Probability that, given a set of uniform random variables, the difference between the two smallest values is greater than a certain value Let $\{X_i\}$  be $n$ iid uniform(0, 1) random variables.  How do I compute the probability that the difference between the second smallest value and the smallest value is at least $c$?
I've messed around with this numerically and have arrived at the conjecture that the answer is $(1-c)^n$, but I haven't been able to derive this.  
I see that $(1-c)^n$ is the probability that all the values would be at least $c$, so perhaps this is related?

There's probably an elegant conceptual way to see this, but here is a brute-force approach.
Let our variables be $X_1$ through $X_n$, and consider the probability $P_1$ that $X_1$ is smallest and all the other variables are at least $c$ above it. The first part of this follows automatically from the last, so we must have
$$P_1 = \int_0^{1-c}(1-c-t)^{n-1} dt$$
where the integration variable $t$ represents the value of $X_1$ and $(1-c-t)$ is the probability that $X_2$ etc satisfies the condition.
Since the situation is symmetric in the various variables, and two variables cannot be the least one at the same time, the total probability is simply $nP_1$, and we can calculate
$$ n\int_0^{1-c}(1-c-t)^{n-1} dt = n\int_0^{1-c} u^{n-1} du = n\left[\frac1n u^n \right]_0^{1-c} = (1-c)^n $$
