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This is a follow-up to this question: Probability of empty bin, where the number of balls is based on another game...

The question is: We flip a fair coin until we obtain our first heads. If the first heads occurs on the kth flip, we are given k balls. We put them into 3 bins labeled 1, 2, and 3 at random. Find the probability that none of the three bins are empty.

In the spirit of answering this question, I want to find the probability that $n$ balls, when placed into 3 bins, will be placed in such a way that no bin is empty? I decided to go down a combinatorics-based route:

You can enumerate the balls as $b_1 ... b_n$

You have $n$ balls, out of which you first choose 3 -> $^nC_3$

You place these three balls into three bins -> 3!

The remaining $n-3$ balls are placed -> $3^{n-3}$

There are a total of $3^n$ possible ways to place n balls into 3 bins.

It's obvious to me this is wrong. It works for the case where we have 3 balls, but as soon as we hit $n=5$ balls, the numerator is bigger than the denominator. But I really can't tell where I have gone wrong here.

Abhay Agarwal
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From your attempted working, it would be apparent (but you should have added) that you have taken both balls and bins to be distinct and that the balls have been distributed to bins uniformly at random.

Given this, the flaw in your approach (too often made by beginners) is that you have fractionated the distribution process, and are thus overcounting. You might have distributed balls 1,2,3 in the first lot and 4,5,6 in the second lot, but another time $\binom{n}{3}$ means you might have distributed balls $1,4,5$ in the first lot and $2,3,6$ in the second lot, and so on and so forth.

Suppose we take the case of $6$ distinct balls into $3$ distinct bins, the first way you might learn is through inclusion-exclusion, which you must have studied by now.

That would give you $3^6 - \binom31 2^6 + \binom32 1^6 = 540$

Another possibility for small problems is to work out for various patterns, using the format

[Lay Down Pattern] x [Permute pattern] which would give

$4-1-1:\; \dfrac{6!}{4!}\dfrac{3!}{2!} = 90$

$3-2-1:\; \dfrac{6}{3!2!}\cdot 3! = 360$

$2-2-2:\; \dfrac{6!}{2!2!2!}\dfrac{3!}{3!} = 90,\;\;$ total $540$

This won't be suitable for larger problems with many patterns

Finally, if you have learnt Stirling numbers of the second kind, which put $n$ distinct balls into $k$ identical non-empty bins, you can use

$k!{n\brace{k}} = 3!*90 = 540$

Of course, finally we would have to divide by $k^n$ to get the probability for each of these methods

true blue anil
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  • I see. So you're saying that the overcounting is happening because I'm counting the balls as distinct?

    When I attempted the original problem, I first tried the stars and bars method to determine the probability of three bins not being empty. However, you can't do that because not every possibility that you get from stars and bars has the same probability. That's why I thought that I had to enumerate the balls - it worked for the case of three balls ($\frac{6}{27}$).

     – Abhay Agarwal May 25 '24 at 19:13
  • No, you have misunderstood. I am not saying that the problem is that you took the balls as distinct, and as you can see, I have not used stars and bars. What I am saying is that you have first put in three balls so that no bin is empty and then put in the remaining. You can not do that, it results in overcounting except through devious countermeasures. I have outlined three ways to correctly count in such cases. You could read the part "Distribution Into Non-empty Bins" from https://brilliant.org/wiki/distinct-objects-into-distinct-bins/ – true blue anil May 25 '24 at 19:33
  • The probability for $6$ balls, $3$ bins will work out to $\dfrac{540}{3^6} = \dfrac{20}{27}$ – true blue anil May 25 '24 at 20:09
  • Okay, thank you very much! I now understand why there is overcounting happening. To confirm my suspicion, I actually wrote out the possible combinations for 2 buckets and 3 balls, and it became crystal clear. Say you have ball 1 and ball 2 in bucket 1, and ball 3 in bucket 2. That is the same thing as ball 2 and ball 1 in bucket 1, and ball 3 in bucket 2. But my method will count those as distinct. Naturally, as the number of balls and buckets increases, this will get immensely complicated. Thanks so much for your help! – Abhay Agarwal May 25 '24 at 20:54
  • You are welcome ! $;;$ :) – true blue anil May 26 '24 at 06:36
  • Since you are new, you should know that if answer{s) satisfy your purpose, you should tick it so as to close the question – true blue anil May 28 '24 at 03:47
  • Got it, thanks so much! – Abhay Agarwal Jun 01 '24 at 17:36
  • Glad to have been of help ! – true blue anil Jun 01 '24 at 18:27