Probability of ending up in the same group

Question

There's 6 groups, I'm in the first, and my friend is in another. The teacher then decides to merge all groups in pairs of 2. So there's gonna be 3 groups. What is the probability that me and my friend ends up in the same new group?

My mind is very split as I read this: Probability of two friends being in the same group

One solution: I thought of is that we have: 6 over 3 as the total combinations: which is 20. Then i can count all the times me and my friend end up in the same group, which should be: 4 over 0, as no other person can join our group. As there's 3 groups, i multiply by 3. So 3/20 total chance of ending up in the same group.

The other solution: I thought that since there's 5 other groups, apart from my, there's 1/5 chance that he ends up in the same group.

So im not sure if either is Correct, or how to better calculate it, any help is appreciated.

If there's any help. It started as 60 students, split into 6 groups of 10. Where i calculated it should be 9/59 chance that we end up in the same group:

Answer 1

Your second argument is correct

Perhaps your first argument should be something like this: you could say there are $\frac1{3!}{6 \choose 2}{4 \choose 2}{2 \choose 2}= 15$ ways of pairing the groups. Of these, there are $\frac1{2!}{4 \choose 2}{2 \choose 2}=3$ with your group and your friend's group together, giving a probability of $\frac3{15}=\frac15$

So the error in your first argument is the number $20$. To convince you that $15$ is correct, here is the full list:

AB,CD,EF    AB,CE,DF    AB,CF,DE    
AC,BD,EF    AC,BE,DF    AC,BF,DE    
AD,BC,EF    AD,BE,CF    AD,BF,CE    
AE,BC,DF    AE,BD,CF    AE,BF,CD    
AF,BC,DE    AF,BD,CE    AF,BE,CD