Is it ever possible to "win" a 50/50 game with a clear goal in mind.

Mathematics Asked by Big Altz on January 3, 2022

Lets say I have 750 dollars and want at least 1250 dollars at the end of a 50/50 game where I can bet any possible value. Is there any way in which I can raise my chances of winning? And if there is, how can I define the best initial bet to be doubled?

(At first I thought problems like these were easy to solve and there was no possible way to "win", but at the same time I have some doubts about it and don’t know the mathematical explanation for it.)

2 Answers

Your optimal probability of getting to $1250 from $750 with fair bets is 60% and you cannot do better. One strategy of doing so is given by @Robert's answer above, called "bold play" in the literature. If you follow his strategy, you find yourself along the following tree:

A tree diagram that shows the potential outcomes of bold play. Starting with a bet of 750, winning outcomes either go 750 to 1250; 750 to 250 to 500 to 1000 to 1250; or 750 to 250 to 500 to 1000 and then back to 750, starting over.

You take each path of this betting tree with probability $1/2$. If your initial probability of winning from $750 is x, then one of three things happen on your way to victory:

  1. You win instantly ($50%$ chance)
  2. After an initial loss, you win 3x in a row ($1/16 = 6.25%$ chance)
  3. After an initial loss and 2 wins, you lose again and get back to your starting fortune of $750 ($1/16 = 6.25%$ chance)

So we have the equation $$x = 0.5 + 0.0625 + 0.0625x,$$ and when solved, we learn that $x = 9/16 + x/16$, or $15x/16 = 9/16$, whence $x = 9/15 = 0.6$. (The other $40%$ probability is that you go broke.)

Why can't we do better than $60%$? Consider your fortune $W_n$ on the $n$th bet, and imagine you make the bet $A_n$ subject to the constraints $A_n = 0$ if $W_n geq 1250$ or $W_n = 0$, $0 leq A_n leq W_n$. Then $W_n$ is a uniformly integrable martingale, because your winning probability on each bet is $1/2$ and $$0 leq W_n leq 2*(1250-1) = 2498$$ is a bounded sequence above and below. This means that $W_n$ converges almost surely and in $L^1$ to a bounded random variable $W_infty$, representing your limiting winnings. The $L^1$ convergence is crucial here, as it means $$Bbb{E}[W_infty] = 750.$$

You have "won" your betting sequence if $W_infty geq 1250$. What is the probability of that? By Markov's inequality, $$Bbb{P}(W_infty geq 1250) leq frac{Bbb{E}[W_infty]}{1250} leq frac{750}{1250} = 60%.$$

(If you started with $x$ dollars instead, the same argument shows you have an $x/1250$ chance of getting to $1250, one way to do this being bold play.)

So bold play actually gives you the "best possible" chance of reaching $1250--although it's also "worst possible" among strategies that achieve $60%$, as the other $40%$ of your time you lose all of your initial $750.

Other betting strategies than bold play may leave you with a better probability of a nonzero limiting fortune $W_infty$ (one example: bet $1/(n+1)$ of your current wealth $W_n$ on bet number $n$ or $1250 - W_n$, whichever is smaller). But no strategy, however clever or intricate, can do better than a $60%$ chance of reaching $1250 from an initial stake of $750.

Answered by Rivers McForge on January 3, 2022

By "50/50 game" I assume you mean you can make any sequence of fair bets at $50/50$ odds, as long as you never bet more than you have at the time. Since these are fair bets, your expected value at the end is the same as your initial fortune, $750$. As long as you ensure that the only possible final outcomes are $0$ and $1250$, you maximize the probability of ending with $1250$. Thus one possible strategy is: whenever you have $x$ where $0 < x < 1250$, bet the minimum of $x$ and $1250 - x$. With probability $1$, you will eventually end up at either $1250$ or $0$.

Answered by Robert Israel on January 3, 2022

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