Given a set of m cards that have values pairwise different with range 1 to m, what is the probability that after shuffling the card, and picking two of them, the first one is larger than the second one?
This is part of a larger problem I'm trying to solve.
Since it is not possible for them to be equal, the problem is symmetric, i.e. the probability that first > second is the same as when second > first, so it must be each one is exactly $1/2$.
On a different level, you are asking if $X,Y \sim \mathcal{U}[1,2, \ldots, m]$ drawn without replacement, what is $\mathbb{P}[X > Y]$? The easiest way is to enumerate the possible drawings. There are exactly $m(m-1)$ choices for the values of the pair $(X,Y)$. Alternatively, note that if $X = k$ then there are $k-1$ choices for $Y$ so that $X > Y$. So the total number of choices where $X>Y$ is $$ \sum_{k=1}^m (k-1) = \sum_{k=0}^{m-1} k = \frac{m(m-1)}{2}. $$ Thus, the probability of $X>Y$ is $\dfrac{m(m-1)/2}{m(m-1)} = \dfrac{1}{2}$.
