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?
from Hot Weekly Questions - Mathematics Stack Exchange
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