Expected value of $2$nd highest draw from uniform dist out of n draws Jane wants to auction off an item, but does not know where to go to find bidders. David offers to find bidders for her, but will charge her $\$10$ per bidder he gets to show up. Each bidder will uniformly value the item between $[500, 1000)$. The highest bidder will win the item and pay the second-highest bidder's price (Vickrey auction). How many bidders should Jane pay David to find?
I need to maximize the difference between $E[\text{second largest bid out of n bidders}]$ and $10n$. I'm not sure how to find $E[\text{2nd largest big}]$
 A: This topic is discussed in depth in Casella and Berger Section 5.4. I've put the relevant text in an imgur post https://imgur.com/a/UFm12aK .
The second largest number is the n-1st order statistic, $X_{(n-1)}$. Theorem 5.4.4 says that $E[X_{(n-1)}]=(n-1)/(n+1)$ for a uniform(0,1), but I believe that if you follow the proof you will find that  $E[X_{(n-1)}]=500+ ((n-1)/(n+1))(1000-500)$ for a unif(500,1000) (I'm assuming you are talking about a continuous uniform distribution, you may mean discrete since you specified which endpoints were included, which would change the answer a bit, but since you just need integer precision in n, the final answer may be the same). 
You are looking to maximize $E[X_{(n-1)}]-10n$. So you can differentiate in $n$ and find the local maxima and check the 2 nearest integer values of $n$ to see which is larger.
Instead of differentiating, you know you want to find where $[(n)/(n+2)-(n-1)/(n+1)]*500=10$ (i.e. adding one person costs 10 dollars but also increases the expected value by of the second highest bid by 10). That happens at n=8.5
For 8 bidders, $E[X_{(7)}]= 500+\frac{7}{9}500=888.89$
For 9 bidders, $E[X_{(8)}]= 500+\frac{8}{10}500=900$
For 10 bidders, $E[X_{(9)}]= 500+\frac{9}{11}500=909.09$
So, the 10th bidder only provided \$9.09 worth of value. Thus you should have 9 bidders and you would expect to make \$890.
