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?

You can calculate the probability by integrating the joint pdf like this
$\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right] = \int\limits_{z = 0}^{z = \infty } {\int\limits_{x = 0}^{x =  - 1 + c + c*z} {\int\limits_{y = 0}^{y = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } }  + \int\limits_{z = 0}^{z = \infty } {\int\limits_{x =  - 1 + c + c*z}^{x = \infty } {\int\limits_{y = 0}^{y =  - 1 + c + c*z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } } $
This is for $c>1$
