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Given $n$ random variables uniformly distributed from $0$ to $1$, what is the probability that no two differ by less than some $x$?

My initial thought was that for $n$ random variables there are $n \choose 2$ pairings, and the probability that for each given pairing the difference is greater than $x$ is given by $({1-x})^2$ so then the answer would be $\Big((1-x)^2\Big)^{n \choose 2}=(1-x)^{n(n-1)}$. But upon inspection this answer is obviously wrong.

For instance consider the case where $n=3$ and $x>0.5$. There exist no solution. This should be obvious since if the smallest number and the middle number differ by more than $x$ and if the middle number and the largest number differ by more than $x$ then that implies the smallest number and the largest number differ by more than $1$ which is impossible. But the formula above gives a non zero answer.

What is the correct answer to this question? Any help would be appreciated

Bertrand Einstein IV
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    This answer gives a derivation as part of a solution to a related problem. – MathIsFun7225 Dec 04 '23 at 04:51
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    Use order statistics: $X_{(1)}, X_{(2)}, \ldots, X_{(n)}$. Then consider the differences of neighboring values: $X_{(2)}-X_{(1)}, X_{(3)}-X_{(2)},\ldots,X_{(n)}-X_{(n-1)}$. The probability of the minimum difference being greater than $x$ is given by $\text{Pr(min difference}>x)=(-n x+x+1)^n$ for $0<x<1/(n-1)$. – JimB Dec 04 '23 at 05:12
  • @MathIsFun7225 What an amazing answer, particularly the analogy for interpreting the probability as volumes of hypercubes – Bertrand Einstein IV Dec 04 '23 at 17:00

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