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I know the title is not descriptive enough, but I don't know how else to say it. I don't know why I can't use the Binomial distribution to get the result I'm looking for. The teacher solved it long ago with a more straightforward method, but I tried today to solve the same problem using Binomial distribution and the result is not the same.

So, the problem is:

A total of $17$ balls are distributed in $20$ baskets, where every basket can hold as many balls as I want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishsable from another. What is the probability of every ball falling into the first basket?

The teacher solved it through combinatorics, using:


I wanted to solve it using Binomial distribution:

$$\dbinom{n}k p^k (1-p)^{n-k} = \dbinom{17}{17} \left(\dfrac1{20}\right)^{17} \left(\dfrac{19}{20} \right)^0$$

Knowing that there are $17$ throws, and I want $17$ successes. And as there are $20$ baskets, and only one is the successful one, the probability of each ball of going into that basket is $1/20$.

The problem is, it doesn't give the same result. And I don't know why, because I think I'm not using the binomial distribution wrong.

Thank you for your time!

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What is the statement of the problem, exactly? I cannot see what the question is in "A total of 17 balls are distributed in 20 baskets, where every basket can hold as many balls as i want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishable from another." – Clement C. May 15 '13 at 3:27
Oh, sorry! I'll edit it now. In the hurry i forgot about the main part! – FDrico May 15 '13 at 3:29
Your answer is completely wrong, and has nothing to do with the problem. You are answering the question "What is the probability that there is 1 bin which has all 17 balls, if the 17 balls are uniformly distributed." – Calvin Lin May 15 '13 at 3:29
Oh... Isn't it the same? I thought i was answering the question: What is the probability of, given 17 trys, have 17 successes, each one having a probability of 1/20. – FDrico May 15 '13 at 3:31
up vote 1 down vote accepted

The key is that the balls are indistinguishable.

Let us consider an example, where we can work out explicitly. Consider $2$ balls and $3$ baskets.

If the balls are distinguishable, say colored red and blue, then the first ball can go into any of the three baskets, while the second ball can also go into any of the three baskets. This thereby gives us a total of $3 \times 3$ options. $$\{\color{blue}1,\color{red}1\},\{\color{blue}1,\color{red}2\},\{\color{blue}1,\color{red}3\},\{\color{blue}2,\color{red}1\},\{\color{blue}2,\color{red}2\},\{\color{blue}2,\color{red}3\},\{\color{blue}3,\color{red}1\},\{\color{blue}3,\color{red}2\},\{\color{blue}3,\color{red}3\} \tag{$\star$}$$ where $\{\color{blue}i,\color{red}j\}$ means blue ball chose basket $i$ and the red ball chose basket $j$. Hence, the probability that both balls fall in the first basket is $\color{green}{\dfrac19}$.

If the balls are non distinguishable, then let us list out the options. $$\{2,0,0\},\{0,2,0\},\{0,0,2\},\{1,1,0\},\{0,1,1\},\{1,0,1\}$$ where $\{m_1,m_2,m_3\}$ means the baskets $i$ has $m_i$ balls. Essentially we are counting $(\star)$, but we now consider $\{\color{blue}1,\color{red}2\}$ and $\{\color{blue}2,\color{red}1\}$ to be the same since there is no color to distinguish between them. Hence, the probability that both balls fall in the first basket is the same as the probability that the first basket contains both the balls, which in this case is $\color{green}{\dfrac16}$.

Hence, in general, if we want to distribute $m$ balls in $n$ (distinguishable) baskets, the total number of ways is

  • $n^m$, if the balls are distinguishable.
  • $\dbinom{n+m-1}{n-1}$, if the balls are indistinguishable.

Clearly, both intuitively and algebraically, we have $\dbinom{n+m-1}{n-1} < n^m$.

Hence, in general, the probability that all the $m$ balls fall in the first basket is

  • $\dfrac1{n^m}$, if the balls are distinguishable.
  • $\dfrac1{\dbinom{n+m-1}{n-1}}$, if the balls are indistinguishable.
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Oh... i think i get it. But, just out of curiosity, as i don't think i read specifically that every try had to be distinguisable from one another when using Bernolli Process: A Bernoulli Process requires binary random variables, independent from one another and identically distributed. From that, do i have any way to know that the throws have to be distinguishables? Thank you for your time! – FDrico May 15 '13 at 4:32
But why on earth would the six outcomes listed have equal probabilities? $\{1,1,0\}$ is twice as likely to happen (according to OP's description of the scenario) as is $\{2,0,0\}$. – alex.jordan May 15 '13 at 4:34
@alex.jordan I read the question as this: Given all possible distribution of $17$ indistinguishable balls in $20$ bins, what is the probability that all the balls are in the first bin? – user17762 May 15 '13 at 4:39
This is akin to saying that when you roll two indistinguishable dice, the probability of rolling a total of 7 is the same as that of a total of 2, both of which are $1/11$, since there are 11 possible values for the total. – alex.jordan May 15 '13 at 4:42
OK. I read it the other way, and it seems inauthentic to have equal probabilities for these ball-distributions rather than for the distinguished ball-distributions. If that's really the original wording of the problem, it's poorly asked to phrase it in terms of a probability. It would seem more appropriate to ask for a count of how many such distributions there are. – alex.jordan May 15 '13 at 4:46

I think you may have some confusion (as do we) about what you mean by the first basket.

If you mean basket number 1 (or 7 or 12 or any other particular number) then your formula will give the correct result $\left(\frac{1}{20}\right)^{17}$.

If you mean the basket that the first ball drops into then you are interested only that the remaining 16 balls fall into that basket $\left(\frac{1}{20}\right)^{16}$.

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With the first basket it means the particular number 1. That's a fair question, i didn't think on the secound possibility. Thank you for you time! – FDrico May 15 '13 at 4:21

Your answer $\left(\frac{1}{20}\right)^{17}$ makes sense to me, given your description of the problem.

I recognize the pattern of the teacher's answer, and that is an answer to a different question. There are $\binom{20+17-1}{17}$ ways to distribute $17$ indistinguishable balls into 20 urns. If all such distributions were equally likely, then that would explain the teacher's answer. However as you describe the problem, not all such outcomes are equally likely. As you describe it, $$16,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ is $17$ times more likely to occur than $$17,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ but your teacher's answer counts these outcomes as equally likely. Perhaps that was an error or perhaps that was somehow stated in the original wording.

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Oh, i feel quite stupid, but i don't undestand why the first result you gave (16,0,0,...) is 17 times more likely than 17,0,0,0,... Maybe my writing is confusing (English is not my prime language) – FDrico May 15 '13 at 4:42
The one ball in the second urn could have been the first, second,...,17th ball that was tossed. That's why it's 17 times more likely of an outcome. – alex.jordan May 15 '13 at 5:23
I think that your teacher was trying to be too clever. I believe that he really wanted you to find the value $\binom{2+17-1}{17}$, and thought he could work it into the denominator of a probability question. But it's inauthentic to suggest that the $\binom{2+17-1}{17}$ outcomes are equally likely. Your answer is a realistic probability answer. It's like having two fraternal twins be born and asking for the probability that both are boys. You say $1/4$; the teacher says $1/3$ because he sees BB, BG, and GG as equally likely outcomes. I feel that is poor use of the term "probability". – alex.jordan May 15 '13 at 5:25

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