1

Given $N$ identical balls I have to distribute them into $n$ boxes. Any number of balls can be in one box.

What is the probability of having $l$ balls in one box?

Is this correct?

$p(l) = \frac{1}{n^{l-1}} \binom{N}{l} = \frac{1}{n^{l-1}} \frac{N!}{(N-l)!l!}$

I asked this one differently, with an explanation of how to reach to the formula: Understanding the $n!$ in the probability of $l$ events from a set of $N$ to happen at the same timestep out of $n$ total but I didn't get any feedback.

And now, based on these two related questions, I think my expression might be correct.

Probability of exactly one empty box when n balls are randomly placed in n boxes.

Distributing n identical balls in k distinct boxes

myradio
  • 149
  • Please distinguish between the following:
    • the boxes are numbered and you want the probability to have $l$ balls in box number one
    • the boxes are not numbered and you want the probability to have $l$ balls in at least one box
    • the boxes are not numbered and you want the probability to have $l$ balls in exactly one box
     – Exodd Feb 22 '18 at 09:43
  • -the boxes are not numbered and I want the probability to have l balls in at least one box – myradio Feb 22 '18 at 09:56
  • I think it is not correct. How did you get to that formula? The big problem when dealing with not numbered boxes is that you have to take count of the repetition of the same partition, and it is very difficult (The number of non-ordered partitions of a number is explored in https://en.wikipedia.org/wiki/Partition_%28number_theory%29#Partition_function) – Exodd Feb 22 '18 at 12:25
  • I can provide some help under certain circumstances: 1) we have n boxes 2) find the probability of putting L balls in exactly one box. The probability of putting one ball into a certain box is 1/n . By using the binomial distribution you get the probability(lets consider it P) of putting L balls into that certain box. Then the probability asked will be P^n (because you have n boxes) – al.al. Feb 22 '18 at 17:28
  • @al.al. Following your reasoning: putting a specific ball in a specific box has probability 1/n . --- In how many ways can I choose l balls from N balls: binomial (N, l). --- Then, 1/n * bin(N, l) is the proba of taking l balls and putting them into one specific box. --- Now, since we have n boxes, we have to SUM this n times, or multiply by n, but I think is incorrect your suggestion multiplying this n times (or put it to the power of n). Assuming the sum that I comment is how I got my expression. – myradio Feb 22 '18 at 19:50
  • @Exodd I think I am already taken that into account with the n! that appears in the denominator. (see my edit making that explicit in the formula) – myradio Feb 22 '18 at 19:51
  • I think you're right about the 'n' as an exponent. I insist though about the probability for a specific box: p = bin(1/n,N). Then the p you ask for will be n*p (and not p^n) – al.al. Feb 22 '18 at 20:46
  • @al.al. ok but then are you saying my formula is correct? – myradio Feb 23 '18 at 09:14

1 Answers1

1

I think there's not a closed form for your question, since it is the number of partitions of $N-l$ balls into $n$ boxes over the partitions of $N$ balls into $n$ boxes.

Your answer is wrong for many reasons, but first of all because it is not less than 1!

For example, if $n=1$, then you have 1 box, and necessarily $p(l)=0$ for every $l\ne N$. Your formula, instead, gives $$p(l) = \binom Nl$$ that is surely distinct from 1 or zero for many values $l$.

Exodd
  • 10,844
  • Yes, that's why I thought it might be wrong and I commented that in my previous question as a normalization issue. What is your k? my l? What part of my reasoning is wrong? I also don't see how this would change if the boxes are numbered. How would you proceed if the boxes are numbered? – myradio Feb 23 '18 at 09:11
  • @myradio The problem is: You never explained your reasoning (as I requested in the comments). If you add it to the post, I can correct it. If the boxes are numbered, then at least you know that the total number of ways of putting N balls in n boxes is $\binom{N+n-1}{N}$, and then you can sum the probabilities of having $l$ balls in the first, second,etc. boxes, that is fairly easy – Exodd Feb 23 '18 at 09:31
  • @myradio I've seen your other question with the explanation(s). I'll answer you there. By the way, the timesteps are numbered, so in your problem the box are also numbered – Exodd Feb 23 '18 at 09:40
  • This is partially a correct answer to the question because the question was just "is this expression correct?". When talking about probabilities the fact that the expression is larger than zero is pretty sound. Nevertheless, the "your answer is wrong for many reasons" is not properly justified IMO. – myradio Feb 27 '18 at 10:07