Probability of (1 + min(X,Y))/(1 + min(X,Z))? I have been trying to derive the probability $\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right]$, where X, Y, and Z are independent and follow exponential distribution with parameters $λ_x$, $λ_y$, and $λ_z$, respectively. c here is a constant. What I did is briefly as follows:
$
\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right] \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\\
 = \Pr \left[ {X < Y,X < Z} \right]\\
 + \Pr \left[ {\frac{{1 + X}}{{1 + Z}} < c} \right]\Pr \left[ {X < Y,X > Z} \right]\\
 + \Pr \left[ {\frac{{1 + Y}}{{1 + X}} < c} \right]\Pr \left[ {X > Y,X < Z} \right]\\
 + \Pr \left[ {\frac{{1 + Y}}{{1 + Z}} < c} \right]\Pr \left[ {X > Y,X > Z} \right]            \;\;\;\;\;\;\;\;(2),$
from which I can obtain a closed-form expression. However, when I do simulation to verify my analytic result, I get

where the red curve is from analysis. I don't know why there is a gap between the results.
Do you think that going from (1) to (2) is problematic?

Let $B=(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c)$. You have define four events:
\begin{align}
A_1&=(X < Y,X < Z)\\
A_2&=(X < Y,X > Z)\\
A_3&=(X > Y,X < Z)\\
A_3&=(X > Y,X > Z)\\
\end{align}
and then wrote \begin{align}
Pr(B)&=\sum_{i=1}^4P(B|A_i)P(A_i)\\
\end{align}
For $i=1$ you then write
\begin{align}
P(B|A_1)P(A_1)&=P(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c|A_1)P(A_1)\\
&=P(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c|X < Y,X < Z)P(X < Y,X < Z)\\
&=P(1 < c|X < Y,X < Z)P(X < Y,X < Z)\\
&=1\times P(X < Y,X < Z)\\
\end{align}
which is correct only if $c>1$ if $c\leq1$ then this first term is zero not one.
