}$ : $\frac{9}{55} \times \frac{2}{5} = \frac{18}{275}$ In the end, jus sum it all: $\frac{1}{55} + \frac{36}{275} + \frac{18}{275} = \frac{5 + 36 + 18}{275} = \boxed{\frac{59}{275}}$

Let's divide the possible pairs of players in 4 categories and count how possible formations there would be:

1) Alfred and Bruce - Only 1 formation

2) Alfred and someone who's not Bruce - 9 formations (Alfred plus someone out of the 9 that's left)

3) Bruce and someone who's not Alfred - 9 formations (same explanation)

4) Two players there are neither Alfred or Bruce - 36 formations = C9,2

Now the probability of the name "Super Boomers" being chosen for each categories is, respectively, 100%, 80%, 40% and 0%. So I would certainly have that name if the pick was of category 1, and I could have it in 7,2 picks of the 2 category, 3,6 picks of the third category and 0 picks of category 4. In other words I could have that name in 11,8 picks of the total 55, thus the probability would be $\frac{11,8}{55} = \frac{59}{275}$ .

Ps: I understand this is not a proper and formal probabilistic solution, but its foundation are on a probabilistic line of thought, and I thought it would serve as such. I also understand that $\frac{11,8}{55}$ is not very clean way to write math. ;)