Odds of getting equal amount of each color in a bag of M&Ms

Question

Assume they were mass produced by Mars in equal amounts of all 6 colors. Assume a bag contains 54 M&Ms. What is the probability that you get a bag with EXACTLY 9 of each color?

Answer 1

Compute the chance of getting exactly $9$ of the first color with $54$ draws with replacement. You have $45$ left that are of $5$ colors. Compute the chance of getting exactly $9$ of the second color within this group of $45$. Keep going.

Answer 2

You have a very big mountain M&M´s. The mountain contains 6 colors of M&M´s, with (approximately) equal proportion. If you pick some M&M´s we can assume that the probability of picking a specific color doesn´t change since the population is very large.

Therefore the probability to pick one color of M&M is constant $\frac16$. For this case you can use the multinomial distribution. The probability to pick exactly $9$ M&M´s from each color is

$P(X_1=9, ..., X_6=9)=\frac{54!}{9!\cdot 9!\cdot 9!\cdot 9!\cdot 9!\cdot 9!}\cdot \left(\frac{1}{6}\right)^9\cdot \left(\frac{1}{6}\right)^9\cdot\left(\frac{1}{6}\right)^9\cdot\left(\frac{1}{6}\right)^9\cdot\left(\frac{1}{6}\right)^9\cdot\left(\frac{1}{6}\right)^{9}$

$=\frac{54!}{9!\cdot 9!\cdot 9!\cdot 9!\cdot 9!\cdot 9!}\cdot \left(\frac{1}{6}\right)^{54}\approx 0.00009651=9.651\cdot 10^{-5}$

You see the probability is not very large. The situation might looks like below

Answer 3

This answer is partially in response to the comment by Michael Lugo above, and partially just a slightly more in depth answer.

Since Mars created equal numbers of each M&M, suppose they created $n$ of each color. You are choosing 54 M&Ms from the collection of $6n$ M&Ms that were created. But, you are looking for the outcome where you have exactly 9 of each color. So, that probability would be:

$$\begin{align*}\dfrac{\dbinom{n}{9}^6}{\dbinom{6n}{54}} & = \dfrac{\dfrac{n^6(n-1)^6(n-2)^6(n-3)^6(n-4)^6(n-5)^6(n-6)^6(n-7)^6(n-8)^6}{(9!)^6}}{\dfrac{6n(6n-1)\cdots (6n-53)}{54!}} \\ & = \dfrac{54!}{(9!)^6}\dfrac{n^{54}+O(n^{53})}{(6n)^{54} + O(n^{53})} \\ & = \dfrac{54!}{(9!)^6}\cdot \dfrac{1}{6^{54}}\left(\dfrac{n^{54}+O(n^{53})}{n^{54}+O(n^{53})}\right)\end{align*}$$

Taking the limit as $n \to \infty$ gets the answer that callculus gave. But, for small $n$, the probability is actually greater that you will get 9 of each color. But this is a decreasing function of $n$. If $n=9$, the probability is 1. Once $n>10000$, the estimate callculus's approach gives is accurate within two decimal places (to $10^{-7}$).