Minimum of Unit-Exponentials Plus Constants Define $e_i$ to be iid random variables drawn from an exponential distribution with parameter $\lambda=1$.
$a_i$ are numerical constants.  I am interested in the probability that
$a_1 + e_1 < a_2 + e_2$
ie, that $a_1 +e_1$ is the min.  The solution is $\exp[- (a_1  - a_2 )]/2$ if $(a_1  - a_2 )>0$, and $1-\exp(a_1 -a_2)/2$ otherwise.  For example, if $a=(1,2)$, then probabilities are $(82\%, 18\%)$.  [Note: I am not interested in the order statistics.]  PS: Formulas and numbers corrected---thanks angryavian below.
What about three (or more) variables?  Again, I want to know the probability that i is the min, given $(a_1 , a_2 , a_3 )$. For example, if $a=(1,2,3)$, then the probabilities I am interested in are $(76.50\%, 17.56\%, 5.94\%)$.
I am guessing there is no analytical generalization, but if there is, I would greatly appreciate pointers.
 A: Without loss of generality suppose $a_1 < \cdots < a_n$.
The probability you seek is
$$P \left(\bigcap_{j\ne i} B_j\right) = E[\mathbf{1}_{\bigcap_{j\ne i} B_j}]
= E\left[\prod_{j \ne i} \mathbf{1}_{B_j}\right]$$
where $B_j := \{e_i + a_i \le e_j + a_j\}$.
If we condition on $e_i$, then the $B_j$ are independent, and we have
\begin{align}
P \left(\bigcap_{j\ne i} B_j\right)
&= E\left[\prod_{j \ne i} \mathbf{1}_{B_j}\right]
\\
&= E\left[E\left[\prod_{j \ne i} \mathbf{1}_{B_j} \mid e_i\right]\right]
\\
&= E\left[\prod_{j \ne i} E[\mathbf{1}_{B_j} \mid e_i]\right]
\\
&= E\left[\prod_{j \ne i} \min\{1, e^{-(e_i + a_i - a_j)}\}\right].
\end{align}
Not quite sure how to compute this analytically in general.

In the case of $n=2$ and $i=1$, some calculation yields
\begin{align}
E[\min\{1, e^{-(e_1 + a_1 - a_2)}\}]
&= (1 - e^{-(a_2  - a_1)}) + e^{a_2 - a_1}\int_{a_j - a_i}^\infty e^{-2x} \mathop{dx}
\\
&= (1-e^{-(a_2 - a_1)}) + e^{a_2 - a_1}\frac{1}{2} e^{-2(a_2 - a_1)}
\\
&= 1 - \frac{1}{2} e^{-(a_2 - a_1)}.
\end{align}
This is a bit different from your reported probability $88.4\%$ and formula $1 - \exp(-(a_2 - a_1))$, but I verified the above formula with simulations:
n <- 2
a <- c(1, 2)
ct <- rep(0, n)
formula <- 0
trials <- 1e5
for (i in 1:trials) {
  e <- rexp(2, 1)
  ct[which.min(a+e)] <- ct[which.min(a+e)]+1
  formula <- formula + min(1, exp(-e[1]-a[1]+a[2]))
}
ct/trials
formula/trials

> ct/trials
[1] 0.81789 0.18211

> formula/trials
[1] 0.8170634

