A "paradox" in the odds of being the minimum variable in a set of IID variables This question arose when I was trying to find the odds that a customer is served before the customer directly ahead of him in an M/M/m queueing system.
For a a RV in a set of IID RV of size N, is the probability of being the minimum value the same as the probability of being less than the minimum value of a set of size N - 1?
Let's say you have 10 IID exponentially distributed random variables.
You pick label one variable as "A" arbitrarily and ask this question: What is the probability that A is the minimum of the set of 10? Obviously because they are IID, the answer is 1/10.
Yet I have an issue. It seems true that the question "Is A the minimum" is equivalent to the question "Is A less than the minimum of the other 9".
The minimum of 9 exponentially distributed variables with parameter lambda has mean
$$\frac {1}{9\lambda} $$
So to find the probability of A being less than the other nine, I took the integral
$$\int_{0}^{\frac {1}{9\lambda}} \lambda e^{-\lambda x} d x $$
But this evaluates to
$$ 1 - e^{\frac{1}{9}} \ne \frac{1}{10}$$
Where have I gone wrong?
 A: The probability of $X_N$ being less than the minimum of $\{X_1, \ldots, X_{N-1}\}$
is simply that: the probably that $X_N$ will be less than whatever the minimum of the other $N-1$ variables happens to be.
The minimum is itself a random variable, not a constant.
If all the variables happen to be iid exponential variables with parameter $\lambda,$
then it is true that the mean of the minimum of the first $N-1$ variable is
$\frac1{(N-1)\lambda}.$
But there is a non-zero probability that $X_N$ is greater than $\frac1{(N-1)\lambda}$
and yet is still smaller than the minimum of the other $N-1$ variables.
There is also a non-zero probability that $X_N$ is less than $\frac1{(N-1)\lambda}$
and yet is greater than the minimum of the other $N-1$ variables.
So $X_N < \min\{X_1, \ldots, X_{N-1}\}$ is not the same event as
$X_N < \frac1{(N-1)\lambda}.$
It should not be surprising that the probabilities of two different events are different.
We can consider a joint distribution of the two variables
$X = X_N$ and $Y = \min\{X_1, \ldots, X_{N-1}\}$
and ask about the probability that $X < Y.$ The answer is
\begin{align}
P(X_N < \min\{X_1, \ldots, X_{N-1}\})
&= \int_0^\infty \int_0^y 
    \lambda e^{-\lambda x} \cdot (N-1)\lambda e^{(N-1)\lambda y}\, dx\, dy \\
&= (N-1)\lambda^2 \int_0^\infty e^{(N-1)\lambda y}
     \int_0^y e^{-\lambda x} \, dx\, dy \\
&= (N-1)\lambda^2 \int_0^\infty e^{(N-1)\lambda y}
     \cdot \frac1\lambda \left(1 - e^{-\lambda y}\right)\, dy \\
&= (N-1)\lambda\int_0^\infty \left(e^{(N-1)\lambda y} - e^{-N\lambda y}\right)\,dy\\
&= (N-1)\lambda \left(\frac1{(N-1)\lambda} - \frac1{N\lambda}\right)\\
&= \frac1N.
\end{align}
