game theory - coin flipping question Lets say 2 players A and B try to have the most money at the end after playing a casino game in which they have a $49\%$ chance to double a wager.
Here are the rules to the bet between A and B:


*

*Both start with $100 

*Player A goes first. Player A plays the casino game as many times as he wants then decides to stop. After that, A can't play the game anymore.

*Then player B plays as many times as she wants to try to achieve a greater total than A.


Obviously the optimal strategy for player B involves playing until she either goes bankrupt or has more money than A (although it's not obvious what bet sizes to use).
What is the optimal strategy for A?
 A: I'll ignore the granularity of money. With arbitrarily small bets allowed, player B only has to reach the same amount as A to win with probability arbitrarily close to $1$. So, let's assume that player B wins if the totals are equal. 
The simple strategy for player A of betting everything once is pretty good, but not optimal. It wins with probability $0.2499$. 
A useful lemma should be that player B might as well play boldly, either betting everything or just enough to win at each step. See Dubbins and Savage, Inequalities for Stochastic Processes: How To Gamble If You Must. The probability of being able to achieve a target $t$ starting from $\alpha t$ is a continuous, increasing function which can be expressed as an infinite sum in terms of the probability of winning each bet and the binary digits of $\alpha$. See exercise 29 of Siegrist, "How To Gamble If You Must."
Some strategies for Player A include aiming for an amount, in which case A might as well bet boldly, too. For example, A could aim for $\$196$ by betting $\$96$, then if that fails trying to double up to $\$128$. From $\$128$ bet $\$68$, etc. The target of $\$196$ lets A win with probability $0.249999385$. 
Suppose A chooses a target which can be achieved with probability $p$. Then A wins with probability $p(1-p)$, which is maximized at the probability $p=1/2$ where A wins with probability $0.25$. This corresponds to a target of $\$195.67803788 = \$\frac{100_{10}}{0.1000001011010011110..._2}$: First bet $\$95.67803788$, and if that fails, try to double up the remaining $\$4.32196212$ $5$ times, then aim for the target again, etc. 
This leaves open the possibility that there is a better strategy which involves stopping at more than one positive value. 
