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.

Quote:ThatDonGuySo 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?

They both go up or down one level based on win/loss.

Quote:MikeSTLThey both go up or down one level based on win/loss.

Got it. In effect, what you're doing is, you're betting 10 on "first and second 12 combined," which pays 1-2 if it wins; you double the bet after each loss, but you halve it after each win.

This is similar to the D'Alembert system, except that in that system, you increase/decrease by the base bet rather than doubling/halving, and "when the number of wins climbs back up to the number of losses," your profit is the base bet times the number of losses.

Again, you don't seem to be "missing" anything. The number of winning sessions will exceed the number of losing sessions. However, your losing sessions will be so large that they will more than cover your wins. Systems like yours "appear" to work because your examples don't let the system run long enough to cover lengthy losing sessions. You can't just say, "The wins and losses cancel each other out," because they don't.

Quote:MikeSTLI full agree there is no system to overcome the house edge with a great enough sample. However, I can’t figure out what I’m missing in the following system for roulette. Any ideas?

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.

Bet two dozens at $5 each and lose: -$10

Bet two dozens at $10 each and lose: -$20 (-$30 total)

Bet two dozens at $20 each and win: +$20 (-$10 total)

So, it is what you think it is. What is it that you think you are missing? If the same dozen hits a bunch of times in a row, or hits an inordinate number of times within a relatively small number of spins, you're screwed.

"Small and medium winning sessions," is a very relative term. You would have various winning, "Sessions," that add up to less than your inevitable devastating losing session(s) would.

It doesn't seem like you're missing anything. Suppose your total bet could go:

10-20-40-80-160-320-640 (half of each bet on each dozen) and then the table limit would come into play.

(14/38)^7 = 0.00092132121

1/0.00092132121 = 1 in 1085.39778434

So, for every 1,085 (roughly) initial spins, you'll expect to have a loss of $1,270 without even winning any. Unfortunately, as you will see below, that's not your only problem.

From Random.org: (37's and 38's represent 0 and 00)

Quote:21 22 3 8 24

3 28 7 2 13

7 17 14 24 18

3 12 5 11 28

30 30 25 9 34

24 37 6 36 16

25 18 23 32 28

8 16 19 38 31

30 23 20 18 25

35 2 30 21 7

16 38 26 27 21

29 5 34 16 4

29 10 29 7 12

15 12 33 24 10

34 34 10 35 34

17 30 35 23 13

28 13 4 30 2

14 15 16 1 4

20 8 11 31 26

23 21 21 24 22

Let's look at this. We're going to pretend that we were betting the first two dozens. Here's how it went:

W $5

W $5

W $5

W $5

L-W-L-W-L-L-W-L-L-W-L-W-L-W-W-W-W $5

L-W-W $5

L-L-W-W-W $5

W $5

L-W-W $5

L-L-W-W-W $5

W $5

W-W-W-W $20

L-W-W $5

W-W $10

L-L-L-L-L-W-L-W-W-W-W-W-W $5

W-W-W $15

L-L-L-W-W-L-W-W-W $5

L-W-L-W-L-W--W $5

W-W $10

L-L-W-L-L-L-W-W-W-W-W $5

L-W-W $5

W- $5

L-W (Unfinished)

Okay, so this particular session would have had $145 in profits. That's on 65 winners and 35 losers (assuming I counted correctly), so pretty good session.

Let's do one more:

Quote:

15 1 18 22 8

17 22 19 12 31

36 24 1 36 18

27 18 33 18 17

15 22 19 18 11

15 14 27 11 32

3 21 37 23 3

28 23 5 24 26

17 21 7 21 13

20 33 38 34 38

34 38 19 6 15

26 20 38 3 17

27 16 27 21 38*

4 13 34 32 28

26 37 35 31 4

8 30 38 11 14

21 28 23 23 30

37 1 38 6 19

5 22 34 11 34

32 36 36 34 35

W-W $10 (2)

L-L-W-W-W $5 (5)

L-W-W $5 (7)

L-L-L-W-L-W-W-L-W-L-W-W-W $5 (14)

Six Wins $30 (20)

L-L-W-W-W $5 (23)

L-L-L-W-W-L-W-W-W $5 (28)

***L-W-L-L-W-W-L-W-L-L-L-L-L-W-L-L-L-W-W-L-W-W-W-W-W-W-L-W-W $5 (43)

W $5 (44)

L-L-W-W-W $5 (47)

W $5 (48)

L-W-L-W-W $5 (51)

W $5 (52)

L-W-L-W-L-W-W $5 (56)

L-L-W-W-L-W-L-L (59)

After this set of 100 spins, you have one that is still unfinished and only made $100 in profits. That's only if we conveniently ignore the fact that you hit the Table Max that I originally proposed (see below). You could not have made another bet as you were seven levels behind, at one point.

The one we didn't finish went L-L-W-W-L-W-L-L-W-W-L-W-W.

I guess the, "Part you're missing," is that the results are not going to distribute perfectly because result distribution does not care to go out of its way to make betting systems work.

Every attempt that takes more than one spin starts with a loss, because it has to. At that point, whatever you're prepared to do bankroll-wise (and the house is prepared to let you do by table limits) is to get to the point where you win one more than you lose. The second session above was completely randomly selected and I did it this way so that I could illustrate to you why this system does not work.

The second session wasn't great in that you only had 59/100 winning spins, but is that really that unusual? 14/38 = 0.36842105263

The answer is: Not really. 41 (or more) losses with 14/38 ways to lose will happen more than 20% of the time in 100 spins.

***HERE is the other problem. You're screwed in this session. The losses don't have to be in a row because your Martingale does not cancel out all losses. In this particular series, you've effectively lost the seventh one in a row and the series would have ended.

Conclusion

Therefore, I conclude if you are missing anything, they are these two things:

1.) It doesn't take seven losses in a row for this system to fail. This system is not a Martingale, even, it's kind of like a Martingale/Labouchere hybrid.

2.) Even your winning sessions aren't going to be distributed such that you consistently get 26 units in profits.

But, you know the important part: That Roulette has a big house edge working against you and that you cannot expect any system to beat the game. I am very happy that your question was more, "Why does this system not work?", as opposed to actually thinking it might work.

I hope my post has helped.

Quote:ThatDonGuyGot it. In effect, what you're doing is, you're betting 10 on "first and second 12 combined," which pays 1-2 if it wins; you double the bet after each loss, but you halve it after each win.

This is similar to the D'Alembert system, except that in that system, you increase/decrease by the base bet rather than doubling/halving, and "when the number of wins climbs back up to the number of losses," your profit is the base bet times the number of losses.

Again, you don't seem to be "missing" anything. The number of winning sessions will exceed the number of losing sessions. However, your losing sessions will be so large that they will more than cover your wins. Systems like yours "appear" to work because your examples don't let the system run long enough to cover lengthy losing sessions. You can't just say, "The wins and losses cancel each other out," because they don't.

Ok I think maybe I see the catch. What do you think of this? So the system will profit necessarily over any number of spins if the last three spins were at base bet level. But under some scenarios even with infinite bankroll you may not have won your way back to base level after the number of sample spins. So you could theoretically be at a loss depending how many wins at base level you had before you entered the final progression.

My approach is looking at the actual, not the probability. If you covered 2 dozens, then looked at how many times it would go to 7,8,9,10,11,12, 13,14,15 in a row without hitting either of them.

There's a point where 10,11,12 happen so rarely that maybe covering 9 or 10 is the sweet spot to profit. ie waiting for 3 in a dozen, then progressing 6 bets from $25. On a 2500 max bet 2:1 table, you could get 6 bets in and net 200-300/hr. The more tables at the casino in play, the less dead time there is.