We throwing $m$ balls to $n$ cells randomly...
At each cell can be more then one ball, or (of course) it can still empty.
What is the expectation of the empty cells?
I'd like to get any help!
Thank you!
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What do you mean with exception? – Mussé Redi Jul 15 '14 at 17:53
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Expectation... Sorry! @MusséRedi My mistake! – CS1 Jul 15 '14 at 17:54
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what did you try? – cirpis Jul 15 '14 at 17:55
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I actually don't know where to start. $X$ is for the balls or for the cells? – CS1 Jul 15 '14 at 17:55
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What you mean with the expectation value of the empty cells ? An expectation value is calculated by summing up the products of the actual values $x$ you get multiplied by the probability $P(x) $. What probabilty (distribution) have you found? – Mussé Redi Jul 15 '14 at 18:03
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@MusséRedi, I think that Bananarama solve it, if I'm understant it right... Thank you!!! – CS1 Jul 15 '14 at 18:06
3 Answers
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The number of balls per cell has a multinomial distribution. But for any one cell, it is binomial. The probability a ball lands in that one cell is $1/n$ and outside it is $(1-1/n). $ The probability that that particular cell remains empty after all balls are thrown is $(1-1/n)^m. $ Now define the Indicator variable $I_k=1$ if cell number $k$ remains empty and $=0$ if cell $k$ is not empty. Then the expected number of empty cells is $$E\sum_{k=1}^n I_k = \sum_{k=1}^nP(\text{cell k is empty})= n(1-1/n)^m $$
Mr.Spot
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Using stars and bars there are $\binom{m+n-1}{n-1}$ ways to distribute the balls allowing for empty cells.
Again by stars and bars there are $\binom{n}{i}\binom{m+n-i-1}{n-i-1}$ ways to leave exactly $i$ cells empty. Since there are $\binom{n}{i}$ ways to select the empty cells and $\binom{m+n-i-1}{n-i-1}$ ways to divide $m$ balls in $n-i$ cells.
Thus the expected number of empty cells is:
$$\sum_{i=1}^{n-1}\dfrac{\binom{n}{i}\binom{m+n-i-1}{n-i-1}i}{\binom{m+n-1}{n-1}}$$
Two years later I realize this is wrong, as for example $(1,1)$ does not have the same probability as $(2,0)$. The answer by Mr.Spot is correct, and more easily computable.
Asinomás
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stars and bars is a method for counting what you want to count: here is a reference E(X) is the final formula. – Asinomás Jul 15 '14 at 18:03
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I see that this problem has been answered, but I thought I'd share my approach anyway.
I find it's most helpful to look at this problem from the perspective of a cell. "If I am a cell, what is the probability that I will remain empty after all $m$ balls are thrown?" Well, each time a ball is thrown, there is a $\frac{n-1}{n}$ chance that it will not land in me. So the probability that I remain empty after all the balls are thrown is $\left(\frac{n-1}{n}\right)^m$, since this event must occur $m$ times. So now we have a formula for the probability that an individual cell will be empty after all of the balls are thrown; to get the expected number of empty cells, we simply multiply this probability by the number of cells, i.e. $n\left(\frac{n-1}{n}\right)^m$.
hexaflexagonal
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