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I am looking at Sextus's answer in Balls are placed into 3 urns. Expected time until some urn has 100 balls..

Essentially, we have $n$ balls that we'd like to place in $m$ buckets. There is a set of constraints on the number of balls in each bucket. Let $k_i$ denote the number of balls in the $i$-th bucket, we require

$$ \sum_{i=1}^m k_i = n \\ 0 \leq k_i < k $$

for some constant $k$.

Then his answer states that "the number of ways to put $k_i$ balls in urn/bucket $i$" is

$$ \frac{n!}{\prod_{i=1}^m k_i!} $$

I am confused. $k_i$ is a constant, so isn't the number of ways to put $k_i$ balls in to bucket $i$ simply 1 since the balls are indistinguishable?

RobPratt
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roulette01
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  • What is meant is the number of distributions of the $n$ balls that for each $i\in{1,\ldots,m}$ have $k_i$ balls in bucket $i$. – Brian M. Scott Apr 17 '21 at 02:51
  • @BrianM.Scott Sorry, I don't quite understand. Is that different from "the number of ways to put $k_i$ balls in bucket $i$? " – roulette01 Apr 17 '21 at 02:55
  • @BrianM.Scott And are the balls here indistinguishable or distinguishable? – roulette01 Apr 17 '21 at 02:59
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    Yes, definitely. It’s the number of ways to put $k_1$ balls in bucket $1$ and $k_2$ balls in bucket $2$ and … and $k_m$ balls in bucket $m$. We imagine putting the balls intu the buckets one at a time, so they are individually identifiable: there is a first ball, a second ball, and so on; that’s why we have $n!$ in the numerator. But we don’t care which $k_i$ balls end up in bucket $i$, so for our purposes those $k_i$ balls are indistinguishable, and we have to divide by $k_i!$. – Brian M. Scott Apr 17 '21 at 02:59
  • @BrianM.Scott Ah, now I understand where the formula came from. However, why do we distinguish between the different order in which we put the balls into the $m$ buckets? (This is what I think you implied?). All the different ordering will give the same eventual result. – roulette01 Apr 17 '21 at 03:02
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    For the same reason that we distinguish HTHTHT from HHHTTT when we’re considering sequences of coin tosses: they are two different ways of arriving at a $3$-$3$ split, so they both contribute to the probability of that split. – Brian M. Scott Apr 17 '21 at 03:05
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    @BrianM.Scott Ah yes, that makes sense when I took a step back and thought about what we were originally trying to solve for. – roulette01 Apr 17 '21 at 03:08
  • @BrianM.Scott I didn't want to create another question on this because it's a simple question still related to this, but in the original answer by Sextus, he/she later states $P(N \leq n) = \frac{1}{m^n} \sum_{\vec{k} \in S} \frac{n!}{\prod k_i!}$. Shouldn't this actually be the expression for $P(N = n)$? – roulette01 Apr 17 '21 at 15:07
  • Actually, I think that it should be $P(N>n)$, if $N$ is the number of balls when the process stops: if I’m reading it correctly, it’s the probability that the stop condition has not yet been reached when $n$ balls have been distributed. – Brian M. Scott Apr 17 '21 at 18:05
  • @BrianM.Scott Yes, $N$ is the number of balls when the process stops. Hmm yes, I agree, but I'm not certain if that makes up the entire $P(N > n)$ or a portion of it. $\frac{1}{m^n} \sum_{\vec{k} \in S} \frac{n!}{k_i!}$, in words, read to me "The probability that each of the $m$ buckets has $\leq k - 1$ balls after distributing $n$ balls." – roulette01 Apr 17 '21 at 18:30
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    Yes, and that is precisely the probability that $N\not\le n$, i.e., that $N>n$. – Brian M. Scott Apr 17 '21 at 18:32

1 Answers1

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I think the reason is the number of ways to place the balls is $1$, however, the number of procedures of putting these $n$ balls is $\frac{n!}{\prod_{i = 1}^{m} k_i!}$. You can consider a sequence $S$ consisting of $k_1$ characters of $1$, $k_2$ characters of $2$, etc. In this sequence, if $S[i] = x$, then you should place a ball in the $x-th$ box on the $i-th$ step. This sequence describes a procedure, thus the number of ways we can create this sequence is the number of procedures. The number of ways you can create this sequence is $P(n; k_1, k_2, k_3, ..., k_m) = \frac{n!}{\prod_{i = 1}^{m} k_i!}$.

Therefore, the number of ways you can place these indistinguishable balls is $1$, while the number of procedures of doing this is $\frac{n!}{\prod_{i = 1}^{m} k_i!}$.

I hope my answer was helpful.

Radin Zahedi
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