Intuitive explanation of $\Pr(X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3)$ where $X_1, X_2, X_3$ are exponentials Let $X_1, X_2, X_3$ be exponential random variables with rates $\lambda_1, \lambda_2, \lambda_3$. We have the following
$$\Pr(X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3) = \frac{\lambda_1 + \lambda_3}{\lambda_1 + \lambda_2 + 2 \lambda_3}$$
Which I found by considering $f(1, 2, 3) = \Pr(X_1 < X_2 < X_3)$ and noticing that the above is equal to $\frac{f(1, 2, 3)}{f(1, 2, 3) + f(2, 1, 3)}$. However, what is a more intuitive explanation of this result? It looks like the probability that $Y_1 < Y_2$, if $Y_1$ is exponential with rate $\lambda_1 + \lambda_3$ and $Y_2$ is exponential with rate $\lambda_2 + \lambda_3$.
 A: We have that 
$$\Pr[X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3] =\frac{\Pr[X_1 < X_2 , \max(X_1, X_2, X_3) = X_3]}{\Pr[\max(X_1, X_2, X_3) = X_3]}$$
$$=\frac{\Pr[X_1 < X_2 < X_3]}{\Pr[X_1<X_3, X_2<X_3) ]}$$
$$=\frac{\Pr[X_1 < X_2 < X_3]}{\Pr[X_1<X_2<X_3]+\Pr[X_2<X_1<X_3]} \quad(1)$$
This looks "intuitive": given $X_3$ is the largest, we are left with two other options (that appear in the denominator) and we have the one that interests us in the numerator. As for calculation,
Let 
$$f_i(x_i) = \lambda_ie^{-\lambda_ix_i},\;\;\; i=1,2,3$$
Then, assuming that the three variables are independent,
$$\Pr[X_1 < X_2 < X_3] = \int_0^{\infty}f_3(x_3)\int_0^{x_3}f_2(t)\int_0^{t}f_1(s)ds\,dt\,dx_3$$
etc, and noting that the distribution function can be written
$F_i = 1- \frac 1{\lambda_i}f_i$.  A quick Monte Carlo verifies that the calculation of the probability by the OP is correct.  
Moreover, the OP correctly noted that the obtained probability is equal to $\Pr(Y_{1|3} < Y_{2|3})$, if $Y_{1|3}$ is Exponential with rate $\lambda_1 + \lambda_3$ and $Y_{2|3}$ is Exponential with rate $\lambda_2 + \lambda_3$.  
Well the following is a well known result for i.i.d Exponentials
$$\min\{X_1,X_3\} = Y_{1|3},   \min\{X_2,X_3\}=Y_{2|3}$$
So we are looking at
$$\Pr [\min\{X_1,X_3\} < \min\{X_2,X_3\}] = \Pr[X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3]$$
...which makes sense, if we start pondering what options does the left-hand-side allows for.
