Let p be the chance of a player being a "baddie". Huge amount of players thus we can approximate that the chances don't shift as you add players to the team.
You are not a baddie, thus your chance of going completely baddie free on your team:
PA(0) := (1 - p)^4
Enemy team has five slots, thus chance of going completely baddie free on enemy team:
PE(0) := (1 - p)^5 = PA(0) * (1 - p) < PA(0)
Chance of getting one baddie on your team:
PA(1) := p * (1 - p) ^ 3 * 4
Chance of getting one baddie on enemy team:
PE(1) := p * (1 - p) ^ 4 * 5 = PA(1) * (1 - p) * 5/4
Chance of getting two baddies on your team:
PA(2) := p ^ 2 * (1 - p) ^ 2 * 43/2
Chance of getting two baddies on enemy team:
PE(2) := p ^ 2 * (1 - p) ^ 3 * 54/2 = PA(2) * (1 - p) * 5/3
Chance of getting three baddies on your team:
PA(3) := p ^ 3 * (1 - p) ^ 1 * 4
Chance of getting three baddies on enemy team:
PE(3) := p ^ 3 * (1 - p) ^ 2 * 5 = PA(3) * (1 - p) * 5/4
Chance of getting four baddies on your team:
PA(4) := p ^ 4 *
Chance of getting four baddies on enemy team:
PE(4) := p ^ 4 * (1 - p) = PA(4) * (1 - p)
Chance of getting five baddies on your team:
PA(5) = 0
Chance of getting five baddies on enemy team:
PE(5) = p ^ 5
Expectation for amount of baddies on your team:
PA(1) + PA(2) * 2 + PA(3) * 3 + PA(4) * 4
Expectation for amount of baddies on enemy team:
(PA(1) * (1 - p) * 5/4) + (PA(2) * (1 - p) * 5/3 * 2) + (PA(3) * (1 - p) * 5/4 * 3)
- (PA(4) * (1 - p) * 4) + (p ^ 5 * 5)
For (1 - p) > 80% it's obvious that the second is larger
For (1 - p) < 80% it's still going to be larger due to the 5*p^5 term (feel free to do the math; though... I don't actually have the time since I'm about to leave)