# 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!

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 $\frac{2}{\binom{20}{10}}$.

• Ok, I am confused because the answer our teacher gave us was 1/((20 choose 10)/2) because she said that by doing (20 choose 10) we are essentially ordering the groups, therefore we need to divide the number of groups we get over 2. This confused me as I thought that (20 choose 10) would be unordered, because it's a combination, not permutation. Is there any reason it would be the second answer instead of the one I gave? Commented Mar 2, 2017 at 20:13
• @user2012813 Apologies, your teacher is correct. See my edit. Commented Mar 2, 2017 at 22:55
• wouldn't the probability be half as much if we are dividing the total set of teams by two, not twice as much? Commented Mar 3, 2017 at 0:10
• @user2012813 See my edit. Commented Mar 3, 2017 at 2:31