[Math] Setting up the probability distribution of Y = the number of empty bowls

probabilityprobability distributions

Each of three balls are randomly placed into one of three bowls. Find the probability distribution for Y = the number of empty bowls.

My first confusion is whether there are 9 balls in total or 3 that we must place into the bowls. Is it in the choice of words? I apologize for this as English is not my first language.

Then, I believe there are three options in which 1 bowl is empty. The first bowl is empty, or the second or the third bowl.

Beyond that, I am not sure how to find the probability of having 2 or 0 empty bowls.

Any help or hint is much appreciated,

Thank you.

Best Answer

You have $3$ bowls and $3$ balls. Let for example $(3, 1, 2)$ be the outcome that the first ball went into bowl $3$, the second one into bowl $1$ and the third into bowl $2$.

We have $27$ possible outcomes: $$ (1, 1, 1)\\ (1, 1, 2)\\ (1, 1, 3)\\ (1, 2, 1)\\ \cdots \\ (3, 3, 3) $$ Let $N$ be the number of empty bowls after all three balls are placed. In order for us to have $N=2$, all balls need to have gone into the same bowl, which corresponds to the events $(1, 1, 1)$, $(2, 2, 2)$, $(3, 3, 3)$, i.e. $3$ events in total. Thus $$ P(N=2) = \frac{3}{27} $$ How many events correspond to $N=0$? Note this is equivalent to finding the number of ways to arrange $(1, 2, 3)$ – look at the factorial for a shortcut to finding this number.

Finally you can apply the fact that $P(N = 1) + P(N=2) + P(N=3) =1$.

That was the combinatorial method. You can do this faster, however, (although less safely, maybe) by multiplying probabilities: How does $N=2$ happen, for example: the first ball goes into any bowl. The second and third have to go into the same as the first one, thus $$ P(N = 2) = 1 \cdot \frac{1}{3} \cdot \frac{1}{3} $$ For $N=0$, the first ball goes into any bowl, the second ball has to go into one of the two remaining empty bowls, and the last ball has to go into the single remaining bowl, i.e. $$ P(N=0) = 1 \cdot \frac{2}{3} \cdot \frac{1}{3} $$

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Probability – Balls in Bins, Probability That Exactly Two Bins Are Empty

Not sure what you did wrong, but I suspect that you ignored something in your casework. Also, the total number of cases should be $n^{n+1}$, not $n^{n+2}$. Here is an example of a much simpler approach.

First, calculate the number of ways there can be two empty bins. The number of ways for the two empty bins to be chosen is $$_{n}C_2=\frac{n(n-1)}{2}$$ then, since there must be exactly two empty bins, we must place one ball in each of the remaining $n-1$ bins so that none of them are empty. We then have one last ball with $n-1$ choices. Thus the number of ways to place the balls and have exactly two empty bins is $$\frac{n(n-1)}{2}*(n-1)$$ $$\frac{n(n-1)^2}{2}$$ And so the probability is $$\frac{\frac{n(n-1)^2}{2}}{n^{n+1}}$$ $$\frac{(n-1)^2}{2n^{n}}$$

EDIT: This is incorrect. By saying that the number of ways to distribute the balls is $n^{n+1}$, I imply that the balls are distinct objects; however, if they are distinct, then the number of ways to place one ball in each bin other than the two designated empty bins is $(n-1)!$, and so the number of ways to place the balls and have two empty bins is $$\frac{n(n-1)}{2}*(n-1)!*(n-1)=\frac{n(n-1)^2(n-1)!}{2}$$ and so the probability is $$\frac{\frac{n(n-1)^2(n-1)!}{2}}{n^{n+1}}$$ $$\frac{(n-1)^2(n-1)!}{2n^n}$$

[Math] Bins and balls problem: expected number of balls placed given number of empty and non-empty bins

Presumably $m=m_e+m_n$.

The likelihood of seeing $m_n$ out of $m$ occupied is $\dfrac{S_2(n,m_n)\, m!}{m^n \;(m-m_n)!}$ where $S_2(x,y)$ is a Stirling number of the second kind. I do not know an easy way of finding the $n$ which maximises this in general but it is easy enough to calculate for given reasonably small $m$.

One possible alternative estimator is $\dfrac{\log(m)-\log(m-m_n)}{\log(m)-\log(m-1)}$, which follows from saying $E[m_n \mid m, n] = m\left(1 -\left(\frac{m-1}{m}\right)^n\right)$. This is neither a maximum likelihood estimator nor an unbiased estimator for $n$, and it is usually not an integer, but it comes fairly close to the maximum likelihood estimator; it is slightly biased upwards when $1 \lt n \lt m$. The table below compares the values when $m=10$ for different $m_n$

 m      m_n   n_MLE    n_alt.est 
10       0     0        0   
10       1     1        1   
10       2     2        2.12  
10       3     3        3.39
10       4     4 or 5   4.85
10       5     6        6.58
10       6     8        8.70
10       7    11       11.43
10       8    15       15.25 
10       9    22       21.85
10      10    infinite  n/a

Best Answer

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Probability – Balls in Bins, Probability That Exactly Two Bins Are Empty

[Math] Bins and balls problem: expected number of balls placed given number of empty and non-empty bins

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