Trying to calculate the probability that one RV exceeds another RV I am running into a silly mistake when trying to calculate the probability that a random variable, $U$ is less than another random variable, $V$. I am hoping that someone can help me spot my mistake.
My setup is as follows:


*

*$U$ ~ $U(0,1)$

*$W = \frac{1}{2}\min U_i$ where $U_i$~$U(0,1)$ for $i = 1 \ldots n$


Given this setup, I believe that the right approach is to condition on the value of $W$ and then integrate across all possible values of $W$. That is,
$P(U<W) = \int_{w=0}^{1/2} P(U < W ~|~ W = w) f(w)dw$
In order to evaluate this expression, we need to first obtain $f(w)$. 
Note that:
$$ \begin{align} F(w) &= P(W<w) \\
&= P( \frac{1}{2}\min U_i < w)\\
&= P( \min U_i < 2w)\\
&= 1-P( \min U_i \geq 2w) \\
&= 1-\prod_{i=1}^n{P(U_i \geq 2w)} \\ 
&= 1-(1-2w)^n
\end{align}$$
Thus, 
$$f(w) = 2n(1-2w)^{n-1} $$
Using this information, I get
$$\begin{align} P(U<W) &= \int_{0}^{1/2} P(U < W ~|~ W = w) f(w)dw \\ 
&\int_{0}^{1/2} wf(w)dw \\ 
&\int_{0}^{1/2} 2nw(1-2w)^{n-1} dw \\ 
\end{align}$$
Unfortunately, however, this integral does not evaluate to a meaningful result.
 A: I believe there is no mistake and the integral is quite meaningful :) You can write 
$$w(1-2w)^{n-1} = \left(w-\frac{1}{2}+\frac{1}{2}\right)(1-2w)^{n-1} = -\frac{1}{2}(1-2w)^n +\frac{1}{2}(1-2w)^{n-1} $$ and continuing from there, the integral evaluates to 
$$\left(\frac{1}{4(n+1)}\cdot (1-2w)^{n+1}-\frac{1}{4n}\cdot (1-2w)^{n}\right)\Bigg\vert_{w=0}^{w=\frac{1}{2}} = \frac{1}{2(n+1)},$$
which intuitively makes sense too: You are considering a sequence of $n+1$ independent uniformly distributed random variables. Conditioned on the set of values you get, any permutation of those values is equally probable. The last element is the smallest with probability of $1/(n+1)$, and now the $1/2$ additional factor accounts for being located in the lower half.
A: It happens that the other conditioning might yield a simpler approach, namely,
$$
P[U\lt W]=\int_0^1P[W\gt u]\mathrm du.
$$
Now, $[W\gt u]=[\min\limits_{1\leqslant i\leqslant n}U_i\gt2u]=\bigcap\limits_{i=1}^n[U_i\gt2u]$ and $P[U_i\gt2u]=(1-2u)^+$ for every $u$ hence, by independence of $(U_i)$,
$$
P[U\lt W]=\int_0^{1/2}(1-2u)^n\mathrm du\stackrel{x=1-2u}{=}\int_0^1x^n\frac{\mathrm dx}2=\frac12\frac1{n+1}.
$$
A: It looks like the integral does evaluate to a meaningful result. I just had a silly algebra mistake. 
We can proceed by substitution. Define $x = 1-2w$, so that $w = \frac{1-x}{2}$ and $dx = \frac{dw}{-2}$. Then:
$$\begin{align} P(U<W) &= \int_{0}^{1/2} 2nw(1-2w)^{n-1} dw \\ 
&= \int 2n \frac{1-x}{2} x^{n-1} \frac{dx}{-2} \\ 
&= \frac{n}{2} \int x^n - x^{n-1} dx \\ 
&= \frac{n}{2} \Bigg(\frac{x^{n+1}}{n+1} - \frac{x^{n}}{n} \Bigg) \\
&= \frac{nx^{n}}{2} \Bigg( \frac{x}{n+1} - \frac{1}{n} \Bigg) \\ 
&= \frac{n(1-2w)^{n}}{2} \Bigg( \frac{(1-2w)}{n+1} - \frac{1}{n} \Bigg) \\ 
\end{align}$$
Plugging in $w=1/2$ and $w=0$ we get:
$$\begin{align} P(U<W) &= \frac{n(1-2w)^{n}}{2} \Bigg( \frac{(1-2w)}{n+1} - \frac{1}{n} \Bigg) \Bigg|_{0}^{1/2}\\ 
&= 0 - \frac{n}{2} \Bigg( \frac{1}{n+1} - \frac{1}{n} \Bigg)\\
&= \frac{n}{2} \Bigg( \frac{1}{n} - \frac{1}{n+1} \Bigg)\\
&= \frac{1}{2(n+1)}
\end{align}$$
