# Trying to create an equation to figure out probability on a 50/50 game using the martingale betting strategy.

Let's say there's a game with exactly 50/50 odds, no minimum or maximum bet, it takes about 1-2 minutes to complete, and you can play it 24/7, 365.

I understand there's a 50% chance i'll win or lose on each play individually..... HOWEVER... am I able to increase my overall odds of doubling my money by having enough to double down on my previous bet x amount of times on each loss before bankruptcy?

To give an example, lets say I'm starting with 511... and want to start off betting with the initial bet of 1. This means I could lose 8 times in a row with bets of 1, 2, 4, 8, 16, 32, 64, and 128 because as long as I win the 9th bet of 256 I would dig myself out of the 255 hole i'm in and actually profit the amount of my initial bet, which in this case is 1.

So in that scenario I would have to win 511 games without losing 9 in a row in order to double my money...

If I started with 1022, I would have to win 1022 games in a row without losing 10 times in a row to double my money.... so I would have to play double the games but have an extra loss.

Start with 2044, and it's 2044 games but I can lose 10 times in a row safely.

..and so on

Do any of these scenarios increase my odds or is it all the same?

IFFFF I can in fact increase my odds... can someone please provide an equation where I could change the variables for number of losses covered before bankruptcy and number of wins needed to figure out my odds for 11 losses, 15 losses, etc until I figure out the amount of losses needed to get close to 99.999% and essentially go on forever. (because again there is no maximum bet)

Thanks so much, and if i'm an idiot please dont be afraid to say so!

• Fundamental principle says no betting strategy will change the odds. Commented Mar 24, 2019 at 21:35
• In this particular case, remember that a longer set of games will have a higher chance of loosing $n$ times in a row. From a different perspective, who would be willing to play with you a game where it exists a strategy that will make them loose money on average? Certainly not a casino nor a firm that accepts bets. Commented Mar 24, 2019 at 22:24

It's a consequence of Doob's martingale stopping theorem that there is no betting strategy that will increase your probability of doubling your initial capital above $$\ \frac{1}{2}\$$ (unless you're willing to go into debt if you run out of capital ). If your strategy is such that whenever you stop, you have either lost all your initial capital, or exactly doubled it, then the probabilities of the two possible outcomes will both be exactly equal to $$\ \frac{1}{2}\$$.
If you start with an initial capital of $$\ 2^{n}-1\$$, and use the so-called "martingale" betting strategy you describe, until you have either doubled your capital to $$\ 2^{n+1}-2\$$, or have insufficient funds to double your previous bet, then you will never be able to double your bet more than $$\ n-1\$$ times. At each stage, you therefore have a probablity $$\ 1-\frac{1}{2^n}\$$ of increasing your capital by $$\ 1\$$, and a probability of $$\ \frac{1}{2^n}\$$ of having to quit with whatever's left of it. To double your initial capital, you have to succeed in increasing it by $$\ 1\$$ for $$\ 2^n\$$ successive times. Your probability of doing this is $$\ \left(1-\frac{1}{2^n}\right)^{2^n}\approx e^{-1}\approx 0.37\$$, considerably less than $$\ \frac{1}{2}\$$.
The probability of success here is less than $$\ \frac{1}{2}\$$ because, as enunciated above, the strategy allows you to walk away with some positive part of your capital left—namely, with probability $$\ \frac{1}{2^n}\left(1-\frac{1}{2^n}\right)^t\$$ you will quit with $$\ t\$$, for each $$\ t\in\left\{1,2,\dots,2^n-2\right\}\$$. Your expected capital on quitting will be $$\left(2^{n+1}-2\right)\left(1-\frac{1}{2^n}\right)^{2^n} +\frac{1}{2^n}\sum_{t=1}^{2^n-2}t\left(1-\frac{1}{2^n}\right)^t\\ =2^n-1\ ,$$ exactly equal to your initial capital, as guaranteed by the martingale stopping theorem.
If you're willing to borrow money whenever you run out of funds, you can increase your probability of doubling your initial capital to any desired probability less than absolute certainty. To have a $$\ 99.999\%\$$ chance of doubling your initial capital, for instance, you merely have to be willing to borrow up to $$\ 99,998\$$ times that initial capital. If your initial capital is $$\ C\$$, and you keep playing until you have either exactly doubled it, or run up a debt of exactly $$\ 99,998\,C\$$, then with probability $$\ 0.99999\$$ you will succeed, and with probability $$\ 0.00001\$$ you will fail, and end up with a debt of $$\ 99,998\,C\$$. Your expected capital when you stop playing will again be: $$0.99999\times 2\ C + 0.00001\times\left(-99,998\,C\right)= C\ .$$