Is there a way to really solve it?

This is a tricky one. I counted the number of allowable strings of length $n$ starting at $n = 1$ and obtained the sequence $1,0,1,0,1,1,1,2,3,4,6,10,15, ....$, which matches this OEIS sequence. So the desired probability of winning is

$P = \dfrac{1}{2} + \dfrac{1}{2^{3}} + \dfrac{1}{2^{5}} + \dfrac{1}{2^{6}} + \dfrac{1}{2^{7}} + \dfrac{2}{2^{8}} + \dfrac{3}{2^{9}} + \dfrac{4}{2^{10}} + \dfrac{6}{2^{11}} + \dfrac{10}{2^{12}} + .....$.

Now the sequence of numerators is Fibonacci-like in that the ratio of successive terms approaches the golden ratio $\phi$. This allows us to estimate that $0.798 \lt P \lt 0.799$.

Edit: As Milo Koumouris has pointed out, my initial calculation was incorrect; the estimate should actually be $P \approx 0.7112$.

I believe that

$P=1-\left(\prod_{n=1}^{\infty }\left(1-\left(\dfrac{1}{2}\right)^n\right)\right)=0.7112119049133975787211002780707692\ldots$

Based on the assumption that the ratio between successive terms of $a(n)$ is monotonically increasing and upper-bounded by $\phi$ for all $n\geq 13$, we can design an approximation technique for $P$:

$\sum_{k=1}^M\left(\dfrac{a(k)}{2^k}\right)+\dfrac{a(M+1)}{2^{M+1}\left(1-\left(\frac{a(M+2)}{2a(M+1)}\right)\right)}<P<\sum_{k=1}^N\left(\dfrac{a(k)}{2^k}\right)+\dfrac{a(N+1)}{2^{N+1}\left(1-\left(\frac{\phi }{2}\right)\right)}$

On the OEIS link, there are only $60$ terms provided, so setting $M=58$ and $N=59$ yields

$\dfrac{3467710474555566700753916705}{4875776756717555040447889408}<P<\dfrac{71576559073\sqrt{5}+819971339679777159}{1152921504606846976}$

which implies

$0.7112119048\ldots <P<0.7112119051\ldots$

and hence we can confirm $P$ rounded to $7$ decimal places as $0.7112119$. Upon searching this value, I found this link to an almost identical problem, with an answer of $0.7112119\ldots$, found by evaluating the product

$P=1-\left(\prod_{n=1}^{\infty }\left(1-\left(\dfrac{1}{2}\right)^n\right)\right)$

So thanks to the much smarter people who actually did the work, came up with this, and proved it, I believe it wouldn't be too hard to use their solutions and verify that this is indeed the answer.

Dunno if this is right but....

Define $P_n$ as the probability of winning starting right after you flipped a total of $n$ tails.

$P_n=\left(\dfrac{1}{2}\right)^{n+1}+\left(1-\left(\dfrac{1}{2}\right)^{n+1}\right)P_{n+1}$

$\Leftrightarrow (2^n-1)P_n=2^nP_{n-1}-1$

Pretty obvious that $P(\infty)=0.$

Aaaand I dunno if this is solvable x'D

0.5

Since there is no way to lose, and it is possible to win given any previous sequence of tosses, the reasoning to solve this problem changes. The only way to 'not win' is to toss an endless sequence, since the game can only terminate in a win. There are clearly infinitely many of these endless games, but since the probability of one of these endless games is infinitesimally small ($(\dfrac{1}{2})^N$ as $N\rightarrow \infty$), it would be sufficient to prove that the infinite number of these endless games is not an exponential divergence. I'm not really sure how to address this...