A Combinatoric Probability Problem, my program seems to give wrong answer say, I have 3 red balls, 4 yellow balls, 5 blue balls. totally 12 balls
now I randomly arrange the balls, a1...a12
a good layout is such:  a1 = a12 or a1 = a11 or a2 = a12 (= means same color)
what's the probability of gaining a good layout?
I believe it is somewhat 66%=0.2879*3-0.06818*2-0.7273+0.01212 but my program (10000 random runs) gives 44%...
Update
My Maths is correct, my program has a bug..., now corrected.

There are ${12\choose 2}=66$ ways to choose a pair of balls, and
${3\choose 2}+{4\choose 2}+{5\choose 2}=19$ ways for them to have
the same color. Thus,
$$\mathbb{P}(a_1=a_{11})=\mathbb{P}(a_1=a_{12})=\mathbb{P}(a_2=a_{12})={19\over 66}.$$
Similarly, we have 
$$\mathbb{P}(a_1=a_{11}=a_{12})=\mathbb{P}(a_1=a_2=a_{12})={3\over 44},$$
and 
$$\mathbb{P}([a_1=a_{11}]\cap [a_2=a_{12}]) = {14\over 165}.$$
Finally, 
$$\mathbb{P}([a_1=a_{11}=a_2=a_{12}]) = {2\over 165}.$$
By inclusion-exclusion, the chance of getting a good configuration is 
$$3\cdot {19\over66}-2\cdot {3\over44}-{14\over 165}+{2\over 165}={36 \over55}=.65454.$$
