I need to find a more rigorous approach to a question which is:
Let $X_1,X_2,X_3,X_3,X_4,X_5$ be i.i.d. continuous random variables having a continuous distribution function. Then $P{(X_1>X_2>X_3>X_4>X_5|X_1=max(X_1,X_2,X_3,X_4,X_5))}$ equals:
I attempt is that we will solve this question in a world where we already have $X_1$ maximum of all. Since all variables are iid then if we get to choose 4 values from the distribution then the total number of combinations to arrange them will be 4! and only 1 event is holding true hence the answer will be $\frac{1}{4!}$.
But does this approach fail in a different situation? What more do I need to prove this answer.
$P(X_1>X_2>X_3>X_4>X_5)=\frac 1 {5!}$ because all the $5!$ permutations of the $X_i$s'have the same distribution (see detals below). The conditional probability is this divided by $P(X_1 =\max \{X_1,X_2,X_3,X_4,X_5\})$. You can evaluate this last probability by conditioning on $X_1$ and the value is $E(F(X_1)^{4})=\int F(x)^{4} dF(X)=\frac 1 3(F(x))^{5}|_{-\infty}^{\infty}=\frac 1 5$. So the answer is $5\frac 1 {5!}=\frac 1 {4!}$. (Here $F$ is the common distribution).
For the claim made in the beginning take any permutation $p$ of $\{1,2,3,4,5\}$. Since the joint distribution of $(X_{p(1)},X_{p(2)},....,X_{p(n)})$ is the same as that of $(X_1,X_2,X_3,X_4,X_5)$ (namely the five fold product of the common induced measure $\mu$) the probability of the event $X_{p(1)}>X_{p(2)}>....>X_{p(n)}$ does not depend on the permutation $p$. The probability of their union is $1$ since $P(X_i=X_j)=0$ for $ i \neq j$. Hence each of them has probability $\frac 1 {5!}$.