Expected value of 2nd-bigger number out of n independent random variables in uniform distribution [r,1], r>0

Question

I am having a problem on which I have 4 players bidding in second price auction with reserved price (r). I need to find the expected value of 2nd bigger number when there are 3 players or 4 bidding over r, bid>r. For example, if 3 players bids over r, I need to find the expected value of the second bigger bid. So far I've found that when 2 players bids over r the expected value is equal to $a + (b-a)/3$ when we have uniform distribution in [a,b]. So in my example in [r,1] will be equal to $ r + (1-r)/3 $ I need the expected value of 2nd bigger value of n values. Haven't thought a way to do so. Found this but didn't help me Expected value of 2nd-smallest out of 3 random variables. Thanks in advance and sorry if my question is amateur.

Answer 1

I think the expected value of the second highest value in $n$ uniform random variables on $[a,b]$ is $a + (b-a)\frac{n-1}{n+1}$.

You can first consider $n$ uniform random variables on $[0,b-a]$, then the pdf of the second highest value is $$f(x) = \frac{n!}{(n-2)!}(\frac{x}{b-a})^{n-2}(1-\frac{x}{b-a}),$$ which has mean value $(b-a)(\frac{n-1}{n+1})$.

Maybe you can see https://i.stack.imgur.com/UFfYC.jpg and Expected value of $2$nd highest draw from uniform dist out of n draws for detailed discussion.