The multinomial theorem is a useful way to count. The counting problems discussed here are generalization to counting problems that are solved by using binomial techniques (see this previous post for an example).
The best way to start is the example discussed in the previous post:

Seven dice are rolled. Find the probability that at least 4 of the dice show the same face.
The solution in the previous post uses the binomial distribution. Binomial solution is possible because in rolling 7 dice, one and only one face can appear 4 or more times. So in this example, there are just two categories to keep track of in rolling a die – is it the value x or a value other than x. Use the binomial distribution to count these possibilities. Then multiply by 6 to get the answer.
If we roll 8 dice instead of 7 dice, this method cannot be used. There can be more than two categories to keep track of. For example, in rolling 8 dice, it is possible that two faces can show up 4 times (e.g. face 1 showing up 4 times and face 2 showing up 4 times). So this is a multinomial counting problem instead of a binomial counting problem. We demonstrate how multinomial theorem is used to do the counting.
Before we do so, we observe that the more dice are rolled, the higher the probability of having at least 4 of the dice showing the same face. As an extreme example, if we roll 100 dice, it is certain that at least 4 of the dice will show the same face. In fact, in rolling 100 dice there is a 100% chance that at least 34 of the dice show the same face.
Working Toward a Multinomial Solution
Here’s the problem. Eight fair dice are rolled. Find the probability that at least 4 of the dice show the same face.
Let’s focus on one specific outcome in rolling 8 dice.

(4, 2, 2, 0, 0, 0)
The above outcome means that 4 dice show the value of 1, 2 dice show the value of 2 and 2 dice shows the value of 3. The number of ways this can happen is a multinomial coefficient.
(1)……
The outcome (4, 2, 2, 0, 0, 0) is one example of 4 dice showing 1 value, 2 dice showing another value and 2 dice showing another value. The above multinomial coefficient says that there are 420 ways the outcome (4, 2, 2, 0, 0, 0) can happen when 8 dice are rolled. In fact, the outcome (0, 0, 0, 2, 2, 4) – 4 dice shows the value of 6, 2 dice show the value of 5 and 2 dice shows the value of 4 – also associates with 420, that there are 420 ways this outcome can happen.
The two outcomes (4, 2, 2, 0, 0, 0) and (0, 0, 0, 2, 2, 4) share something in common. That is, 1 of the value of the die appearing one time, 2 values of the die appearing two times, and the remaining 3 values of the die not appearing at all. How many ways can the 6 values of the die permute in this way? The answer is another multinomial coefficient.
(2)……
Multiplying the two multinomial coefficients together gives the number of ways, in rolling 8 dice, the result “4 dice shows one value, 2 dice show another value and 2 dice shows another value” can happen.
(3)……
Let’s summarize what we have done so far. We start with the outcome (4, 2, 2, 0, 0, 0), which is short hand for 4 of the dice showing the value of 1, 2 of the dice showing the value of 2 and 2 of the dice showing the value of 3. The number of ways this can happen is 420, which is the multinomial coefficient calculated in (1). The multinomial coefficient calculated in (2) is the number of the 6 positions (6 values of the die) can permute according to the criterion: 1 of the values appearing one time, two of the values appearing two times each and three of the values do not appear at all. The product of (1) and (2) is the number of ways 4 dice show one value, 2 dice show another value and 2 dice show another value when rolling 8 dice.
The multinomial counting process discussed here is a double application of multinomial coefficients, with the first one on the rolls of the dice and the second on on the 6 values of the die.
A Multinomial Solution
The key in solving the full problem is to list out all the different outcomes in addition to (4, 2, 2, 0, 0, 0). The following is the listing of all the possibilities.
Outcome  

A  (4, 3, 1, 0, 0, 0) 
B  (4, 2, 2, 0, 0, 0) 
C  (4, 2, 1, 1, 0, 0) 
D  (4, 1, 1, 1, 1, 0) 
E  (5, 3, 0, 0, 0, 0) 
F  (5, 2, 1, 0, 0, 0) 
G  (5, 1, 1, 1, 0, 0) 
H  (6, 2, 0, 0, 0, 0) 
I  (6, 1, 1, 0, 0, 0) 
J  (7, 1, 0, 0, 0, 0) 
K  (8, 0, 0, 0, 0, 0) 
L  (4, 4, 0, 0, 0, 0) 
The table shows 12 possible outcomes. Note that the outcome (4, 2, 2, 0, 0, 0) discussed in the preceding section is outcome B in the table. The first 11 items in the table are outcomes that have only one face showing 4 or more times. The last item is the outcome that has two faces showing 4 times each. As shown in the preceding section, we calculate two multinomal coefficients and multiply them together. The first multinomial coefficient, as demonstrated in (1), is the number of ways the particular outcome can happen. The second multinomial coefficient, as demonstrated in (2), is the number of ways the 6 values of the die can permute similar to that specific outcome. The following table gives the calculated results.
Outcome  Count  

A  (4, 3, 1, 0, 0, 0)  33600 
B  (4, 2, 2, 0, 0, 0)  25200 
C  (4, 2, 1, 1, 0, 0)  151200 
D  (4, 1, 1, 1, 1, 0)  50400 
E  (5, 3, 0, 0, 0, 0)  1680 
F  (5, 2, 1, 0, 0, 0)  20160 
G  (5, 1, 1, 1, 0, 0)  20160 
H  (6, 2, 0, 0, 0, 0)  840 
I  (6, 1, 1, 0, 0, 0)  3360 
J  (7, 1, 0, 0, 0, 0)  240 
K  (8, 0, 0, 0, 0, 0)  6 
L  (4, 4, 0, 0, 0, 0)  1050 
Total  307896 
In rolling 8 dice, how many possible outcomes are there? The answer is . This is due to the fact that each roll of a die has 6 outcomes. According to the multiplication principal, rolling 8 dice would have in total outcomes. Out of 1,679,616 outcomes, 307,896 of them are such that at least 4 of the dice show the same face. Thus in rolling 8 fair dice, the probability that at least 4 of the dice show the same face is:
When rolling a 8 fair dice, there is roughly an 18.33% chance that at least 4 of the dice showing the same face.
Practice Problem
To reinforce the concept of using a double application of multinomial coefficients, it is a good idea to work a practice problem. Naturally, we can just up the dice count by one. Here’s the problem: Nine fair dice are rolled. Find the probability that at least 4 of the dice show the same face.
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Answer to Practice Problem
The desired probability is:
The answer is based on a total of 16 outcomes.
Outcome  

A  (4, 3, 2, 0, 0, 0) 
B  (4, 2, 2, 1, 0, 0) 
C  (4, 2, 1, 1, 1, 0) 
D  (4, 1, 1, 1, 1, 1) 
E  (5, 3, 1, 0, 0, 0) 
F  (5, 2, 2, 0, 0, 0) 
G  (5, 1, 1, 1, 1, 0) 
H  (6, 3, 0, 0, 0, 0) 
I  (6, 2, 1, 0, 0, 0) 
J  (6, 1, 1, 1, 0, 0) 
K  (7, 2, 0, 0, 0, 0) 
L  (7, 1, 1, 0, 0, 0) 
M  (8, 1, 0, 0, 0, 0) 
N  (9, 0, 0, 0, 0, 0) 
O  (5, 4, 0, 0, 0, 0) 
P  (4, 4, 1, 0, 0, 0) 
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