Method: pick any two competing dozens and place equal bets. So let’s say $5 on first and second 12. If you win, repeat. If you lose martingale. But unlike some martingales you don’t automatically go back to the base bet. You go down one level since the martingale isn’t a win, rather a previous loss eraser (because of the hedge with the second dozen). So, example, lose at 5, bet 10, lose and bet 20, win and go back to 10, win and now you’re back to 5.

So the idea is you cancel out losses (as long as you can avoid losing it all with the martingale) and every win at the base level is one bet. So on American roulette with 100 average spins, you would win 63 and lose 37. Let’s say 37 of those wins have to be dedicated to canceling out the 37 losses at break even, so you are left with 26 units profit. Or say it’s a less lucky night and it’s 55-45. 45 of the wins give no profit due to cancelling so you have 10 units profit.

I know I must be missing some simple issue here, but I can’t figure out what it is. It seems like it would be the case you’d often have small to medium winning sessions and occasional total loss sessions when you lose at the top end of the martingale.

Quote:MikeSTLHowever, I can’t figure out what I’m missing in the following system for roulette. Any ideas?

...

as long as you can avoid losing it all with the martingale...

And there you have it. What you are missing

"as long as you can avoid losing it all with the martingale..."

How do you do that?

https://wizardofvegas.com/member/oncedear/blog/6/#post1370

Quote:OnceDearAnd there you have it. What you are missing

"as long as you can avoid losing it all with the martingale..."

How do you do that?

/member/oncedear/blog/6/#post1370

Maybe I should restate my confusion. In the real world, I think you would often run into losing sessions of your total buy in. What I’m not seeing is in the theoretical example why it turns out to be in the player’s favor. For example, I can clearly see why betting a single number that pays 35-1 but will only hit an average of once every 38 spins is a theoretical loser. But in the example of the system above I don’t see that house edge so clearly displayed. That’s why I feel like I’m missing something. Otherwise an infinite theoretical bankroll would get a positive result when I’m aware it should instead end up at the house edge.

Quote:MikeSTLI know I must be missing some simple issue here, but I can’t figure out what it is. It seems like it would be the case you’d often have small to medium winning sessions and occasional total loss sessions when you lose at the top end of the martingale.

What OnceDear said - this statement pretty much describes how all Martingales work; numerous small wins against a small number of really large losses.

You are also assuming that each winning spin either "cancels" a losing spin or makes a profit. This isn't the case.

Assume you start with 5 on first 12 and 5 on second 12.

The first spin is a 1; the first 12 bet wins, so that stays at 5, while the second 12 bet loses, so now that's 10.

The next spin is a 2; the first 12 bet wins again, so that stays at 5, while the second 12 bet loses, so now that's 20.

The next spin is a 3; the first 12 bet wins 10, but the second 12 bet loses 20 - you just lost 10 on a "winning" spin. Of course, if the third spin is 13 instead of 3, you win 35 on the spin.

Quote:MikeSTLOr restated with the original example, it’s clear that with a flat bet over 100 spins you win 315 but lose 370. But it’s less clear why it would be a loser if the wins and losses can cancel each other, then any wins left would be profit it seems like to me.

The entire principle behind Martingale, D'Alembert, and other "bet more when you lose" systems is, wins and losses don't "cancel each other" because since you are betting more after losses, the wins win more than the losses lose...until you get to the point where either you're out of money or you're out of time, and you have to leave with a massive loss.

Also, as I said earlier, losses and wins don't necessarily "cancel each other."

Quote:MikeSTLJust to be clear, you would martingale both or neither on this system. So if one of your dozens hits then both stay at minimum. If multiple losses in a row, then your next win only cancels previous loss which is why you have to “win” your way back down to base bet and only then can you profit again with a win.

So let's say the first bet is 5 on, say first 12 and 5 on second 12

If either wins, both stay at 5; if both lose, both are now 10

If one of the two bets of 10 wins, do they both go back down to 5, or does one go down to 5 while the other stays at 10?

Five spin example:

1. Two dozens at $5 each, $10 total bet. Win, up $5

2. Repeat $10 total bet. Lose down $5

3. Double both $20 total bet. Win up $5

4. Back to $10 total bet. Lose down $5.

5. Double both again $20 total. Win up $5

That’s what I mean where if you make your way back down to base bets you are even with the losses. So any wins above losses would be profit of one unit.