MHB How Does a Markov Chain Model the Distribution of Stickers Among Children?

[rachelro]

A teacher leaves out a box of N stickers for children to take home as treats. Children form a queue and look at the box one by one. When a child finds k⩾1 stickers in the box, he or she takes a random number of stickers that is uniformly distributed on {1,2,…,k}.

1- What is the expectation of the number of stickers taken by the second child, as a function of the initial number of stickers N?

2- If E_N denotes the expected number of children who take at least one sticker from the box given that it initially contained N stickers. How can I compute a formula to represent E_N+1 in terms of E_1+⋯+E_N. Also, how E_N can be expressed in terms of k?

 

[MarkFL]

Hello and welcome to MHB, Sarah! :D

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[chisigma]

A teacher leaves out a box of N stickers for children to take home as treats. Children form a queue and look at the box one by one. When a child finds k⩾1 stickers in the box, he or she takes a random number of stickers that is uniformly distributed on {1,2,…,k}.

Very interesting problem!... in order to understand correcly however let's do an example: the first child finds N stichers... that meants that he/she takes a random number of stickers uniformly distributed on {1,2,...,N}?... and that meants that, if the first child selects k stichers, the number of stickers selected by the second child is uniformly distributed in {1,2,...,N-k}?... Kind regards $\chi$ $\sigma$

 

[chisigma]

A teacher leaves out a box of N stickers for children to take home as treats. Children form a queue and look at the box one by one. When a child finds k⩾1 stickers in the box, he or she takes a random number of stickers that is uniformly distributed on {1,2,…,k}.

1- What is the expectation of the number of stickers taken by the second child, as a function of the initial number of stickers N?

Under the assumption that what I written in post #3 is true, let's try to answer to the question 1. The mean value of object selected among k is...

$\displaystyle E\{n\} = \sum_{n=1}^{k} \frac{n}{k} = \frac{k\ (k+1)}{k\ 2} = \frac{k+1}{2}$ (1)

If $n_{1}$ is the number of objects selected by the first child, then the second child can choose among $N-n_{1}$ objects so the mean numbers of object choosen by ther second child is...

$\displaystyle E \{n_{2}\} = \frac{1}{2\ N}\ \sum_{n=1}^{n} (N-n+1) = \frac{1}{2\ N}\ \frac{N\ (N+1)}{2} = \frac{N+1}{4}$ (2)

Kind regards

$\chi$ $\sigma$

 

[rachelro]

Thank you for your comment, but I think the distribution for the first child should be between [1,N]. I was thinking to go ahead with the following argument:

n1 is the number of objects taken by the first child and it is uniformly distributed in [1, N], but n2 is uniformly distributed in [1, N-n1]. so we have to evaluate:
E(n2)= E( E(n2|N-n1))