Say you have a game with change $p$ of winning. The reward for winning is $+1$, losing costs you $-1$. When you win 3 or more times in a row, the reward for winning becomes $+2$.
I am interested in what you average result $E(p)$ would be for playing a game (when you assume that game to be part of a series which all have change $p$ of winning). I also am interested in what $p$ you would need to have to have an average reward of $0$.
From simulations I that $E(\frac{1}{2}) \approx \frac{1}{8}$ and $E(0.455) \approx 0$. Is there any way to calculate the values?
For the game, we can set up simultaneous equations:
$$ \begin{align} E_0&=p\, (E_1+1)-q\\ E_1&=p\, (E_2+1)-q\\ E_2&=p\, (E_3+1)-q\\ E_3&=p\, (E_3+2)-q \end{align} $$
where $q=1-p$ and the required solution is:
$$E_0=\dfrac{{p}^{4}+2\,p-1}{1-p}$$
Hence, for $p=\frac{1}{2}$, $E_0=\frac{1}{8}$
For $E_0$ to be $0$, solve ${p}^{4}+2\,p-1=0$
The CAS maxima gives:
$ p=\frac{\sqrt{3}\,\sqrt{12\,\sqrt{2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}}-\sqrt{{\left( 2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}\right) }^{\frac{2}{3}}-4}\,\left( {3}^{\frac{1}{4}}\,{\left( 2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}\right) }^{\frac{2}{3}}-4\,{3}^{\frac{1}{4}}\right) }-{3}^{\frac{5}{8}}\,{\left( {\left( 2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}\right) }^{\frac{2}{3}}-4\right) }^{\frac{3}{4}}}{2\,{3}^{\frac{7}{8}}\,{\left( 2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}\right) }^{\frac{1}{6}}\,{\left( {\left( 2\,\sqrt{43}+2\,{3}^{\frac{3}{2}}\right) }^{\frac{2}{3}}-4\right) }^{\frac{1}{4}}}\\ p\approx 0.47462661756261 $