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Today I dropped my hand into a bag of m&m's and found that, without looking, I had pulled out 6 orange ones and one yellow one. This struck me as pretty unusual, so I started wondering what the exact chances would be. It's been a while since I mathed regularly, though, and probability was never my strong point. My rough guess, backed up by some rudimentary modeling in Python, is "really unlikely, but probably more likely than drawing six out of six identically colored m&m's". That's not a very interesting answer, though, so I was hoping someone here could help me out.

Here's the problem more precisely stated: given a bag of m&m's with equal numbers of each of the six colors (red, yellow, green, blue, brown, and orange), what is the likelihood that a randomly chosen handful of seven will contain at least six of the same color? We may assume that the bag is infinite if that helps simplify things; I doubt it would make a noticeable difference except for those dumb "fun-sized" bags (if I'm mistaken here, though, please point it out!)

Bonus points for:

  • Showing the probability that exactly six out of seven will be of the same color
  • Comparing to the probability of drawing six out of six identical m&m's (no "wild" m&m)
  • Generalizing for arbitrary numbers of colors, m&m's drawn, and "wild" (extra) m&m's
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We could assume that the bag is itself is filled at random from an extremely large population where each colour has the same fraction of the population. The answer will be based on the binomial distribution. –  Henry Sep 5 '12 at 23:54

2 Answers 2

up vote 2 down vote accepted

Make the simplifying assumption that we are drawing with replacement. Since it is unlikely that anyone would place back an M&M, more informally suppose we are sampling from a huge bag with equal numbers of each colour.

So we have $7$ trials (picks with replacement). Call getting a yellow on any trial a success. The number of successes has Binomial Distribution, with $n=7$, and probablity $p$ of success on any trial given by $p=\frac{1}{6}$.

Thus the probability of $6$ or $7$ successes (yellows) is $$\binom{7}{6}\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^1+\binom{7}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^0.$$ For your probability, multiply by $6$, since there are $6$ colours. The answer turns out to be $(1/6)^4$. Small, about $7.7\times 10^{-4}$.

If we have $n$ varieties instead of six, a very similar calculation gives a probability of $(7n-6)/n^6$.

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Can you please explain or provide reference for how you reached that probability? Why isn't the probability of getting six of the same color just 11 choose 6, for instance (from Wikipedia)? Also, I don't understand why the probability of getting six yellow or seven yellow is multiplied by 6. Wouldn't it instead be the probability of six yellow, multiplied by six, then adding the probability of seven yellows? –  Malcolm Sep 7 '12 at 22:01
    
@MalcolmMcCrimmon: To the probability of $6$ or $7$ yellow, we must add the probability of $6$ or $7$ green, of $6$ or $7$ red, and so on. There are $6$ colours in all. That's why I multiplied by $6$. As to the probability of $6$ or $7$ yellow, it is the probability of $6$ or $7$ "successes" when success means yellow. This is a straight Binomial distribution calculation, $7$ trials, proability of success on any trial is $1/6$. Will add appropriate Wikipedia link in a few minutes. –  André Nicolas Sep 7 '12 at 22:15
    
Thanks! That made things much clearer. –  Malcolm Sep 7 '12 at 22:28

Given that there are $n$ of each type in the bag, there are $35n\binom{n}{6}$ ways of picking exactly 6 of a given color (red). So the probability of picking up some 6 of the same color out of 7 is $\dfrac{210n\binom{n}6}{\binom{6n}{7}}$, since if you pick 6 of one color, you can't pick 6 of another, so the probabilities add up. As $n \to \infty$, this probability tends to $\frac{35}{6^6}$.

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