$X$, $Y$, and $Z$ are continuous IID, $\text{P}(X<Y<Z)$? So we have 3 continuous IID random variables $X$, $Y$ and $Z$. I want to find $p = \text{P}(X&lt;Y&lt;Z)$.    

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.
Let $F$ be the common distribution function. Then
\begin{align*}
P(X &lt; Y &lt; Z) &amp;= E\big( E\left(\mathbb{I}_{X &lt; Y} \mathbb{I}_{Y &lt; Z}\mid Y, Z\right)\big)\\
&amp;= E\big(F(Y)\mathbb{I}_{Y &lt; Z} \big)\\
&amp;=E\Big(E\big(F(Y) \mathbb{I}_{Y &lt; Z} \big) \mid Y\Big)\\
&amp;=E\left(F(Y)-F(Y)^2\right)\\
&amp;=\int_{-\infty}^{\infty}F(y)dF(y) - \int_{-\infty}^{\infty}F(y)^2dF(y)\\
&amp;=\frac{1}{6}.
\end{align*}
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.