Quote:

**RiotKiddington**:

10 people per game, the max number of troll on your team is 4 (because the last person is you). The chance of max troll is 44%. then, the chance of been pick for captain is 20% (1 out of 5 people). So the question is what is the max chance of been a troll AND picked to be the captain.

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. More formally, if events A and B are independent, then the probability of both A and B occurring is:

P(A and B) = P(A) x P(B)

where P(A and B) is the probability of events A and B both occurring, P(A) is the probability of event A occurring, and P(B) is the probability of event B occurring.

so in our case, it is (4/9) * (1/5) = 4 / 45 = 8.89%

Did I made a mistake?

yes you did makle a mistake but not the one he tried to call you one

your mistake was that you did the probability for one troll to be captain not if there are 4 trolls on your team

also it still may be a bit confusing to the non-probability-knowledgeable potion of the player based to understand your first paragraph, may I suggest you get someone who failed around 8th (I think) grade math to help you get it more user friendly if you really want to have this make superb sense.

back to what your mistake may have been

the way to do it for 4 trolls on your team is 44.4% times 4/5 giving you about a 35% chance of a troll being captain at MOST*

*with the data available this is the most accurate (and realistic) "max" achievable, to be totally accurate you need to know the portion of player base that is a troll and how often they troll - even more accurate is if you include times of trolling, plus a host of other factors i am to sleep deprived to find specifically right now.