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I have a bag of jelly beans with approx 1190 Jelly Belly's in it. There are 50 different flavors. Assuming the amount of Jelly Belly's per flavor are equal (so, 23.8 of each bean):

If I pull 6 Jelly Beans from my unopened bag, what are the odds that 3 of them will be the same color?

I'm asking because I got a bag of jelly beans for Christmas and got 3 of the same color in the handful I just pulled out... and it's been many many years since my statistics class in college.

Thank you for the help. It's driving me batty and nobody at work cares about my Jelly Belly question except me =(

I would have tried to figure this out on my own as it seems very easy, but I don't even know where to begin looking.

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23.8 of each bean??? –  Rasmus Jan 9 '12 at 17:43
    
Can we be even more approximate and say the bag has 1200 beans? –  David Mitra Jan 9 '12 at 17:44
    
Sure? Does it even matter though since there are only 50 colors? You can be as approximate as you want. I won't know the difference. ..and yeah, one of the beans was missing a corner, hence the .8! –  Bryan Jan 9 '12 at 17:45
    
@Rasmus Some people decide they don't like the flavor of a bean after all and put it back in the bag :) –  David Mitra Jan 9 '12 at 17:45
    
ps jelly beans don't have corners. –  Bryan Jan 9 '12 at 17:50
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2 Answers 2

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Let's pretend the bag is very large, so drawing one bean of a flavor doesn't change the probabilities. There are $50^6$ possible draws, all the same probability. The ways to get exactly $3$ of a flavor can be counted as $\binom 63=15$ choices for which $3$ will match times $50$ choices for which flavor to match times $49^3$ for the other $3$. There is a tiny error as we double count the case you get two three of a kinds. So the chance is $\frac {15*50*49^3}{50^6}=352947/62500000 = 0.005647152$ or about $1$ in $177$.

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You sir are my hero. Thank you. I can now get back to work. –  Bryan Jan 9 '12 at 18:02
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It's called "probability without replacement". The answer is (23.8/1190) x (22.8/1189) x (21.8/1188), so the probability of that happening is one in 142,094.

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This would be the probability of getting three of a specified color in three draws, not what OP asked. –  Ross Millikan Jan 9 '12 at 19:31
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