Probability of ending up in a group A friend of mine asked me this question, but I'm not an expert in discrete probability or combinatorics.

40 people are divided into four groups of 10 with equal probability. Of the 40 people there are three friends. What would the probability be that exactly two out of the three friends end up in the same group?

I have tried a couple of methods but can't convince myself that any of them are correct.
Attempt: One of the three friends are placed in any of the groups. The probability of the second friend ending up in the same group is $\frac{9}{39}$ and the probability that the last of the friends ends up in another group is $\frac{30}{38}$. Since there are three friends we multiply with 3 and we get
$$P(\text{exactly two friends in same group})=3\cdot\frac{9}{39}\cdot\frac{30}{38}\approx 0.55.$$
Is this correct?
 A: You are correct.
Here is a different way to approach the problem. You can count the number of ways to choose teams that result in two friends together and the third apart, and then divide by the total number of ways. I will assume that the order within a team does not matter, and that the teams themselves are unordered (there is no first/second/third/fourth team).

*

*There are $\binom32$ ways to choose the pair of friends that will be together.


*There are $\binom{37}8$ ways to choose the other eight people on the team with the pair of friends.


*There are $\binom{29}{9}$ ways to choose the other nine people on the team with the lonely friend.


*There are two teams of ten remaining to choose. This can be done in $\frac12\binom{20}{10}$ ways; you need to divide by $2$, because in our counting method, the order of the teams does not matter. Alternatively, you can choose some particular person $x$ from the reamining $20$ people, and argue there are $\binom{19}{9}$ ways to choose the teammates of $x$ (note that $\binom{19}9=\frac12\binom{20}{10}$).
We then multiply these factors all together, and divide by the total number of ways to divide $40$ people into four unlabeled teams of ten, which is $\frac1{4!}\frac{40!}{(10!)^4}$. Again, we divide by $4!$ because the teams are unordered. The result is equal to $135/247$, exactly matching your computation. Wolfram|Alpha confirmation.
