Jack is in a different team than Carl and Carl is in a different team than Andy. What is the probability that Jack and Andy are in the same team? 27 players are divided randomly into 3 teams. Jack is in a different team than Carl and Carl is in a different team than Andy. What is the probability that Jack and Andy are in the same team?
What I think this problem is saying is that the players are equally divided into 3 teams of 9 players each and that the teams have already been made before asking the question. Therefore I started solving in the following way:
There are three teams (A,B and C). Carl can be in either one of them, so we have 3 choices . Jack can be in one of the remaining two, depending on which team Carl is on. Andy can be either in the last remaining team or in Jack's team. Suppose Carl is in team A; Jack could be in team B and Andy in C or viceversa; Jack and Andy could both be in team B or both in C. So in total 4 possible cases if Carl is in team A.
But at the beginning Carl could be in any of the three teams, and for each team he could be in there are 4 possible cases. So, total cases equals 4 times 3 = 12.
How many of those possible cases have Jack and Andy in the same team?
2 for each team Carl could be in, so 2 times 3 = 6.
Therefore I think the answer to problem must be 6/12 = 1/6 = 16.67% chance that Jack and Andy are on the same team.
I'm not sure, because in my logic I totally discarded the fact that there are 27 players (9 per team); I just considered the three teams, because I think the answer remains the same (even if there were let's say 5 players on each team) when we think team-wise, which imo is how one should approach this problem.
Any thoughts?
 A: Actually, instead of counting, it is easier solved using probability.
Imagine $3$ groups of $9$ slots totalling $27$
Carl could be in any slot, Andy is known to be in any of the $18$ slots in "other" groups , thus for Jack to be in the same group as Andy,
$Pr = \Large\frac8{17}$

ADDED
Pehaps more easily understandable as C, A, J being sequentially placed fulfilling conditions as
Pr = $\Large\frac{27}{27}\frac{18}{18}\frac{8}{17}$
A: Your mistake is well-exposed in the comments on your question.
There are $18$ persons that are not in the team that contains Carl.
These $18$ persons are split up in two groups of $9$ persons.
One of them is Jack and next to him there are $17$ others (Andy is one of them) of which $8$ end up as team members of Jack.
So the probability that Andy ends up as a team member of Jack equals $\frac8{17}$
A: If you insist on approaching by counting principles, let us look at what we are talking about more carefully.  Let $A$ be the event that Jack and Andy are on the same team.  Let $B$ be the event that Jack is on a different team than Carl and that Andy is also on a different team than Carl.
We are looking for $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}$

"the total number of possible outcomes would be 27C9 = 4686825, right?"

No.  The random processes involved have us completing the assignment of people to all of the teams.  $\binom{27}{9}$ would be the number of ways of picking who goes to the first team.  There still remains the question of who among the remaining 18 people go to the second team for a total of $\binom{27}{9}\cdot\binom{18}{9}$ or $\binom{27}{9,9,9}$ or $\dfrac{27!}{9!9!9!}$ ways (they are all the same number, represented differently).
Using this as our sample space, finding $|B|$ would be done by first picking which team Carl is on (three choices).  Then picking where Jack goes (must be different).  Finally, break into cases based on if Andy goes to the same team as Jack or not.  Lastly, fill out the remaining players on each team noting that in the case that Jack and Andy are on the same team they need only seven additional teammates while if they were on different teams they would need eight.  This gives a total of $3\cdot 2\cdot (\binom{24}{8}\binom{16}{8}\binom{8}{8}+\binom{24}{8}\binom{16}{7}\binom{9}{9})$
Finding $|A\cap B|$ we would approach similarly except when breaking into cases based on if Jack and Andy were on different teams there is only the one case.
This gives our final probability as:
$$\dfrac{~~~~~\frac{3\cdot 2\cdot \binom{24}{8}\binom{16}{7}\binom{9}{9}}{\binom{27}{9}\binom{18}{9}\binom{9}{9}}~~~~~}{\frac{3\cdot 2\cdot (\binom{24}{8}\binom{16}{8}\binom{8}{8}+\binom{24}{8}\binom{16}{7}\binom{9}{9})}{\binom{27}{9}\binom{18}{9}\binom{9}{9}}} = \dfrac{8}{17}$$
Ugly to look at, isn't it?  There is good reason why we try to teach you the shorter method.
