What is the probability that in a box of a dozen donuts picked from 14 flavors there’s no more than 3 flavors in the box?

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Problem

Dave’s Donuts offers 14 flavors of donuts (consider the supply of eachflavor as being unlimited). The “grab bag” box consists of flavorsrandomly selected to be in the box, each flavor equally likely for eachone of the dozen donuts. What is the probability that at most three flavors arein the grab bag box of a dozen?

Solution

For this we will need the multinomial distribution, which is a discreteprobability distribution. In a string of characters there are characterspossible to fill one position of the string, which is characters long. Therandom variable counts the number of occurences of character 1 in thestring, the number of occurences of character 2, and so on until .Let be the individual probability each of the characterscould appear in a position of the string; each position is filled independentlyof the characters in other positions. Let such that . Then

Here, , , and . So we can say

(1)

We will say

Compute each of those probabilities separately.

If and , there is exactly one flavor in the box.(1) shows the probability this happens is . Sincewe could pick an and there were 14 ways to make this decision, we cansay

(1)

Let’s now compute . We start by fixing. We get

(2)

Unfortunately (2) includes cases where there’sactually only one flavor present in the box, so compute

		(3)
		(4)

Of course we could have picked different variables to fix at zero, andthere were ways to pick the variables to fix at zero (orequivalently, pick the variables to not fix at zero), finally yielding

(5)

Now to compute . Again we start by fixing and compute

(6)

We could try and use tricks to compute (6) or we can acknowledge that we’rebusy people and ask SymPy to do it. Check that the following Python code iscorrect:

from sympy import init_session, binomial
init_session()

def multinomial(params):
    if len(params) == 1:
        return 1
    return binomial(sum(params), params[-1]) * \
                    multinomial(params[:-1])

l1 = list()
for i in range(1, 10 + 1):
    v = sum([multinomial([i, j, (12 - i - j)]) for j in range(1,
                                                    11 - i + 1)])
    l1.append(v)

sum(l1)/14**12    # Solution

The resulting probability is . We could havepicked different flavors to fix, and there were ways to pickthe flavors to fix, so we get

(7)

We can write (1) and (5) as and , respectively.Summing these probabilities yields

(8)

Related Question

This is the proper way to obtain the probability that there are at most threeflavors in the “grab bag” box, but how many boxes exist in which there are atmost three flavors when we discount the number of ways there are to arrange thedonuts in a box?

If there’s exactly one flavor, then we pick it and fill the box with thatflavor; there’s 14 ways to pick one flavor. If there’s exactly two flavors inthe box, we’ll call them Flavor 1 and Flavor 2. There is at least one donut ofFlavor 1 and one of Flavor 2. Now pick the rest of the donuts’ flavors, orderdoesn’t matter, there is replacement; there are ways to do that. Then pick the two flavors: there’s ways to do that, and thus boxes with exactly twoflavors. Similarly, for exactly three flavors, there are ways for there to be exactlythree flavors. Sum these numbers. (See https://math.stackexchange.com/q/3230011.) There are 21,035 such boxes.

Special thanks to Math Stack Exchange user wavex for his help with this problem! He provided the following R script for simulating it:

total = 0
for (y in 1:10000000){
  x = rmultinom(1,12,c(1/14,1/14,1/14,1/14,1/14,
                       1/14,1/14,1/14,1/14,1/14,1/14,1/14,1/14,1/14))
  x <- c(x)

  count = 14
  for(i in x){
    if(i==0){
     count = count -1
   }   

  }
 if(count <= 3){total = total + 1}
}
sprintf("%.20f", total / 10000000)

In his run of the code this event occured only 27 out of 10,000,000 times. A rare event indeed!

About this document …

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