Check your intuition

It's easier to calculate the probability that Beatrice doesn't win.

Beatrice loses when the coin always comes up tails. This happens with probability $\prod_{n=1}^{\infty}\frac{2^n+1}{2^n+2}$ . We claim this infinite product converges to $\frac{1}{2}$ .

Proof. We proceed by induction on partial sums to show that $\prod_{n=1}^k\frac{2^n+1}{2^n+2}=\frac{1}{2^{k+1}}+\frac{1}{2}.$

This holds when $k=1$ , since $\frac{2^1+1}{2^1+2}=\frac{3}{4}=\frac{1}{2^2}+\frac{1}{2}$ . Assume the claim holds when $k=m$ . Then $\prod_{n=1}^{m+1}\frac{2^n+1}{2^n+2}=\left(\frac{1}{2^{m+1}}+\frac{1}{2}\right)\cdot\frac{2^{m+1}+1}{2^{m+1}+2}$ $=\frac{(2^{m+1}+2)(\frac{1}{2}+\frac{1}{2^{m+2}})}{2^{m+1}+2}=\frac{1}{2^{m+2}}+\frac{1}{2},$ so the claim holds by induction. Then $\prod_{n=1}^{\infty}\frac{2^n+1}{2^n+2}=\lim_{k\to\infty}\frac{1}{2^{k+1}}+\frac{1}{2}=\frac{1}{2}.$ QED.

Thus, the probability that Beatrice doesn't win is $\frac{1}{2}$ , so the probability that Beatrice does win is $\boxed{\frac{1}{2}}$ .