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If I want to distribute 50 identical candies to 100 children, what is the expected number of candies a child with at least one candy has?

For example, if I give 24 candies to child A and 26 candies to child B, and don't give any candy to other 98 children, since only two children has nonzero candies, if this way the only way how I can distribute 50 identical candies to 100 children, the quantity I'm looking for would be 25.

I tried for an hour, and in the end came up with the following "solution":

Let $Q(k)$ be the average number of candies a child with at least one candy has provided that we distribute the candies only to $k$ children. Then I assumed that the average number of candies a child from this set has $50/k$, and there are $\binom{100}{k}$ different ways of selecting these set of children, so doing a weighted average, I got

$$ \frac{ \sum_{k=1}^{50} 50*(100!) / (k * (k!) * (100-k)!)} { \sum_{k=1}^{50} 100! / ((k!) * (100-k)!)} \approx 1.08481. $$

Is my solution correct? If not, could you provide me with an detailed answer about how you solved it?

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    $\begingroup$ I think you should give more info about how the distribution is happening. $\endgroup$
    – cgss
    Commented Sep 29, 2020 at 10:07
  • $\begingroup$ I think the question should be: What is the expected number of candies a child with at least one candy has? $\endgroup$ Commented Sep 29, 2020 at 10:49
  • $\begingroup$ @AdamRubinson I think the question is to find the expected average of candies with children with at-least $1$ candy. The average number of candies with children with at-least one candy is dependent on the sample, so it only makes sense to calculate the expectation of the average. $\endgroup$ Commented Sep 29, 2020 at 10:51
  • $\begingroup$ Yes, you're right. $\endgroup$ Commented Sep 29, 2020 at 10:52
  • $\begingroup$ You want to find $E[50/S]$ where $S$ is the number of students who got at-least one candy. The general approach is to find the distribution of $S$ and use it to find $50E[1/S]=50\sum_{s=1}^{50}\frac1sf_S(s)$. $\endgroup$ Commented Sep 29, 2020 at 16:52

2 Answers 2

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Pick any child, and let's say the number of candies he receives is $X$, so we want to find $E(X|X>0)$, i.e. $$E(X|X>0) = \sum_{x>0} x P(X=x | X >0)$$

Evidently $X$ has a Binomial distribution with $n=50$ and $p = 0.01$, so $$P(X = x) = \binom{50}{x} 0.01^x 0.99^{50-x}$$ for $0 \le x \le 50$. Now $$P(X=x | X>0) = \frac{P(X=x)}{P(X > 0)}$$ for $1 \le x \le 50$, and $P(X>0) = 1- P(X=0) = 1 -.99^{50}$, so $$E(X|X>0) = \sum_{x=1}^{50} \frac{x \binom{50}{x} 0.01^x 0.99^{50-x}}{1-.99^{50}}$$ We also have $$\sum_{x=1}^{50} x \binom{50}{x} 0.01^x 0.99^{50-x} = \sum_{x=0}^{50} x \binom{50}{x} 0.01^x 0.99^{50-x}$$ which is the expected value of a Binomial distribution, so the sum is $np = 50 \cdot 0.01 = 0.50$. Hence $$E(X | X>0) = \frac{0.50}{1-.99^{50}} = \boxed{1.26584}$$

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  • $\begingroup$ "Evidently X has a Binomial distribution": why ? $\endgroup$
    – Our
    Commented Sep 29, 2020 at 20:49
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    $\begingroup$ @onurcanbkts Each candy has a fixed probability of 0.01 of being given to the child, independently of all the other pieces of candy, and the total number of candies is fixed at 50. $\endgroup$
    – awkward
    Commented Sep 29, 2020 at 22:44
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The probability that any given child would not get any candy is (99/100)^50, about .6050, so 100*(1-.6050), 39.499, would get 1 or more: on average 50/34.999 or 1.26584.

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    $\begingroup$ Usually $E\left[\frac{50}{X}\right]\neq\frac{50}{E\left[X\right]}$. $\endgroup$
    – David K
    Commented Sep 29, 2020 at 16:44

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