MHB Probability of Not Getting Candy

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The probability of a nephew not receiving any candy when c candies are distributed randomly among n kids is calculated as (1 - 1/n)^c. A participant suggests that the correct formula should be (n-1/n)^c, but this is equivalent to the original expression. When n equals c, the probability approaches (1 - 1/n)^n, which approximates e^-1 as n becomes large. This means that as the number of kids and candies increases, the probability stabilizes around 0.37. The discussion clarifies the mathematical reasoning behind the probability calculation and its limit.
alfred2
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Hi I've found this exercise but I do not understand the solution:
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In a party c candies are given randomly to n kids. ¿Which is the probablility that my nephew doesn't receive any candy? We supose the candies and the kids are numerated. Each of the candies can be given to any of the n kids, so there are n^c possible cases and the unfavorable ones for my nephew are all the manners to distribute the candies between the (n-1) remaining kids i.e. (n-1)^c. So the probability is (1-1/n)^c. If n=c the probability is practically independent of n, being aproximately equals to e^-1=0.37
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First of all I think there's a mistake and the probability is (n-1/n)^c and not (1-1/n)^c. Then what's the meaning of "If n=c the probability is practically independent of n"? What does independent mean in this case? And how do we obtain e^-1? Thanks! If anyone know a book as "probability and statistics for dummies" please let me know! Thanks ;)
 
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alfred said:
Hi I've found this exercise but I do not understand the solution:
_________________________________________________________________
In a party c candies are given randomly to n kids. ¿Which is the probablility that my nephew doesn't receive any candy? We supose the candies and the kids are numerated. Each of the candies can be given to any of the n kids, so there are n^c possible cases and the unfavorable ones for my nephew are all the manners to distribute the candies between the (n-1) remaining kids i.e. (n-1)^c. So the probability is (1-1/n)^c. If n=c the probability is practically independent of n, being aproximately equals to e^-1=0.37
_________________________________________________________________
First of all I think there's a mistake and the probability is (n-1/n)^c and not (1-1/n)^c. Then what's the meaning of "If n=c the probability is practically independent of n"? What does independent mean in this case? And how do we obtain e^-1? Thanks! If anyone know a book as "probability and statistics for dummies" please let me know! Thanks ;)

Welcome to MHB, alfred! :)You seem to have left out a couple of parentheses.
The probability is $$\frac {(n-1)^c}{n^c} = \left(\frac {n-1}{n}\right)^c = \left(1 - \frac {1}{n}\right)^c$$.If n=c, the probability becomes $$\left(1 - \frac {1}{n}\right)^n$$.
If n is large enough this will approximate its limit for $n \to \infty$.Since $$\frac 1 e=\lim_{n \to \infty} \left(1-\frac 1 n\right)^n$$ (see for instance wiki), the probability approximates $e^{-1}$.
 
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