Are you sure this is fair?

Being $p$ the probability of getting heads and Henry the one who wins with heads, let's calculate Henry's chances of winning, which we want to be $P=\frac{1}{2}$ since the game has to be fair. For Henry to win, he must win at the 2nd turn after Tom loses the first, or at the fourth after Tom loses the third, and so on. But notice that the third turn is like starting the game over. This allows us to skip the infinite sum and write a simple equation for $P$ :

$P=(1-p)+p\cdot(1-p)\cdot P$

For $P=\frac{1}{2}$ we get:

$p^2+p-1=0$

The solution to this equation is the reciprocal of the golden ratio! $\quad \dfrac{1}{\phi}=\dfrac{\sqrt{5}-1}{2}=\boxed{0.618033988...}$

The probability of Tom winning is $\begin{aligned} P(\text{Tom}) &= \overbrace{(1-p)}^{\color{#3D99F6}\text{Tails flip }1} + \overbrace{p\times(1-p)\times(1-p)}^{\color{#3D99F6}\text{Heads flip }1\text{, Tails flip }2\text{, Tails flip }3} +p\times(1-p)\times p\times(1-p)\times(1-p) + \cdots \\ &= \sum_{i=0}^\infty (1-p)^{n+1}p^n \\ &= (1-p)\sum_{i=0}^\infty (p-p^2)^n \\ &= (1-p)\left(\frac{1}{1-(p-p^2)}\right) \qquad\qquad {\color{#3D99F6}\text{Since }0<p-p^2<1} \\ &= \frac{1-p}{p^2-p+1} \end{aligned}$

Now, if both players have the same probability of winning the game, we have that $P(\text{Tom})=\frac{1}{2}$ . So $\begin{aligned} \frac{1-p}{p^2-p+1} &= \frac{1}{2} \\ p^2 - p + 1 &= 2-2p \\ p^2 + p - 1 &= 0 \\ p &= \frac{-1 \pm \sqrt{5}}{2} \\ p &= \frac{\sqrt{5} - 1}{2} && \color{#3D99F6}\text{Taking the positive solution} \\ p &= \phi - 1 && \color{#3D99F6}\text{Where }\phi\text{ is the Golden Ratio} \\ p &\approx \boxed{0.618} \end{aligned}$