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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?

edited Jan 22 at 14:22

Douglas Zare
13.1k●3●36●58

asked Jan 22 at 5:37

nishikun
31●3

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Hi nishikun: I think this is a cute question, but it is not obvious why it is "research-level". It might be, in which case you should explain why/how — motivation is a very important part of writing mathematics. But most likely you would be better off at math.stackexchange, which has a broader range of questions. Please look over mathoverflow.net/faq for more information. – Theo Johnson-Freyd Jan 22 at 6:34

thanks, i posted it there just now – nishikun Jan 22 at 7:10

Douglas Zare · Accepted Answer · 2013-01-22 14:09:46Z UTC

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.

game theory - coin flipping question

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