TITLE: Probability of $2$ people being together out of a group of $4$.
QUESTION [7 upvotes]: There were $2$ people who wanted to be in a group together. Those $2$ people were in a pool of $4$ people. Someone would randomly select $2$ people in a row and those people would be in a group together. The remaining $2$ would also be in a group. 
What the probability of those $2$ people who wanted to be in the same group actually being in the same group?
What I tried
Lets assume that the $2$ who want to be together are person A and B respectively and person C and D are the other $2$.
The probability of choosing person A and B is $\dfrac 14 \cdot \dfrac 13 = \dfrac 1{12}.$

REPLY [22 votes]: We are placing the four people in two groups of two people.  Notice that person $A$ must be in a group with one of the other three people, of whom only one is person $B$.  Hence, the probability that persons $A$ and $B$ are placed in the same group is $1/3$.

REPLY [4 votes]: Your mistake is that you are considering only one possibility where A and B end up in the same group. What are the others?
Another approach can be direct counting. There are total of $4!$ possible arrangements, out of which $2\cdot 1$ that begin with AB and $2\cdot 1$ that begin with BA, but also $2\cdot 1$ that end in AB and $2\cdot 1$ that end in BA, so the probability is $\dfrac{2+2+2+2}{24} = \dfrac 13$.