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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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  • $\begingroup$ 23.8 of each bean??? $\endgroup$
    – Rasmus
    Jan 9, 2012 at 17:43
  • $\begingroup$ Can we be even more approximate and say the bag has 1200 beans? $\endgroup$ Jan 9, 2012 at 17:44
  • $\begingroup$ 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! $\endgroup$
    – Bryan
    Jan 9, 2012 at 17:45
  • $\begingroup$ @Rasmus Some people decide they don't like the flavor of a bean after all and put it back in the bag :) $\endgroup$ Jan 9, 2012 at 17:45
  • $\begingroup$ ps jelly beans don't have corners. $\endgroup$
    – Bryan
    Jan 9, 2012 at 17:50

3 Answers 3

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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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  • $\begingroup$ You sir are my hero. Thank you. I can now get back to work. $\endgroup$
    – Bryan
    Jan 9, 2012 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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  • $\begingroup$ This would be the probability of getting three of a specified color in three draws, not what OP asked. $\endgroup$ Jan 9, 2012 at 19:31
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The accepted answer is incorrect (logic is right) but 6C3 = 20 (not 15) , hence the final answer would be .0077 not .0056

Also another way to look at this is plain old binomial probability : nCr p^r (1-p)^n-r Here we have 6 tries (n=6), to get 3 success (same color, r=3 ) , probability of success won't change in any meaningful way as beans are too many (1190) . Calculating p (probability of success is bit tricky ) Its 24/1190 ( as 23.8 are of same color) . So for a one specific color

Pr(x=3 same color) = 6C3*(24/1190)^3 (1166/1190)^3 = .000154. 
But this could be true for any of the 50 flavor possible , so 50*0.000154 = .0077
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  • $\begingroup$ Why the downvote, for poiting out 6C3 should be 20 , not 15? Or giving binomial way of looking at the problem? $\endgroup$
    – Yash
    Feb 14, 2019 at 13:41

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