TITLE: Dividing colored balls between children
QUESTION [5 upvotes]: The following was an exercise I solved:

We have 8 numbered balls - two blue, two red, two green, and two yellow. When dividing them between 4 children, 2 balls each, what is the probability at least one child will get two balls of the same color?

I solved it using the exclusion-inclusion principle, and got the end result $3/7$. My question is, since this is such a nice fraction, was there any way of solving the problem such that I could've arrived at the fraction directly?
(This is the calculation I did: http://www.wolframalpha.com/input/?i=C(4%2c1)*(4*C(6%2c2)*C(4%2c2)C(2%2c2))%2f2520-C(4%2c2)(4*3*C(4%2c2)C(2%2c2))%2f2520%2bC(4%2c3)(4*3*2*C(2%2c2))%2f2520-C(4%2c4)*(4*3*2*1)%2f2520&incParTime=true)

REPLY [2 votes]: I don't think there is a one-line computation of this probability or its complement; but you can do without the inclusion-exclusion principle. In order to compute the probability $p$ that no child receives two balls of the same color one may argue as follows:
At the beginning the eight balls are in an urn. The first child has 28 possibilities of drawing two balls without replacement,  4 of which are bad. Therefore we have a probability of ${6\over7}$ that the first child gets two balls of different colors, say $\{a,b\}$. 
The six balls left have a color distribution $\{a,b,c,c,d,d\}$. The second child has 15 possibilities of drawing two balls out of these, 2 of which are bad. With probability ${1\over15}$ she picks $\{a,b\}$. The remaining four balls $\{c,c,d,d\}$ admit three pairings, and one of these is bad. With probability ${2\cdot 4\over15}$ the second child picks $\{a,c\}$ (or similar). The remaining four balls $\{b,c,d,d\}$ again admit three pairings, one of them bad. Finally, with probability ${2\cdot 2\over15}$ the second child picks $\{c,d\}$, in which case four balls of different colors remain which can be paired in any way.
Putting it all together we see that the probability $p$ is given by
$$p\ =\ {6\over7}\cdot\Biggl({1\over15}\cdot{2\over3}+{8\over15}\cdot{2\over3}+{4\over15}\cdot1\Biggr)={4\over7}\ .$$