0
$\begingroup$

If there are 100 tickets in an urn, and I'm only trying to see 10 of them while drawing with replacement, how do I find the result for that?

My intuition makes me want to say it's $n\sum_{i=1}^k\frac{1}{i}$ where $n$ is the total number of tickets, and $k$ is the number of tickets you're actually trying to collect.

$\endgroup$
2
  • $\begingroup$ Do you want the expected value of the number of tries until you see the $10$ tickets? Another possibility would be to determine the distribution of the number of tries. $\endgroup$
    – madprob
    Commented Nov 6, 2017 at 23:22
  • $\begingroup$ @madprob The former. Thanks for any help! $\endgroup$
    – JShoe
    Commented Nov 6, 2017 at 23:40

1 Answer 1

1
$\begingroup$

You can use the usual technique. Let $N_{i}$ denote the number of tickets until you see the $i$-th new ticket among the $k$. Also, let $N$ denote the number of tickets untill all $k$ tickets are seen. Since $N=\sum_{i=1}^{k}{N_i}$, conclude that \begin{align*} E[N] &= E\left[\sum_{i=1}^{10}{N_i}\right] \\ &= \sum_{i=1}^{10}E[N_i] \end{align*} Note that $N_i$ follows a geometric distribution with parameter $\frac{k+1-i}{n}$. Therefore $E[N_i]=\frac{n}{k+1-i}$ and $E[N] = n\sum_{i=1}^{k}{\frac{1}{k+1-i}} = n\sum_{i=1}^{k}{\frac{1}{i}}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .