$X$, $Y$, and $Z$ are continuous IID, $\text{P}(XSo we have 3 continuous IID random variables $X$, $Y$ and $Z$. I want to find $p = \text{P}(X<Y<Z)$.    
The easy way to do it is to use symmetry and thus we have $3!$ equally likely orderings and thus $p = 1/3!$ 
I was thinking about doing it the hard way and go for the integral of the joint pdf over a specific region. This is where I'm stuck. I'm not sure about the limits of the integral for each one. Any help or guidance is appreciated. 
 A: Let $F$ be the common distribution function. Then
\begin{align*}
P(X < Y < Z) &= E\big( E\left(\mathbb{I}_{X < Y} \mathbb{I}_{Y < Z}\mid Y, Z\right)\big)\\
&= E\big(F(Y)\mathbb{I}_{Y < Z} \big)\\
&=E\Big(E\big(F(Y) \mathbb{I}_{Y < Z} \big) \mid Y\Big)\\
&=E\left(F(Y)-F(Y)^2\right)\\
&=\int_{-\infty}^{\infty}F(y)dF(y) - \int_{-\infty}^{\infty}F(y)^2dF(y)\\
&=\frac{1}{6}.
\end{align*}
A: Here is a brutal way. Let $F$ be the CDF of the common distribution of $X$, $Y$ and $Z$. Then starting @Did's expression for $p$ and using the relation $F' = f$ triple times,
\begin{align*}
p
&=\int_{-\infty}^{\infty} f(z) \bigg( \int_{-\infty}^z f(y) \bigg( \int_{-\infty}^y f(x) \, \mathrm{d}x \bigg) \mathrm{d}y \bigg) \mathrm{d}z \\
&=\int_{-\infty}^{\infty} f(z) \bigg( \int_{-\infty}^z f(y)F(y) \, \mathrm{d}y \bigg) \mathrm{d}z \\
&=\int_{-\infty}^{\infty} f(z) \left[ \frac{1}{2}F(y)^2 \right]_{y=-\infty}^{y=z} \, \mathrm{d}z \\
&=\int_{-\infty}^{\infty} \frac{1}{2} f(z) F(z)^2 \, \mathrm{d}z \\
&=\left[ \frac{1}{6}F(z)^3 \right]_{z=-\infty}^{z=\infty} \\
&=\frac{1}{6}.
\end{align*}
A: If you want to consider the joint integral, maybe you could start out by reducing the problem a little. Due to continuity with $F$ denoting the distribution function, 
$P(X<Y<Z)=P(F(X) < F(Y) < F(Z))=P(U_1<U_2<U_3)$,
where the $U_i$ are iid $\text{Unif}(0,1)$-distributed, $i=1,2,3$.
Then (just calculate):
$P(U_1<U_2<U_3)
=
\int_{u_1<u_2<u_3} 1_{(0,1)^3}(u_1,u_2,u_3) \, \mathrm{d}(u_1,u_2,u_3)
=
\int_0^1 \int_0^{u_3} \int_0^{u_2} \mathrm{d}u_1 \mathrm{d}u_2 \mathrm{d}u_3
=
\int_0^1 \int_0^{u_3} u_2 \, \mathrm{d}u_2 \mathrm{d}u_3
=
\int_0^1 \frac{u_3^2}{2} \, \mathrm{d}u_3
=
\frac{1}{6}.
$
The idea to transform such problems into the uniform case is very typical.
A: Comment. Your intuitive method seems sound.
Here are simulations with standard uniform (on $(0,1)$),
standard normal (on the real line), and standard 
exponential (on the positive half line). With a million
simulated realizations of each random variable, you 
should expect about two-place accuracy.
# std unif
m = 10^6; x = runif(m); y = runif(m);  z = runif(m)
cond = (x < y) & (y < z)
mean(cond)
## 0.166949

# std norm
m = 10^6; x = rnorm(m); y = rnorm(m);  z = rnorm(m)
cond = (x < y) & (y < z)
mean(cond)
## 0.167202

# std exponential
m = 10^6; x = rexp(m); y = rexp(m);  z = rexp(m)
cond = (x < y) & (y < z)
mean(cond)
## 0.166133

As a start towards a demonstration using integrals, I suggest you
do it first in 2-D with just random variables $X$ and $Y$. Make sure the density function
is defined using an indicator function to restrict it to the support
of its distribution. Then use symmetry.
In the plot below (for standard uniform random variables and 100,000 iterations), 
49,988 points with $X < Y$ are plotted. And 16,833 of the points with $X < Y < Z$ are
plotted in red. Maybe the figure will help you set limits for the 
appropriate region in 2-D space (subsequently in 3-D).

m = 10^5; x = runif(m); y = runif(m);  z = runif(m)  
cond = (x < y) & (y < z)
plot(x[x < y], y[x < y], pch=".", col="blue")
points(x[cond], y[cond], pch=".", col="red")

A: Assuming they have pdfs,
$$\int_{-\infty}^{\infty}\int_{-\infty}^{z}\int_{-\infty}^{y} f_{X,Y,Z}(x,t_y,t_z)dxdt_ydt_z$$
$$ = \int_{-\infty}^{\infty}\int_{-\infty}^{z}\int_{-\infty}^{y} f_X(x)f_Y(t_y)f_Z(t_z)dxdt_ydt_z$$
$$ = \int_{-\infty}^{\infty}f_Z(t_z)\int_{-\infty}^{z}f_Y(t_y)\int_{-\infty}^{y} f_X(x)dxdt_ydt_z$$
To begin let
$$u=\int_{-\infty}^{z}f_Y(t_y)\int_{-\infty}^{y} f_X(x)dxdt_y$$
$$dv = f_Z(t_z)dt_z$$
Can you take it from here?
Otherwise, what you said.
