Martingale Strategy

I was discussing this strategy with my friend Bevio considering at the casino.

I learned the formal term from this guy

The strategy is explained here:

The original martingale strategy:

• If you lose, double your previous bet size and bet that.

So your bet sizes are $1,2,4,8,16,\dots ,2^{k-1}$ (up to the $k$-th round).

• More generally, if you lose on the $k$-th round, your total loss so far is ${\sum }_{i=1}^k{2}^{i-1}=2^k-1$.
• Else, you will win 1 pound lol

My idea variation

Play roulette, if you lose, bet twice the total amount you lost. if you win, stop. you are guaranteed to make money and your ev is positive.

https://chatgpt.com/share/67775392-23a4-8002-94c0-44ff87eaee6a

How much until you bust?

Let’s say that we have $N$ dollars. The most of rounds we can lose in a row $k$ is: $N=2^k-1$ $k={\log }_2(N+1)$ N = 100 → 7 losses N = 1000 → 10 losses N = 10,000 → 14 losses … N = 1,000,000 → 20 losses

How to calculate EV?

It ends up being a net 1 if you have infinite money.

Let’s explore: EV is given by EV = probability of winning first one * reward + probability of winning second time * reward + probability of winning third one * reward …

$0.5\ast 1+0.5^2\ast 1+0.5^3+\cdots ={\sum }_{k=1}^{\infty }0.5^k$ This is a Geometric Series that converges: ${\sum }_{k=1}^{\infty }a{r}^k=\frac{a}{1-r}-a$ ${\sum }_{k=1}^{\infty }0.5^k=\frac{1}{1-0.5}-1=1$

Other explorations

$P(doublingmoney)={\left(1-\frac{1}{N}\right)}^N$

• This is Euler Number as $N$ goes to infinity

What about my variation of the game? What’s the EV?

Martingale seems annoying because you only win 1$ each time. What if we bet proportional to how much we lose? If we win, we always double the money that we just lost.

My idea variation

Play roulette, if you lose, bet twice the total amount you lost. if you win, stop. you are guaranteed to make money and your ev is positive.

So these would be the bet sizes: $1,2,6,18,54,...,2\ast 3^{k-1}$

• So actually, it’s just tripling the money that you just put last round, The profits are: $1,1,3,9,27,\dots ,max(1,3^{k-2})$

If i wanted my profits to be after winning on round $k$ to be $2^{k-1}$, like the following: $1,2,4,8,16,\dots ,2^{k-1}$ What would my bet sizes have to be? $1,3,8,20,48,\dots ,(k+1)2^{k-2}$

• kind of complicated to solve it algebraically

What about EV?

$EV={\sum }_{k=1}^{\infty }(\text{Profit if first win is on round k})\times (\text{Probability of first win on round k})$ $EV={\sum }_{k=1}^{\infty }((k+1)2^{k-2})\times (0.5^k)$

EV: 0.5 * 1 + 0.5 * (3) + 0.5 *