Unfair Coin

Since I flip first and Paul flip second. I flip all the odd number turn and Paul, even. For Paul to win he must flip a head at the last turn, which is even, and all flips before the last must be tail. The probability for that to happen is as follows:

$\begin{aligned} P & = {\color{#D61F06} \frac 56} \cdot {\color{#3D99F6}\frac 16} + {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#3D99F6}\frac 16} + {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#D61F06} \frac 56} \cdot {\color{#3D99F6}\frac 16} + \cdots \\ & = \sum_{n=0}^\infty {\color{#3D99F6}\frac 16} \left({\color{#D61F06}\frac 56}\right)^{2n+1} = \frac 16 \cdot \frac 56 \sum_{n=0}^\infty \left(\frac {25}{36}\right)^n = \frac 5{36}\left(\frac 1{1-\frac {25}{36}} \right) = \boxed{\dfrac 5{11}} \end{aligned}$

Let the probability that I win be $P.$ If I flip a heads on my first toss, then the coin goes to him, so the game "restarts" with Paul flipping the coin. Therefore, the probability that Paul wins is $\frac 56 P.$ Since either one of us needs to win, we have $P+\frac 56 P=1 \implies P=\frac 6{11}.$ Therefore, the probability of Paul winning is $\frac 56 P=\frac 5 {11}.$