There are 12 players that are randomly distributed to 2 teams, so there are 6 players on each team.
Question 1: What is the probability of two specific players being on the same team?
Using combinations, the answer is
$$2\times\frac{\binom{10}{4}}{\binom{12}{6}}=\frac{5}{11}\approx45\%$$
because you have 12 players to pick from to fill 6 spots for the first team for the denominator, and 10 players to pick from to fill the 4 remaining spots after fixing player 1 and 2 for the numerator. You multiply by 2 because it could happen for either team.
I would assume the answer is 50%. Why isn't it?
Suppose there are two teams A and B, and two players 1 and 2. The probability of player 1 being on team A is 1/2, and the probability of player 2 being on team A is also 1/2. Thus, the odds of player 1 and 2 being on team A is 1/4 and probability of both of them being on team B is also 1/4. Adding those two up gives 1/2. Thus, there's a 50% chance of both players being on team A or team B.
Question 2: Why isn't the answer exactly 50% when doing it via combinations? Even intuitively, it doesn't make sense why it's not 50%. The odds of two players being on two different teams is ~ 55%?? Why? I don't understand.
Furthermore, as teams get bigger (e.g., 50 players, so 2 teams of 25 players), the odds of 2 players being on the same team becomes 49%. I would assume the odds to stay the same, yet it's approaching 50%. Why?
Best Answer
The answer to the first question: Assume the first player has joined one of the two teams at random. Then there are 11 players left, 5 more spots at the first players team and 6 more spots at the other team. The result is that the second player is more likely to be placed in the other team because there are more spots. The first player "blocks" a spot in the first team.