Answer:
[tex]1 - p[/tex]
Step-by-step explanation:
Consider the [tex](\Omega, \mathcal{F}, \mathbb{P})[/tex] where [tex]\mathcal{F}[/tex] is sigma algebra and [tex]\mathbb{P}[/tex] is probabilistic measure. Denote [tex]A \subset \Omega[/tex] where Paul wins. By additivity of measure we know that
[tex]\mathbb{P}(A) + \mathbb{P}(\Omega \setminus A) = \mathbb{P}(\Omega) = 1[/tex].
So
[tex]\mathbb{P}(\Omega \setminus A) = 1 - \mathbb{P}(A) = 1 - p[/tex].
But [tex]\Omega \setminus A[/tex] is exactly the set where Paul does not win. Q.E.D.