Six Numbers, One Inequality

The answer for both questions is $0.5$ and this can be obtained without any calculations. Indeed, any placement of the digits that satisfies the inequality is paired with another one that does not. The two are obtained from each other by swapping the two digits adjacent to the inequality sign. Nicu Mihalache was the first to suggest that approach on twitter.

We can do the counting as well. The sample space consists of $6!$ elements. There are ${6\choose 2}$ ways to select a pair adjacent to the inequality symbol and only one way to place them so the inequality is satisfied. The remaining four numbers can be placed in $4!$ ways, giving the total number of "successful" combinations as ${6\choose 2}\cdot 4!$ and the corresponding probability as $\displaystyle \frac{{6\choose 2}\cdot 4!}{6!}=\frac{1}{2}.$

These problems have been inspired by Donald Knuth's puzzles from the multi-author collection Puzzle Box, v 2.