Basic Probability Question Calculating Possible Team Combinations Twenty people, numbered from {1,2,3,..,20} have to form two teams, each containing 10 people.  People are randomly arranged into teams.
What is the probability that one of the teams consists of players 1,2,3,4,5,6,7,8,9,  and 10?  
I know that doing (20 choose 10) will give me all possible subsets of the teams, but I'm not sure how to then select the one team. Would it be 1/(20 choose 10)? 
Any help is appreciated, thanks! 
 A: Yes: each of the $\binom{20}{10}$ possible teams of ten are equally likely, so your answer is correct.

I guess there is some ambiguity in the question, but both of us interpreted it incorrectly.
We have answered the following question:

Suppose we randomly form two teams A and B with ten people each. What is the probability that A contains players 1-10?

However, the question is asking for the probability that either A or B has players 1-10. So, we are wrong, and the probability is twice as much as your original answer.

Interpretation 1: there are $\binom{20}{10}$ ways to select a team of 10, so there are $\binom{20}{10}/2$ ways to divide the 20 people into two (unlabeled) groups of 10. (For example, selecting $\{1,\ldots,10\}$ as a team would provide the same two teams as selecting $\{11,\ldots,20\}$ as a team.) Each of the $\binom{20}{10}/2$ outcomes are equally likely and exactly one of them corresponds to the event "one of the teams is $\{1,\ldots,10\}$." So, the probability is $\frac{1}{\binom{20}{10}/2}$.
Interpretation 2: there are $\binom{20}{10}$ ways to select a team of $10$ to be "team A." The event "one of the teams is $\{1,\ldots,10\}$" corresponds will occur when either $\{1,\ldots,10\}$ or $\{11,\ldots,20\}$ is selected to be team A. So, the probability is $\frac{2}{\binom{20}{10}}$.
