(perhaps) A Stupid Question about Poker first time to ask a question here so I hope this would be an interesting one.
Suppose me and a friend of mine are playing poker (1 vs 1), and at a certain time during the heads up we get bored and decide to go all in every hand without looking at our cards until a winner is declared. 
We suppose (of course) that before dealing cards the change of one of of us winning the all in is perfect equal: 0.5 me and 0.5 my friend.
BUT we starts the heads up with different stacks, say that I start with P percentage of the total chips in play, and my friend have 1-P percentage of the chip in play.
So the question here is : what is the probability of me winning the heads up??
My first answer that I think it's correct is : P.
I made the calculations for a particular case in which i start the heads up with 80 and my friend with 20 and came out that my chance of winning are exactly 0.8.
I attached a photo with my solution (hope the graph it's clear and the calculations also : the chance of winning the first time is 0.5, the second time 0.25 and then with probability 1/16 we return to initial stacks which is the top of the tree which i marked with a star)
Hope you guys like the question and thank you in advance for the help

 A: This doesn't seem to be about poker anymore. It looks more like a stupid coin-tossing game.
You're correct, but it takes more than just calculating separate examples to prove that. I'll assume that coins are indefinitely splittable.
Assume that we have the probability $p(x)$ for winning where $x$ is the amount of the total coins you have. You have certainly that $p(1)=1$ and $p(0)=0$ (since you've already won or lost in those cases). Furthermore if we consider your stack after a single toss:
$$\begin{matrix}
& \text{win} & \text{loose} \\
\hline
x \le 1/2 & 2x & 0 \\
x \ge 1/2 & 1  & 2x - 1 \\
\end{matrix}$$
This means that we can see that this gives us a way to give $p(x)$ certain properties. If $x\le1/2$ the chance of winning is $1/2$ of the chance of winning with $2x$ of stack (as you have $1/2$ chance of reaching that in the next toss and otherwise you loose). We can do similar calculation for $x\ge 1/2$ and get (and of course the $p(1/2) = 1/2$ by symmetry):
$$p(x) = \begin{cases}
p(2x)/2 & \text{ if } x < 1/2 \\
1/2 & \text { if } x = 1/2 \\
1/2 + p(2x-1)/2 & \text{ if } x > 1/2 \\
\end{cases}$$
We can immediately see that $p(x)=x$ would satisfy that condition.
To see that this is indeed the only solution one would note that this property of $p$ could be used to determine $p(j2^{-k})$ recursively and that $p(j2^{-k}) = j2^{-k}$. Then you can also see from the property that $p(x)$ is continuous and since we have proven that $p(x)=x$ on a dense set then $p(x)=x$ always.

Another way is to consider how the table looks if you express $x$ in binary. $2x$ means to shift the bits one step to the left and so does $2x-1$ except that one discards the most significant (fractional) bit. In your example you would have $x = 0.8_{(10)} = 0.110011001100\dots_{(2)}$. Those are the tosses that makes the game continue. The probability for a win is when the left most fractional bit is $1$ and the toss ends the game which has the probability $2^{-j}$ for the $j$th toss. This means that you in the example can win on the first or second toss, the fifth or sixth etc.
