Conditonal probability symmetry Let $X_1,X_2.....X_6$ be six iid rv.

What is the probability of 
  $P(X_6>X_2|X_1=\max(X_1,X_2...X_5))$?

It's a question of sheldon ross. The writer tried to solve it using the following method  that is given in the following image. How he apply symmetry in question to give the probability of
$P(X_6>X_2|X_1=\max(X_1,...,X_5),X_6<X_1)=1/2$.

 A: If we assume that $\mathbb{P}(X_i = X_j) = 0$ when $i=j$ so that the distinction between strict and non-strict inequalities disappears, then it is clear that
$$
\bigl\{X_1 = \max(X_1,\dots,X_5)\, \text{and}\, X_6 < X_1\bigr\} = \bigl\{X_1 = \max(X_1,\dots,X_6)\bigr\}
$$
So you need to work out
$$
\mathbb{P}\bigl( X_6 > X_2 | X_1 = \max(X_1,\dots,X_6)\bigr).
$$
Is the symmetry any clearer?
A: If $X_1,X_2,\ldots, X_6$ are i.i.d. and have probability $0$ of any equality, then they can come in $6!=720$ possible orders


*

*In a fifth of these cases, i.e. $144$ of them, you have $X_1=\max(X_1,X_2,\ldots,X_5)$ 

*and in a sixth of those, i.e. $24$ of them, have $X_6 > X_1 = \max(X_1,X_2,\ldots,X_5) > X_2$ 

*while the other $120=5!$ cases have $X_6 < X_1 = \max(X_1,X_2,\ldots,X_5)$, i.e. $X_1 = \max(X_1,X_2,\ldots,X_5, X_6)$ 


*

*and precisely half of those, i.e. $60$, have $X_6 < X_2 < X_1 = \max(X_1,X_2,\ldots,X_6)$   

*and the other half, i.e. also $60$, have $X_2 < X_6 < X_1 = \max(X_1,X_2,\ldots,X_6)$
It is this final point which is the symmetry.
From this, the desired probability can alternatively be stated with a counting argument as $$P(X_6 > X_2 \mid X_1 = \max(X_1,X_2,\ldots,X_6)) = \dfrac{24+60}{144}=\dfrac7{12}$$
