Probability of having the two best football players in different teams

Question

I have problems understanding the solution for this lesson:

From $22$ football players we create two teams with $11$ players each. What is the probability of having the two best players in different teams?

I am pretty sure that the number of overall possible outcomes is $\binom{22}{11}$, which will divide the favorable number of outcomes. However I don't know how to get that. As far as I know the solution is $\frac{11}{21}$.

Thanks in advance!

Answer 1

The first "best" player is on one team. There are 10 other available slots on that team, and 11 available slots on the other, so the chance the second "best" player is on the second team is 11/(10+11) = 11/21.

Answer 2

Well all dividing of a players is given by $$n={1\over 2}{22\choose 11}$$ and all good outcomes are given by $$m={20\choose 10}$$ so $$ P = {m\over n} = {2\cdot 20!\cdot11!\cdot11!\over 22!\cdot10!\cdot10!} ={11\over 21}$$

Answer 3

We'll name the teams A and B. We'll count the unwanted possibility (call it P), then 1-P will be the answer.

2 ways to put these two best players in same team (A or B.) we should choose another 9 players for this team.

let us say C(n, r) means the number of ways to choose r people from n people.

P = 2*C(20, 9) / C(22, 11)

(C(22, 11) is the number of overall possible outcomes as you just mentioned)

The answer is 1-P = 11/21.