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The wikipedia page for the multinomial distribution says it can represent the probability of counts for each side of a $k$-sided dice rolled $n$ times. But this StackExchange answer says the same quantity is counted using stars and bars. Which is right, or how are the quantities being counted differently in a way I'm not seeing?

To me it seems that the counts for each side of a $k$ sided dice rolled $n$ times should be done using stars and bars as per the logic of the SE answer, and not the multinomial coefficient.

The application of the multinomial coefficient $\frac{n}{x_1!x_2!...x_k!}$ that I'm aware of is counting counting the number of strings with repeated letters, e.g. the number of rearrangements of MISSISSIPPI is $\frac{11!}{1!4!4!2!}$.

If think of the $i$th letter as indicating the outcome of the $i$th trial, with $k$ possible letters for $k$ different possible outcomes, then these strings represent sequences of outcomes of each trial. The multinomial coefficient counts these sequences, but shouldn't we be counting the number of occurrences of each outcome while ignoring the order? For example, the multinomial coefficient distinguishes between the sequences $abca$ and $aabc$. To be invariant to the order, we need to use stars and bars?

xojfqa
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  • The wp page later says multinomial distribution gives the probability of any particular combination of numbers, so the opening statement is a bit ambiguous. It's looking for a specific combination of numbers when rolling multiple k-sided dice – Barry Carter Aug 04 '22 at 13:18
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    How many ways are there for me to play a game of monopoly with my family? Two ways, either I win or I lose. Or we could have said four ways: I win, my dad wins, my sister wins, or my mom wins. Or we could have said sixteen ways, I won and went first, I won but went second, I won but went third... My mom won but went last. Or we could have said a few hundred ways: I won and my first property I bought was Mediterranean etc... or I could have said it was a stupendously large number: the exact turn sequence was ____. All of these are perfectly fine ways of counting the number of possibilities – JMoravitz Aug 04 '22 at 13:47
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    What information and what level of specificity we choose to use depends on context. Are we just trying to keep a loose idea of a leader board to determine who the better player is? Or are we trying to train an AI to play monopoly as well as a human and so knowing which properties to purchase to help it win is important? If we are wanting to ask probability questions about this then choosing a sample space that lends itself well to calculations is important. There is a great deal of freedom here and while there may be wrong answers, there are many right answers too. – JMoravitz Aug 04 '22 at 13:50

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Note that the Wikipedia article talks about the probability of counts for each side of a $k$-sided dice rolled $n$ times. When you roll a die, its chances of landing in each of $1-6$ is equiprobable, and the multinomial coefficient will help to give the probability for a particular pattern of throws.

For example, a pattern of $2,1,2,1,1,0,1,2,$ will have a probability of $\dfrac{\dbinom{10}{2,1,2,1,1,0,1,2}}{6^{10}}$

Stars and bars instead gives all possible ways of filling $1-6$, but these are not equiprobable, eg you can understand that if you roll a die $10$ times, getting all ten as $6's$ is much less probable than a more even distribution

true blue anil
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Consider a very simple example, $k = n = 2.$ This is equivalent to tossing a coin twice, where we consider a coin to be a "two-sided" die with a $1$ on the tails side and $2$ on the heads side.

According to the stars-and-bars interpretation, there are only three outcomes to be counted: two tails, two heads, or one tail and one head.

According to the multinomial interpretation, there are four outcomes to be counted:

  • Tails on both tosses.
  • Tails on the first toss, heads on the second toss.
  • Heads on the first toss, tails on the second toss.
  • Heads on both tosses.

The difference is that the stars-and-bars interpretation is only counting the number of times each face showed without regard to when it showed.

In practice, a coin flipped two times has a $1/4$ probability to show heads both times, not a $1/3$ probability. This is consistent with the notion that each face is equally likely on the first toss, each face is equally likely on the second toss, and the face that shows on the second toss is independent of the face that shows on the first toss. Hence if we want to count equally-likely outcomes, we use the multinomial interpretation.

I note that you got your "stars and bars" from the Stack Exchange page How many ways can we write $N$ as a sum of $K$ strictly positive numbers? It's worth pointing out that the information on this page is completely correct: stars and bars give you the number of ways you can write $N$ as a sum of $K$ strictly positive integers. What is not correct is your inference that this has anything to do with the probabilities of a set of $K$ dice rolled some number of times. For example, there is exactly one way to make the sum $2$ and exactly one way to make the sum $3,$ but when you roll two six-sided dice the probability of the sum $3$ is twice the probability of the sum $2$.

David K
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  • Thank you, this makes sense. The first answer on that linked SE thread gives the setting "Suppose a D6 is rolled ten times with all 6 numbers appearing at least once" and uses stars and bars to answer it, hence my inference that this is relevant to dice rolls. But I think they were only counting the number of possible sequences of outcomes, whereas you're saying that we don't want to just count each sequence once, we want to count each sequence in proportion to the number of ways it can arise, since that's precisely what we need in a probability setting? Have I misrepresented your argument? – xojfqa Aug 04 '22 at 14:17
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    Usually I would say stars and bars doesn't even count the sequences of the numbers that show on the die. Applying it to a coin tossed three times, stars and bars would summarize the three sequences TTH, THT, and HTT as a single outcome, "two tails and one head". If you define the "sequence" as "number of tails, number of heads", however, then TTH, THT, and HTT are all the same sequence, $(2,1),$ so yes, in that sense stars and bars counts the sequence once whereas it can occur in three ways. – David K Aug 04 '22 at 15:59