This answer only examines the betting strategy the OP advocates (always bet half of what you have) and some of its variants. With this strategy:
One never gets broke (one's fortune is always positive). One doubles up almost surely if $p$ is large enough but with a probability which is less than $1$ if $p$ is small enough.
Note that there is no theorem saying that every positive submartingale reaches every level almost surely, hence arguments based on averages of increments only, cannot suffice to reach a conclusion.
In the strategy the OP advocates, the fortune performs a multiplicative random walk whose steps from $x$ are to go to $\frac32x$ and to $\frac12x$ with probability $p$ and $1-p$ respectively. Thus, the logarithm of the fortune performs a usual random walk with steps $\log\frac32$ and $\log\frac12$, whose constant drift is
$m(p)=p\log\frac32+(1-p)\log\frac12$ hence $m(p)=p\log3-\log2$ has the sign of $p-p^*$ where $p^*$ solves the equation $3^{p^*}=2$, hence $p^*=\frac{\log2}{\log3}=.6309...$
If $p\geqslant p^*$, $m(p)$ is nonnegative hence the logarithm of the fortune reaches the set $[C,+\infty)$ almost surely for every $C$. In particular, one doubles up almost surely.
If $p<p^*$, $m(p)$ is negative hence the logarithm of the fortune has a positive probability $q$ to never visit the set $[\log2,+\infty)$, for example. This means that with probability $q$, one will bet an infinite number of times without ever getting broke nor doubling up. In effect the fortune at time $k$ will go to zero like $a(p)^k$ when $k\to+\infty$, with $a(p)=\frac123^p<1$.
There is no easy formula for the probability $q$ to never double up but the probability to double up $n$ times decreases exponentially like $\exp(-b(p)n)$ when $n\to\infty$, where $b(p)$ is the unique positive solution of the equation $p(3^b-1)=2^b-1$.
If $p<p^*$, the strategy where one always bets half of what one has fails. What happens with the strategy where one always bets a proportion $r$ of what one has? The same analysis applies and shows that one doubles up almost surely, for every $p\geqslant\wp(r)$, where $p=\wp(r)$ solves the equation $(1+r)^{p}(1-r)^{1-p}=1$ (note that $\wp(\frac12)=p^*$).
The other way round, for every given $p>\frac12$, a strategy which wins almost surely is to bet a proportion $r$ of what one has, provided $p\geqslant\wp(r)$. Since $\wp(r)\searrow\frac12$ when $r\searrow0$, one gets:
For each $p>\frac12$, the strategy where one always bets a proportion $r$ of what one has wins almost surely for every small enough positive value of $r$ (for example, $r\leqslant 2p-1$).
Once $A$ has finished playing, ending with an amount $a$, the strategy for $B$ is simple and well-known: Use bold play.
That is, aim for a target sum of $a+\epsilon$ and bet what is needed to reach this goal exactly or bet all, whatever is less. As seen for example here, the probability of $B$ reaching this target is maximized by this strategy and depends only on the initial proportion $\alpha:=\frac{100}{a+\epsilon}\in(0,1)$. (Of course, $B$ wins immediately if $a<100$). While the function $p(\alpha)$ that returns $B$'s winning probability is fractal and depends on the dyadic expansion of the number $\alpha$, we can for simplicyity (or a first approximate analysis) assume that $p(\alpha)=\alpha$: If the coin were fair, we would indeed have $p(\alpha)=\alpha$, and the coin is quite close to being fair.
Also, we drop the $\epsilon$ as $B$ may chose it arbitrarily small. (This is the same as saying that $B$ wins in case of a tie).
In view of this, what should $A$ do?
If $A$ does not play at all, $B$ wins with probability $\approx 1$.
If $A$ decides to bet $x$ once and then stop, $B$ wins if either $A$ loses ($p=0.51$) and $B$ wins immediately or if $A$ wins $p=0.49$ and then $B$ wins (as seen above) with $p(\frac{100}{100+x})\approx \frac{100}{100+x}$. So if $A$ decides beforehand to play only once, she better bet all she has and thus wins the grand prize with probaility $\approx 0.49\cdot(1-p(\frac12))\approx \frac14$.
Assume $A$ wins the first round and has $200$. What is the best decision to do now?
Betting $x<100$ will result in a winning probability of approximately
$$0.49\cdot(1-\frac{100}{200+x})+0.51\cdot(1- \frac{100}{200-x}) $$
It looks like the best to do is stop playing (with winning probability $\approx\frac12$ now).
Alernatively, let us assume instead that $A$ employs bold play as well with a target sum $T>100$. Then the probability of reaching the target is $\approx \frac{100}{T}$, so the total probability of $A$ winning is approximately
$$ \frac{100}T\cdot(1-\frac{100}T)$$
and this is maximized precisely when $T=200$.
This repeats what we suspect from above:
The optimal strategy for $A$ is to play once and try to double, resulting in a winning probability $\approx \frac14$.
Admittedly, the optimality of this strategy for $A$ is not rigorously shown and especially there may be some gains from exploiting the detailed shape of $B$'s winnign probability function, but I am pretty sure this is a not-too-bad approximation.
Best Answer
Expanded comment to answer:
In general, to calculate the net amount earned, you would subtract the expected losses from the expected winnings. Let's denote this as $E[W]-E[L]$ where $E$ denotes the expected value.
First, solving for $E[W]$ gets us
$$E[W] = P[W]*x_1 $$, where $x_1$ denotes the amount gained on a win.
Similarly, solving for $E[L]$ gives us:
$$E[L] = P[L]*x_2$$
Then, we simply apply the numbers to the given problem. For your first example, we have
$$E[W] = \frac{5}{6}*1$$
$$E[L] = \frac{1}{6}*1$$
$$E[W]-E[L] = \frac{5}{6}-\frac{1}{6}=\frac{4}{6} = \frac{2}{3}$$. This value denotes the amount you are expected to win per round of playing. Accordingly, finding the expected winnings for multiple rounds gives us $$\frac{2}{3} * \text{number of rounds}$$
Additionally, plugging in our numbers for the even/odd game gives:
$$E[W] = \frac{3}{6}*1$$
$$E[L] = \frac{3}{6}*1$$
$$E[W]-E[L] = \frac{3}{6}-\frac{3}{6}=\frac{0}{6} = 0$$, as you had calculated originally
If you want to make a simple yet profitable game, I'd recommend making a game that has so many combinations that it's hard to track the probabilities (such as blackjack, which, even with only one player and a dealer, gives over 6 million possible starting hands, assuming no card repetition (and that doesn't include the added combinations from "hitting"). Alternatively, choose a game like roulette, which is quite simple and has up to a 47.4% (almost 50-50) chance of winning if you only bet on one color. Dice are somewhat harder to make these types of games with since there are only 36 possible rolls of a pair of dice (6 per die), but if you add several rounds or rules to one game, you can complicate things significantly harder to calculate, while the game can be easy to understand.