Probability of being Second Highest Suppose there are $n-1$ draws from a uniform distribution $[0, 1]$, then I get a draw from the distribution.  What is the probability that if I shout out an $x$, then $x$ will be exactly the second highest of all the draws?  I have a feeling its something like $n x^{n-1} (1-x)$, but I can't show it rigorously.
There are $n-1$ draws, then you draw $x$ on the $n$-th draw.

The probability that the $n$-th draw is the second-last order statistic of the $n$ draws, when given the value of that draw is $x$, is:

$$\frac{(n-1)!}{(n-2)!\;1!}x^{n-2}(1-x) \\[4ex] = (n-1)(1-x)x^{n-2}$$

Reasoning:&nbsp; Of the $n-1$ draws preceding yours, the favoured space is formed by $n-2$ of them being no greater than yours (individual probability $x$ each), and $1$ of them being no less (probability $1-x$), and there are ${n-1\choose n-2}$ ways of arranging these favoured results.
The problem is equivalent to finding the probability that $x$ is the second last order statistic.
In that case, out of the remaining $(n-1)$ draws, there should be $n-2$ draws which are smaller than $x$ and 1 draw is bigger than $x$. The probability of a draw being smaller than $x$ is $F(x)=x$. There are in total $n$ independent draws, out of which $n-1$ are smaller than or equal to $x$ so each has probability $F(x)=x$, and the last draw is greater than $x$ having probability $1-F(x)=1-x$. Note that these draws form a Binomial distribution with parameters $n$ and $F(x)$.

So you have the required probability as $\dfrac{n!}{(n-1)!1!}$$x^{n-1}$$(1-x)$