## A multinomial example

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)……$\displaystyle \frac{8!}{4! \ 2! \ 2!}=420$

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)……$\displaystyle \frac{6!}{1! \ 1! \ 3!}=60$

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)……$\displaystyle \frac{8!}{4! \ 2! \ 2!} \times \frac{6!}{1! \ 1! \ 3!}=420 \times 60=25200$

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 $6^8=1679616$. 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 $6^8$ 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:

$\displaystyle \frac{307896}{1679616}=\frac{12829}{69984}=0.1833$

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.

.

.

.

.

.

.

.

.

.

.

.

.

The desired probability is:

$\displaystyle \frac{2136336}{6^9}=\frac{2136336}{10077696}=\frac{44507}{209952}=0.212$

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)

Dan Ma statistical

Daniel Ma statistical

Dan Ma practice problems

Daniel Ma practice problems

Daniel Ma mathematics

Dan Ma math

Daniel Ma probability

Dan Ma probability

Daniel Ma statistics

Dan Ma statistics

Dan Ma mathematical

Daniel Ma mathematical

$\copyright$ 2019 – Dan Ma

## Practice problems for order statistics and multinomial probabilities

This post presents exercises on calculating order statistics using multinomial probabilities. These exercises are to reinforce the calculation demonstrated in this blog post.

_____________________________________________________________________________________

Practice Problems

Practice Problems 1
Draw a random sample $X_1,X_2,\cdots,X_{11}$ of size 11 from the uniform distribution $U(0,4)$. Calculate the following:

• $P(Y_4<2
• $P(Y_4<2

$\text{ }$

Practice Problems 2
Draw a random sample $X_1,X_2,\cdots,X_7$ of size 7 from the uniform distribution $U(0,5)$. Calculate the probability $P(Y_4<2<4.

$\text{ }$

Practice Problems 3
Same setting as in Practice Problem 2. Calculate $P(Y_7>4 \ | \ Y_4<2)$ and $P(Y_7>4)$. Compare the conditional probability with the unconditional probability. Does the answer for $P(Y_7>4 \ | \ Y_4<2)$ make sense in relation to $P(Y_7>4)$?

$\text{ }$

Practice Problems 4
Same setting as in Practice Problem 2. Calculate the following:

• $P(Y_4<2
• $P(2
• $P(2
• Does the answer for $P(2 make sense in relation to $P(2?

$\text{ }$

Practice Problems 5
Draw a random sample $X_1,X_2,\cdots,X_6$ of size 6 from the uniform distribution $U(0,4)$. Consider the conditional distribution $Y_3 \ | \ Y_5<2$. Calculate the following:

• $P(Y_3 \le t \ | \ Y_5<2)$
• $f_{Y_3}(t \ | \ Y_5<2)$
• $E(Y_3 \ | \ Y_5<2)$
• $E(Y_3)$

where $0. Compare $E(Y_3)$ and $E(Y_3 \ | \ Y_5<2)$. Does the answer for the conditional mean make sense?

$\text{ }$

Practice Problems 6
Draw a random sample $X_1,X_2,\cdots,X_7$ of size 7 from the uniform distribution $U(0,5)$. Calculate the following:

• $P(Y_4 > 4 \ | \ Y_2>2)$
• $P(Y_4 > 4)$
• Compare the two probabilities. Does the answer for the conditional probability make sense?

$\text{ }$
_____________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$
_____________________________________________________________________________________

Practice Problems 1

• $\displaystyle \frac{11550}{177147}$
• $\displaystyle \frac{18480}{177147}$

$\text{ }$

Practice Problems 2

• $\displaystyle \frac{11088}{78125}$

$\text{ }$

Practice Problems 3

• $\displaystyle P(Y_7>4 \ | \ Y_4<2)=\frac{11088}{22640}$
• $\displaystyle P(Y_7>4)=\frac{61741}{78125}$

$\text{ }$

Practice Problems 4

• $\displaystyle P(Y_4<2
• $\displaystyle P(2
• $\displaystyle P(2

$\text{ }$

Practice Problems 5

• $\displaystyle P(Y_3 \le t \ | \ Y_5<2)=\frac{-10t^6+84t^5-300t^4+400t^3}{448}$
• $\displaystyle f_{Y_3}(t \ | \ Y_5<2)=\frac{-60t^5+420t^4-1200t^3+1200t^2}{448}$
• $\displaystyle E(Y_3 \ | \ Y_5<2)=\frac{55}{49}$
• $\displaystyle E(Y_3)=\frac{84}{49}$

$\text{ }$

Practice Problems 6

• $\displaystyle \frac{3641}{12393}$
• $\displaystyle \frac{2605}{78125}$

$\text{ }$

_____________________________________________________________________________________

$\copyright \ 2015 \text{ by Dan Ma}$

## Calculating the occupancy problem

The occupancy problem refers to the experiment of randomly throwing $k$ balls into $n$ cells. Out of this experiment, there are many problems that can be asked. In this post we focus on question: what is the probability that exactly $j$ of the cells are empty after randomly throwing $k$ balls into $n$ cells, where $0 \le j \le n-1$?

For a better perspective, there are many other ways to describe the experiment of throwing $k$ balls into $n$ cells. For example, throwing $k$ balls into 6 cells can be interpreted as rolling $k$ dice (or rolling a die $k$ times). Throwing $k$ balls into 365 cells can be interpreted as randomly selecting $k$ people and classifying them according to their dates of birth (assuming 365 days in a year). Another context is coupon collecting – the different types of coupons represent the $n$ cells and the coupons being collected represent the $k$ balls.

_____________________________________________________________________________________

Practice Problems

Let $X_{k,n}$ be the number of cells that are empty (i.e. not occupied) when throwing $k$ balls into $n$ cells. As noted above, the problem discussed in this post is to find the probability function $P(X_{k,n}=j)$ where $j=0,1,2,\cdots,n-1$. There are two elementary ways to do this problem. One is the approach of using double multinomial coefficient (see this post) and the other is to use a formula developed in this post. In addition to $X_{k,n}$, let $Y_{k,n}$ be the number of cells that are occupied, i.e. $Y_{k,n}=n-X_{k,n}$.

Practice Problems
Compute $P(X_{k,n}=j)$ where $j=0,1,2,\cdots,n-1$ for the following pairs of $k$ and $n$.

1. $k=5$ and $n=5$
2. $k=6$ and $n=5$
3. $k=7$ and $n=5$
4. $k=6$ and $n=6$
5. $k=7$ and $n=6$
6. $k=8$ and $n=6$

We work the problem for $k=7$ and $n=5$. We show both the double multinomial coefficient approach and the formula approach. Recall the $k$ here is the number of balls and $n$ is the number of cells.

_____________________________________________________________________________________

Example – Double Multinomial Coefficient

Note that the double multinomial coefficient approach calculate the probabilities $P(Y_{7,5}=j)$ where $Y_{7,5}$ is the number of occupied cells when throwing 7 balls into 5 cells. In throwing 7 balls into 5 cells, there is a total of $5^7=$ 78125 many ordered samples. To calculate $P(Y_{7,5}=j)$, first write down the representative occupancy sets for the event $Y_{7,5}=j$. For each occupancy set, calculate the number of ordered samples (out of 78125) that belong to that occupancy set. Then we add up all the counts for all the occupancy sets for $Y_{7,5}=j$. This is best illustrated with an example. To see the development of this idea, see this post.

Fist, consider the event of $Y_{7,5}=1$ (only one cell is occupied, i.e. all the balls going into one cell). A representative occupancy set is (0, 0, 0, 0, 7), all the balls going into the 5th cell. The first multinomial coefficient is on the 5 cells and the second multinomial coefficient is on the 7 balls.

(0, 0, 0, 0, 7)

$\displaystyle \frac{5!}{4! 1!} \times \frac{7!}{7!}=5 \times 1 =5$

Total = 5

So there are 5 ordered samples out of 16807 that belong to the event $Y_{7,5}=1$. Note that the first multinomial coefficient is the number of to order the 5 cells where 4 of the cells are empty and one of the cells has 7 balls. The second multinomial coefficient is the number of ways to order the 7 balls where all 7 balls go into the 5th cell.

Now consider the event $Y_{7,5}=2$ (all 7 balls go into 2 cells). There are three representative occupancy sets. The following shows the calculation for each set.

(0, 0, 0, 1, 6)

$\displaystyle \frac{5!}{3! 1! 1!} \times \frac{7!}{1! 6!}=20 \times 7 =140$
(0, 0, 0, 2, 5)

$\displaystyle \frac{5!}{3! 1! 1!} \times \frac{7!}{2! 5!}=20 \times 21 =420$
(0, 0, 0, 3, 4)

$\displaystyle \frac{5!}{3! 1! 1!} \times \frac{7!}{3! 4!}=20 \times 35 =700$

Total = 140 + 420 + 700 = 1260

Here’s an explanation for the occupancy set (0, 0, 0, 3, 4). The occupancy set refers to the scenario that 3 of the 7 balls go into the 4th cell and 4 of the 7 balls go into the 5th cell. But we want to count all the possibilities such that 3 of the 7 balls go into one cell and 4 of the 7 balls go into another cell. Thus the first multinomial coefficient count the number of ways to order 5 cells where three of the cells are empty and one cell has 3 balls and the remaining cell has 4 balls (20 ways) The second multinomial coefficient is the number of ways to order the 7 balls where 3 of the 7 balls go into the 4th cell and 4 of the 7 balls go into the 5th cell (35 ways). So the total number of possibilities for the occupancy set (0, 0, 0, 3, 4) is 20 times 35 = 700. The sum total for three occupancy sets is 1260.

We now show the remaining calculation without further elaboration.

Now consider the event $Y_{7,5}=3$ (all 7 balls go into 3 cells). There are four representative occupancy sets. The following shows the calculation for each set.

(0, 0, 1, 1, 5)

$\displaystyle \frac{5!}{2! 2! 1!} \times \frac{7!}{1! 1! 5!}=30 \times 42 =1260$
(0, 0, 1, 2, 4)

$\displaystyle \frac{5!}{2! 1! 1! 1!} \times \frac{7!}{1! 2! 4!}=60 \times 105 =6300$
(0, 0, 1, 3, 3)

$\displaystyle \frac{5!}{2! 1! 2!} \times \frac{7!}{1! 3! 3!}=30 \times 140 =4200$
(0, 0, 2, 2, 3)

$\displaystyle \frac{5!}{2! 2! 1!} \times \frac{7!}{2! 2! 3!}=30 \times 210 =6300$

Total = 1260 + 6300 + 4200 + 6300 = 18060

Now consider the event $Y_{7,5}=4$ (all 7 balls go into 4 cells, i.e. one empty cell). There are three representative occupancy sets. The following shows the calculation for each set.

(0, 1, 1, 1, 4)

$\displaystyle \frac{5!}{1! 3! 1!} \times \frac{7!}{1! 1! 1! 4!}=20 \times 210 =4200$
(0, 1, 1, 2, 3)

$\displaystyle \frac{5!}{1! 2! 1! 1!} \times \frac{7!}{1! 1! 2! 3!}=60 \times 420 =25200$
(0, 1, 2, 2, 2)

$\displaystyle \frac{5!}{1! 1! 3!} \times \frac{7!}{1! 2! 2! 2!}=20 \times 630 =12600$

Total = 4200 + 25200 + 12600 = 42000

Now consider the event $Y_{7,5}=4$ (all 7 balls go into 5 cells, i.e. no empty cell). There are two representative occupancy sets. The following shows the calculation for each set.

(1, 1, 1, 1, 3)

$\displaystyle \frac{5!}{4! 1!} \times \frac{7!}{1! 1! 1! 1! 3!}=5 \times 840 =4200$
(1, 1, 1, 2, 2)

$\displaystyle \frac{5!}{3! 2!} \times \frac{7!}{1! 1! 1! 2! 2!}=10 \times 1260 =12600$

Total = 4200 + 12600 = 16800

The following is the distribution for the random variable $Y_{7,5}$.

$\displaystyle P(Y_{7,5}=1)=\frac{5}{78125}=0.000064$

$\displaystyle P(Y_{7,5}=2)=\frac{1260}{78125}=0.016128$

$\displaystyle P(Y_{7,5}=3)=\frac{18060}{78125}=0.231168$

$\displaystyle P(Y_{7,5}=4)=\frac{42000}{78125}=0.5376$

$\displaystyle P(Y_{7,5}=5)=\frac{16800}{78125}=0.21504$

3.951424

Remarks
In throwing 7 balls at random into 5 cells, it is not like that the balls are in only one or two cells (about 1.6% chance). The mean number of occupied cells is about 3.95. More than 50% of the times, 4 cells are occupied.

The above example has small numbers of balls and cells and is an excellent example for practicing the calculation. Working such problems can help build the intuition for the occupancy problem. However, when the numbers are larger, the calculation using double multinomial coefficient can be lengthy and tedious. Next we show how to use a formula for occupancy problem using the same example.

_____________________________________________________________________________________

Example – Formula Approach

The formula we use is developed in this post. Recall that $X_{k,n}$ is the number of empty cells when throwing $k$ balls into $n$ cells. The formula calculates the probabilities $P(X_{7,5}=j)$ where
$0 \le j \le 5$. The first step is to compute the probabilities $P(X_{7,m}=0)$ for $m=5,4,3,2,1$. Each of these is the probability that all m cells are occupied (when throwing 7 balls into $m$ cells). These 5 probabilities will be used to calculate $P(X_{7,5}=j)$. This is a less direct implementation of the formula but gives a more intuitive explanation.

\displaystyle \begin{aligned} P(X_{7,5}=0)&=1-P(X_{7,5} \ge 1) \\&=1-\sum \limits_{j=1}^5 (-1)^{j+1} \binom{5}{j} \biggl[ 1-\frac{j}{5} \biggr]^7 \\&=1-\biggl[5 \biggl(\frac{4}{5}\biggr)^7-10 \biggl(\frac{3}{5}\biggr)^7+10 \biggl(\frac{2}{5}\biggr)^7 -5 \biggl(\frac{1}{5}\biggr)^7 + 0 \biggr] \\&=1-\frac{81920-21870+1280-5}{78125} \\&=\frac{16800}{78125} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,4}=0)&=1-P(X_{7,4} \ge 1) \\&=1-\sum \limits_{j=1}^4 (-1)^{j+1} \binom{4}{j} \biggl[ 1-\frac{j}{4} \biggr]^7 \\&=1-\biggl[4 \biggl(\frac{3}{4}\biggr)^7-6 \biggl(\frac{2}{4}\biggr)^7+4 \biggl(\frac{1}{4}\biggr)^7 - 0 \biggr] \\&=1-\frac{8748-768+4}{16384} \\&=\frac{8400}{16384} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,3}=0)&=1-P(X_{7,3} \ge 1) \\&=1-\sum \limits_{j=1}^3 (-1)^{j+1} \binom{3}{j} \biggl[ 1-\frac{j}{3} \biggr]^7 \\&=1-\biggl[3 \biggl(\frac{2}{3}\biggr)^7-3 \biggl(\frac{1}{3}\biggr)^7+0 \biggr] \\&=1-\frac{384-3}{2187} \\&=\frac{1806}{2187} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,2}=0)&=1-P(X_{7,2} \ge 1) \\&=1-\sum \limits_{j=1}^2 (-1)^{j+1} \binom{2}{j} \biggl[ 1-\frac{j}{2} \biggr]^7 \\&=1-\biggl[2 \biggl(\frac{1}{2}\biggr)^7-0 \biggr] \\&=1-\frac{2}{128} \\&=\frac{126}{128} \end{aligned}

$P(X_{7,1}=0)=1$

Each of the above 5 probabilities (except the last one) is based on the probability $P(X_{7,m} \ge 1)$, which is the probability that there is at least one cell that is empty when throwing 7 balls into $m$ cells. The inclusion-exclusion principle is used to derive $P(X_{7,m} \ge 1)$. The last of the five does not need calculation. When throwing 7 balls into one cell, there will be no empty cells.

Now the rest of the calculation:

$\displaystyle P(X_{7,5}=0)=\frac{16800}{78125}$

\displaystyle \begin{aligned} P(X_{7,5}=1)&=P(\text{1 empty cell}) \times P(\text{none of the other 4 cells is empty}) \\&=\binom{5}{1} \biggl(1-\frac{1}{5} \biggr)^7 \times P(X_{7,4}=0) \\&=5 \ \frac{16384}{78125} \times \frac{8400}{16384} \\&=\frac{42000}{78125} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,5}=2)&=P(\text{2 empty cells}) \times P(\text{none of the other 3 cells is empty}) \\&=\binom{5}{2} \biggl(1-\frac{2}{5} \biggr)^7 \times P(X_{7,3}=0) \\&=10 \ \frac{2187}{78125} \times \frac{1806}{2187} \\&=\frac{18060}{78125} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,5}=3)&=P(\text{3 empty cells}) \times P(\text{none of the other 2 cells is empty}) \\&=\binom{5}{3} \biggl(1-\frac{3}{5} \biggr)^7 \times P(X_{7,2}=0) \\&=10 \ \frac{128}{78125} \times \frac{126}{128} \\&=\frac{1260}{78125} \end{aligned}

\displaystyle \begin{aligned} P(X_{7,5}=4)&=P(\text{4 empty cells}) \times P(\text{the other 1 cell is not empty}) \\&=\binom{5}{4} \biggl(1-\frac{4}{5} \biggr)^7 \times P(X_{7,1}=0) \\&=5 \ \frac{1}{78125} \times 1 \\&=\frac{5}{78125} \end{aligned}

Note that these answers agree with the ones from the double multinomial coefficient approach after making the adjustment $Y_{7,5}=5-X_{7,5}$.

_____________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$
_____________________________________________________________________________________

Problem 1
5 balls into 5 cells

$\displaystyle P(X_{5,5}=0)=P(Y_{5,5}=5)=\frac{120}{3125}$

$\displaystyle P(X_{5,5}=1)=P(Y_{5,5}=4)=\frac{1200}{3125}$

$\displaystyle P(X_{5,5}=2)=P(Y_{5,5}=3)=\frac{1500}{3125}$

$\displaystyle P(X_{5,5}=3)=P(Y_{5,5}=2)=\frac{300}{3125}$

$\displaystyle P(X_{5,5}=4)=P(Y_{5,5}=1)=\frac{5}{3125}$

Problem 2
6 balls into 5 cells

$\displaystyle P(X_{6,5}=0)=P(Y_{6,5}=5)=\frac{1800}{15625}$

$\displaystyle P(X_{6,5}=1)=P(Y_{6,5}=4)=\frac{7800}{15625}$

$\displaystyle P(X_{6,5}=2)=P(Y_{6,5}=3)=\frac{5400}{15625}$

$\displaystyle P(X_{6,5}=3)=P(Y_{6,5}=2)=\frac{620}{15625}$

$\displaystyle P(X_{6,5}=4)=P(Y_{6,5}=1)=\frac{5}{15625}$

Problem 3
7 balls into 5 cells. See above.

Problem 4
6 balls into 6 cells

$\displaystyle P(X_{6,6}=0)=P(Y_{6,6}=6)=\frac{720}{46656}$

$\displaystyle P(X_{6,6}=1)=P(Y_{6,6}=5)=\frac{10800}{46656}$

$\displaystyle P(X_{6,6}=2)=P(Y_{6,6}=4)=\frac{23400}{46656}$

$\displaystyle P(X_{6,6}=3)=P(Y_{6,6}=3)=\frac{10800}{46656}$

$\displaystyle P(X_{6,6}=4)=P(Y_{6,6}=2)=\frac{930}{46656}$

$\displaystyle P(X_{6,6}=5)=P(Y_{6,6}=1)=\frac{6}{46656}$

Problem 5
7 balls into 6 cells

$\displaystyle P(X_{7,6}=0)=P(Y_{7,6}=6)=\frac{15120}{279936}$

$\displaystyle P(X_{7,6}=1)=P(Y_{7,6}=5)=\frac{100800}{279936}$

$\displaystyle P(X_{7,6}=2)=P(Y_{7,6}=4)=\frac{126000}{279936}$

$\displaystyle P(X_{7,6}=3)=P(Y_{7,6}=3)=\frac{36120}{279936}$

$\displaystyle P(X_{7,6}=4)=P(Y_{7,6}=2)=\frac{1890}{279936}$

$\displaystyle P(X_{7,6}=5)=P(Y_{7,6}=1)=\frac{6}{279936}$

Problem 6
8 balls into 6 cells

$\displaystyle P(X_{8,6}=0)=P(Y_{8,6}=6)=\frac{191520}{1679616}$

$\displaystyle P(X_{8,6}=1)=P(Y_{8,6}=5)=\frac{756000}{1679616}$

$\displaystyle P(X_{8,6}=2)=P(Y_{8,6}=4)=\frac{612360}{1679616}$

$\displaystyle P(X_{8,6}=3)=P(Y_{8,6}=3)=\frac{115920}{1679616}$

$\displaystyle P(X_{8,6}=4)=P(Y_{8,6}=2)=\frac{3810}{1679616}$

$\displaystyle P(X_{8,6}=5)=P(Y_{8,6}=1)=\frac{6}{1679616}$

_____________________________________________________________________________________

$\copyright \ 2015 \text{ by Dan Ma}$

## A Problem of Rolling Six Dice

Problem
Suppose that we roll 6 fair dice (or equivalently, roll a fair die 6 times). Let $X$ be the number of distinct faces that appear. Find the probability function $P(X=k)$ where $k=1,2,3,4,5,6$.

Equivalent Problem
Suppose that we randomly assign 6 candies to 6 children (imagine that each candy is to be thrown at random to the children and is received by one of the children). What is the probability that exactly $k$ children have been given candies, where $k=1,2,3,4,5,6$?

___________________________________________________________________________

Discussion
Note that both descriptions are equivalent and are refered to as occupancy problem in [1]. The essential fact here is that $n$ objects are randomly assigned to $m$ cells. The problem then asks: what is the probability that $k$ of the cells are occupied? See the following posts for more detailed discussions of the occupancy problem.

Each of these posts presents different different ways of solving the occupancy problem. The first post uses a counting approach based on the multinomial coefficients. The second post developed a formula for finding the probability that exactly $k$ of the cells are empty.

The first approach of using mulltinomial coefficients is preferred when the number of objects $n$ and the number of cells $m$ are relatively small (such as the problem indicated here). Otherwise, use the formula approach.

___________________________________________________________________________

Using the approach of multinomial coefficients as shown in this post (the first post indicated above), we have the following answers:

$\displaystyle P(X=1)=\frac{6}{6^6}=\frac{6}{46656}$

$\displaystyle P(X=2)=\frac{930}{6^6}=\frac{930}{46656}$

$\displaystyle P(X=3)=\frac{10800}{6^6}=\frac{10800}{46656}$

$\displaystyle P(X=4)=\frac{23400}{6^6}=\frac{23400}{46656}$

$\displaystyle P(X=5)=\frac{10800}{6^6}=\frac{10800}{46656}$

$\displaystyle P(X=6)=\frac{720}{6^6}=\frac{720}{46656}$

For more practice problems on calculating the occupancy problem, see this post.
___________________________________________________________________________

Reference

1. Feller, W., An Introduction to Probability Theory and its Applications, Vol. I, 3rd ed., John Wiley & Sons, Inc., New York, 1968