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Suppose that $X_1, X_2, X_3$ are iid random variables. I have seen this fact many times that $$\mathbb{P}(X_1<X_2<X_3)=\frac{1}{6}$$ but I want to know that why every permutation of $X_1, X_2, X_3$ is equally likely and also is this true for both discrete and continuous case and can we generalize this fact for all $n \in \mathbb{N}, n \geq 2$. If yes then how to prove it?

Xi'an
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user671269
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    $\mathbb{P}(X_1<X_2<X_3)=\frac{1}{6}$ is not correct for iid discrete random variables, as there is a positive probability of equality, though you would still have $\mathbb{P}(X_1<X_2<X_3)=\mathbb{P}(X_3<X_1<X_2)=\mathbb{P}(X_2<X_1<X_3)$ etc. by symmetry/exchangeability – Henry Apr 04 '23 at 08:56
  • $\mathbb{P}(X_1<X_2<X_3)=\frac{1}{6}$ Is it true if $X_1, X_2, X_3$ are iid continuous random variables – user671269 Apr 04 '23 at 10:13

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It is a matter of notations, nothing more. For instance, define$$Y_1=X_2,Y_2=X_3,Y_3=X_1$$ Then $$(Y_1,Y_2,Y_3)\sim(X_1,X_2,X_3)$$ (meaning both triplets share the same distribution) and $$\mathbb{P}(X_1<X_2<X_3)=\mathbb{P}(Y_1<Y_2<Y_3)=\mathbb{P}(X_2<X_3<X_1)$$

Xi'an
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