# 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.

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What do you get for just the chance that a1=a12? –  Ross Millikan Aug 23 '11 at 16:06
my maths told me it's 28.8% –  colinfang Aug 23 '11 at 16:07
I agree with that. Then .06818 is the chance of a1=a11=a12, which I also agree with. Maybe you could check your program for the individual contributions. –  Ross Millikan Aug 23 '11 at 17:36

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 $$3\cdot {19\over66}-2\cdot {3\over44}-{14\over 165}+{2\over 165}={36 \over55}=.65454.$$

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your simulation program might be wrong. My simulation shows that the probality on 10000 round is 65.6%.

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