Colored balls into groups If 3 red, 3 blue, and 3 green balls are randomly divided into three groups of three balls each, what is the probability that none of the groups have all the balls of the same color?
My attempt: 
Let $A$ be the event that all groups have all balls of the same color. There is 1 way that this can occur, irrespective of inter-group ordering and considering same-colored balls to be indistinguishable.
Let $B$ be the event that exactly one group has all balls of the same color. One group of identically-colored balls can be chosen $\binom{3}{1}=3$ ways. The remaining two groups are totally determined by the initial identically-colored group, as they both must consist of two balls of one of the remaining two colors and one ball of the other remaining color. This again is irrespective of the ordering of groups or of balls within groups.
So there are $3+1=4$ ways that $A\cup B$ can occur. 
There are 8 total possible outcomes: 4 outcomes belonging to $A \cup B$, 1 outcome in which all three groups contain one ball of each color, and 3 outcomes in which one group contains all differently-colored balls and the other groups contain the remaining balls. 
Thus, the probability of $(A \cup B)^C$ is $\frac{1}{2}$.
Am I correct? And, is there a more generalized and efficient way to solve this sort of problem? Took me far too long...
 A: The probability depends on how the random selection occurs.
An obvious randomization process is that we put all the balls in an urn, mix them thoroughly, and then blindly draw out one at a time.
The first three balls become one group of three, the next three balls are another group of three, and the last three balls also are a group of three.
Now let's consider the probability that the first three balls drawn will all be the same color.
The first ball we draw can be any color; call it color X.
We now want to draw a second ball of the same color,
but there are now $8$ balls in the urn with $2$ of color X,
so we have a $\frac14$ chance to draw color X again.
Finally, we want to draw another ball of color X.
There is only $1$ such ball out of $7$ in the urn, so we have a $\frac17$
probability of drawing it.
The probability that the first three balls are all the same color is
therefore $\frac14 \times \frac17 = \frac{1}{28}.$
Now in order to have three same-colored groups (all red, all green, or all blue), the first three balls we draw must all be one color.
We just found that has probability $\frac{1}{28}.$
But you think we have a $\frac18$ probability ($1$ case out of the $8$ you listed) that all groups have all balls the same color?
It seems you have chosen $8$ separate cases that are not equally likely
under a reasonable interpretation of "randomly divided,"
but you are assuming they all have equal probability.
You can of course construct such a probability by making a spinner with
eight equal-sized sectors, label each sector with one of your cases,
spin the spinner to select a case, and then arrange the balls so they match
the pattern selected by the spinner. But if we resort to that kind of
"randomly divided," we can make the probability of three-of-a-kind
anything we want ($0,$ $1,$ or anything in between).

As a hint of how to get back on track, assuming you consider my drawing-from-the-urn scheme to be sufficiently random,
I have already calculated the probability that the first group will be all one color.
You can continue the process to find the probability that all three groups will be the same color.
From this you can get the probability that only the first group will be one color. 
Now consider the probability that only the second group will be all one color, and the probability that only the third group will all be one color.
Those two probabilities may seem more difficult to calculate at first, but consider the deeper symmetries of the problem:
the process gives you a random permutation of the balls with all permutations equally likely.
