Let $U, V \sim U(0,1)$ be independent. What is $P(U \leq V)$? Let $U, V \sim U(0,1)$ be independent. What is $P(U \leq V)$?
My attempt:
We are looking for $P(U-V \leq 0)$. For a given $t \in [0,1]$ this is equivalent to $$P(U \leq t, V \gt t)= P(U \leq t)P(V \gt t)=P(U \leq t)(1-P(V\leq t))=t(1-t)=t-t^2$$
Then summing over $t$ on $[0,1]$, I get:
$$\int_0^1 t-t^2 \, dt = \frac{1}{6}$$
Which seems like a reasonable value, but I'm not exactly sure whether my approach with doing this in terms of t was correct, can anybody please validate my solution?
 A: Your attempt:

We are looking for $P(U-V \leq 0)$. For a given $t \in [0,1]$ this is equivalent to $$P(U \leq t, V \gt t)= P(U \leq t)P(V \gt t)=P(U \leq t)(1-P(V\leq t))=t(1-t)=t-t^2$$

Your error is in not considering cases where $V$ is less than a given $t$ but still greater than $U$.
What you should have attempted was to integrate the product of the probability density function of $U$ and the conditional cumulative probability distribution of $V$ greater than that value. 
$$\begin{align}
\mathsf P(U\leq V) & = \int_0^1 { f_U(t) }\cdot{ \mathsf P(t\leq V\mid U=t) } \;\operatorname d t 
\\ & = \int_0^1 f_U(t)\cdot\mathsf P(V\geq t)\,\operatorname d t & \text{by independence of }U,V
\\ & = \int_0^1 1\cdot(1-t)\,\operatorname d t & \text{since uniformly distributed} 
\\ & =\frac 1 2
\end{align}$$
Which is in agreement with the symmetry argument's result.
A: The distribution of $(U,V)$ is the same as the distribution of $(V,U)$.  Therefore any statement about $U$ and $V$ that follows from a specification of that joint distribution is true if $U$ and $V$ are interchanged.  That means if we can say $P(U\le V)=c$ then it is also true that $P(V\le U)=c$.  If you know that $P(U=V)=0$, then that tells us that $P(U<V)=P(V<U)=c$ and $P(U<V)+P(V<U)=1$.  So $c+c=1$.
That's probably the simplest way to do it, but you can also notice that the pair $(U,V)$ is uniformly distributed in the unit square.  Draw the square, and look at the graph of the part of it where $U\le V$, and that should tell you the answer.
