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?

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    $\begingroup$ 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. $\endgroup$ – Theo Johnson-Freyd Jan 22 '13 at 6:34

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.

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  • $\begingroup$ So it is a big advantage to win at a tie. If $A$ wins a tie then the "do not bet at all" strategy wins with probability $51\%$ which, from what you say, might be slightly more than twice as good as any strategy with actual betting. So perhaps the modification of no one wins a tie is more interesting, although then the strategic play might be that no one bets. $\endgroup$ – Aaron Meyerowitz Jan 22 '13 at 18:57
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    $\begingroup$ Aaron, that's not right (and that must be what Douglas means in his first paragraph). If I reach $\$200$, I can arrange to reach $\$200+\epsilon$ with extremely high probability; first I try betting $\epsilon$, then if I lose that one, $2\epsilon$, then if I lose that one, $4\epsilon$, etc. If $\epsilon$ is small, I have an enormous number of chances to win before I run out of funds. So the rule for breaking ties is not important. $\endgroup$ – James Martin Jan 22 '13 at 19:16
  • $\begingroup$ This strategy ("Bold play to the target $t$ reachable with probability $\frac12$") indeed seems optimal for $A$. A strategy for $A$ leads to a probability distribution of stopping values $X$. Assume a strategy has $P(0<X<100)>0$; then replacing it with a strategy that switches to bold play with target $101$ whenever $0<X<100$ improves the expected winning probability. Likewise, if $P(a<X<a+\epsilon)>0$ with $100\le a <t$ and $\epsilon$ small, then appending bold play towards $t$ whenever $a<X<a+\epsilon$ is an improvement (check!). [continued] $\endgroup$ – Hagen von Eitzen Jul 10 '13 at 13:32
  • $\begingroup$ [continued] Remains to show that $P(X>t)>0$ also allows an improvement; there I suppose one should look at the bet leading to an amount $>t$ for the first time ... $\endgroup$ – Hagen von Eitzen Jul 10 '13 at 13:33

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