Question by J.J.: Would this type of betting system produce a profit on a roulette game?
If you were to combine the martingale system with the D’Alembert system wouldn’t you be able to theoretically beat a roulette game? Theoretically being because you would need an immense bankroll to be able to play this way, but the system itself would produce a win. The reason being that by combining the two systems you would make enough profit to cover any losses, therefore keeping the game in the black. Whereas, the martingale itself does not produce big enough profits to cover the losses, eventually you will lose your bankroll, even if your bankroll was enormous. But, combining these two systems produces such massive profits that it will cover the losses, and you won’t lose your starting bankroll, you’ll be playing in the black.

There is one more detail that would have to be added to this system. You would have to rebet the zero. When a zero came up you would lose that bet and bet the same amount again. The reason for this is that this would keep the progression from going to Infinity. Rebetting the zero would make the progression 50/50. If you didn’t do that the progression would run to Infinity.

Just to clarify, combining the D’Alembert and Martingale systems you would be doubling the bet after each loss and instead of going back to 1 unit after a win, you would just go down 1 bet. Like this.
1-2-4-8-16-32-16-8-16-32-64-32-64-32-16-8-etc. Each time you lose you bet double and each time you win you bet half of the previous bet. And, rebet the same amount if a zero comes up.

Now, this obviously can’t be used for a Casino or anything like that, but for scientific purposes, I can see it being very useful.

Best answer:

Answer by Chase W
No. Game of chance.

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