How to compute $P(Z_2 \le Z_3 \le Z_4| Z_2=Z_1 )$? where we have an i.i.d. sequence of standard normal $Z_1,Z_2,Z_3,Z_4$. Suppose that we have an i.i.d. sequence of standard normal $Z_1,Z_2,Z_3,Z_4$. How to compute
\begin{align}
P(Z_2 \le Z_3 \le Z_4| Z_2=Z_1 )
\end{align}
My idea
\begin{align}
P(Z_2 \le Z_3 \le Z_4| Z_2=Z_1 ) =P( Z_1 \le Z_3 \le Z_4 ) =\frac{1}{3!}
\end{align}
The last step follow because there are $3!$ ways that arrange $Z_1,Z_3,Z_4$ and normal is symmetric.
However, I am not sure if the second step is fine and we can drop the conditioning?
 A: Partial answer that is too long for a comment:

Consider the joint distribution of $(Z_1-Z_2, Z_2)$. This is a bivariate normal distribution with means $0$ and $0$, variances $2$ and $1$, and correlation $-\frac{1}{\sqrt{2}}$. Thus it has the same distribution as $(2U, -\frac{1}{\sqrt{2}} U +\frac{1}{\sqrt{2}} V)$ where $U$ and $V$ are i.i.d. standard normal. (You can verify this by checking that the means, variances, and correlation are the same.)
So, the conditional distribution of $Z_2$ given $Z_1=Z_2$ is the same as the conditional distribution of $-\frac{1}{\sqrt{2}} U + \frac{1}{\sqrt{2}} V$ given $2U=0$. By plugging in $U=0$, we immediately see that the conditional distribution is $N(0, 1/2)$, as Brian Tung mentioned in the comments.

Thus you can rewrite your original probability as
$$P(\frac{1}{\sqrt{2}}V \le Z_3 \le Z_4)$$
where $V, Z_3, Z_4$ are i.i.d. standard normal. Some sort of volume/symmetry argument might allow you to compute this, but I'm not sure.
A: Here's a more direct approach. Let $E=\{Z_1=Z_2\}$ and $h(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Notice $(Z_1,Z_2,Z_3,Z_4)\sim f$ where $$f(x,y,z,w)=h(x)h(y)h(z)h(w)$$ The conditional distribution of $(Z_1,Z_2,Z_3,Z_4)$ given $E$, denoted by $f_{E}$, equals $$f_{E}(z_2,z_3,z_4)=\frac{f(z_2,z_2,z_3,z_4)\sqrt{2}}{\iiint_{\mathbb{R}^3}f(z_2,z_2,z_3,z_4)\sqrt{2}dz_4dz_3dz_2}$$ The probability you're trying to compute is $$P(Z_2 \leq Z_3 \leq Z_4|Z_1=Z_2)=\int_{-\infty}^{\infty} \int_{z_2}^{\infty} \int_{z_3}^{\infty} f_{E}(z_2,z_3,z_4)dz_4dz_3dz_2$$ This is approximately 15.2%
Remark: The $\sqrt{2}$ represents the "surface area component" for the parametric surface $\vec{r}(u,v,w)=(u,u,v,w)$ whose trace is $$\Big\{(z_1,z_2,z_3,z_4)\in \mathbb{R}^4\Big|z_1=z_2\Big\}$$
