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?
Best Answer
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 = $\Large\frac{27}{27}\frac{18}{18}\frac{8}{17}$