You can solve this combinatorially, without using calculus. All you need to look at is the probability that the first $n$ samples are in a certain order, and for any particular order this is simply $1/n!$

The game ends after exactly $n$ steps if and only if the first $n-1$ samples are in increasing order, and the last sample is not. The last sample can occupy any of the $n$ positions except the highest, so there are $n-1$ such sequences; hence the probability that the game ends after exactly $n$ steps is $\frac{n-1}{n!}$.

And $A$ wins if the game ends after an even number of steps, so $A$'s probability of winning is $$\begin{align} \sum_{n=1}^\infty\frac{2n-1}{(2n!)} & = \sum_{n=1}^\infty\left(\frac{1}{(2n-1)!}-\frac{1}{(2n!)}\right) \\ & = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n!} \\ & = 1-\sum_{n=0}^\infty\frac{(-1)^n}{n!} \\ & = 1 - \frac{1}{e} \end{align}$$

You can solve this combinatorially, without using calculus. All you need to look at is the probability that the first $n$ samples are in a certain order, and for any particular order this is simply $1/n!$

The game ends after exactly $n$ steps if and only if the first $n-1$ samples are in increasing order, and the last sample is not. The last sample can occupy any of the $n$ positions except the highest, so there are $n-1$ such sequences; hence the probability that the game ends after exactly $n$ steps is $\frac{n-1}{n!}$.

And $A$ wins if the game ends after an even number of steps, so $A$'s probability of winning is $$\begin{align} \sum_{n=1}^\infty\frac{2n-1}{(2n!)} & = \sum_{n=1}^\infty\left(\frac{1}{(2n-1)!}-\frac{1}{(2n!)}\right) \\ & = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n!} \\ & = 1-\sum_{n=0}^\infty\frac{(-1)^n}{n!} \\ & = 1 - \frac{1}{e} \end{align}$$

This assumes nothing about the particular distribution of the samples, except that it is continuous. So the answer is the same whatever the distribution.