I am having trouble finding the expectation value of the following.
We play a betting game on the weather, $80\%$ of the days it rains, $20\%$ of the days it doesn't rain. In this case betting on rain is the optimal strategy, say I start with money $m=1$ and with this strategy I would obviously bet a fraction $b$ of my money every time. I am getting mixed answers on the expectation value of my money after $n=30$ games $\ldots$
My first guess is that I expect to win $0.8\cdot30=24$ times, thus expect to have $$E[m]=(1+b)^{24}(1-b)^6$$ after $30$ games, taking the derivative would allow me to find my optimal betting fraction.However I also reason that I can use the following model to get my expectation value
\begin{align} E[m] & = \sum_{k=0}^{30}(1+b)^k(1-b)^{30-k}{30 \choose k}0.8^k0.2^{30-k} = (1+0.6b)^{30} \end{align}
These values are obviously not the same, this one is maximised for $b = 1$, which makes no sense since one loss would bankrupt me. But here I just used $E[m] = \sum m(x) P[X=x]$. With $m(x)$ the money I would have with $x$ wins, and $P$ the probability of $x$ wins.
Which of the two values is right, and where is my reasoning going wrong? I would expect both approaches to reach the same conclusion... Common sense tells me that the first answer is right, but I cant figure out why...