Betting on the series one game at a time

Idea. This is the replication of a derivative on a binomial tree. At every score, the money you hold must be exactly what you need afterwards, whichever team wins the next game. That pins down both the money and the bet at every score, working backwards from the end of the series.

Step 1: one game. Suppose that after the next game you need $u$ if A wins it and $d$ if A loses it. With $x$ now and a bet $b$ on A, you get $x+b$ or $x-b$. So $$x=\frac{u+d}{2},b=\frac{u-d}{2}.$$

Step 2: the value at each score. Write $V(i,j)$ for the money needed when A has won $i$ games and B has won $j$. At the end, $V(4,j)=100$ and $V(i,4)=-100$. Averaging backwards is the same as asking: if the remaining games were fair coin flips, what would A's chance of winning the series be? With $p(i,j)$ that chance, $V(i,j)=100(2p(i,j)-1)$.

Step 3: after game 1. If A wins game 1, the score is 1 to 0 and B must win 4 of the next 6 games before A wins 3. Counting with fair coin flips over 6 more games, A wins the series unless B wins at least 4 of them: $$\begin{array}{rl}p(1,0)& =1-\frac{\left(\genfrac{}{}{0ex}{}{6}{4}\right)+\left(\genfrac{}{}{0ex}{}{6}{5}\right)+\left(\genfrac{}{}{0ex}{}{6}{6}\right)}{2^6}\\ & =1-\frac{22}{64}=\frac{21}{32}.\end{array}$$ By symmetry $p(0,1)=\frac{11}{32}$.

Step 4: the first bet. $$\begin{array}{rl}V(1,0)& =100\left(2\cdot \frac{21}{32}-1\right)=31.25,\\ V(0,1)& =100\left(2\cdot \frac{11}{32}-1\right)=-31.25,\\ b& =\frac{1}{2}(V(1,0)-V(0,1))=\mathrm{31.25.}\end{array}$$

So you bet 31.25 dollars on A in game 1. Your starting money $V(0,0)$ is 0, as it must be: you are replicating a fair bet.

Remark. The answer does not depend on how likely A is to win each game. Replication only needs the possible outcomes, not their probabilities, exactly as in option pricing.