Fair Game

If $p_A$ is Alice's probability and $p$ is the probability of flipping head,

$p_A=p+(1-p)p^2+(1-p)^2p^3+...=\displaystyle\sum_{x=0}^\infty(1-p)^xp^{x+1}$

because Alice wins if occurs H or THH or THTHH and so on, where H means Head and T means Tail.

So let's solve the series:

$\displaystyle\sum_{x=0}^\infty(1-p)^xp^{x+1}=p\displaystyle\sum_{x=0}^\infty(1-p)^xp^x=p\displaystyle\sum_{x=0}^\infty((1-p)p)^x=p\frac{1}{1-(p-p^2)} \equiv\frac{1}{2}$

And now let's solve for $p$ :

$p^2-3p+1=0$

$p_1=\frac{3-\sqrt5}{2}\;\vee \; p_2=\frac{3+\sqrt5}{2}$

The only valid solution is $p_1$ because $p_2>1$ . Therefore $p=\frac{3-\sqrt5}{2}\simeq\boxed{38.19\%}$

Condition the problem on the outcome of the first one or two tosses. If Alice throws H, she wins. If Alice throws H and Charlie T, he wins. If Alice throws H and Charlie throws H as well, we are back at the original state, with Alice about to roll again. Thus, if the probability of throwing H is $p$ and if the probability that Alice wins is $q$ , then $\begin{aligned} q & = \; p \times 1 + (1-p)(1-p)\times0 + (1-p)p \times q \\ q & = \; \frac{p}{1 - p(1-p)} \end{aligned}$ Since we want $q=\tfrac12$ we obtain $p^2 - 3p + 1 = 0$ , and so $p = \tfrac12(3 \pm\sqrt{5})$ . Only $p = \tfrac12(3-\sqrt{5})$ is a valid probability, making the answer \boxed{38.19660113\} %.

If $p$ is the probability fo heads,

Alice's probability is $p+p^{2}(1-p)+p^{3}(1-p)^{2}+...=\frac{1}{2}$

Charlie's probability is $(1-p)^{2}+p(1-p)^{3}+p^{2}(1-p)^{4}+...=\frac{1}{2}$

Multiply all of Charlie's probability by $\frac{p}{(1-p)^{2}}$ to get $p+p^{2}(1-p)+p^{3}(1-p)^{2}+...=\frac{p}{2(1-p)^{2}}$

This is the same as Alice, so $\frac{1}{2}=\frac{p}{2(1-p)^{2}}$

$p^{2}-3p+1=0$

$p=\frac{3 \pm \sqrt{5}}{2}$

The minus solution is about $p=0.381966=\boxed{38.1966}\%$

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Incidentally $p=1-\frac{1}{\phi}$