Modified Martingale system for even money betting on Roulette

To start out, I know all systems are flawed. Even so it seems you can improve your odds by following some simple rules. I have used this system and won more than lost. I have lost, but usually because of the human element and not following the rules, and on the rare run when 10-13 times in a row happens – just bad timing.

So, I play all 3 outside even bets, red-black, odd-even, and 1:18-19:36.
I watch each spin until any of them come in 3 times in a row and that’s when I start to bet 5 then 10 then 20. After the 20 loses, I concede and go back to 5 then 10 to 20 again. With this method, if it comes in on the 4th, 5th, or 6th spin I win 1 unit ($5 in most cases). If it comes in on the 7th, 8th, or 9th spin I lose $30. If it comes in on the 10th, 11th, or 12th spin, I lose $65.
Now with that said, most times it comes in on the 4th or 5th or even 6th in which case I win 1 unit. I only need to win 6 of these to cover the loss of 6 units when the occasional 7th – 9th run happens.
Now the way I see it, the odds of it coming in on the 4th 5th or 6th is 8 times more likely than coming in on the 7th 8th or 9th.
So in theory, if 4-5-6 comes in 8 times for every 1 time of 7-8-9 then I should be up 2 units.
I know I am not counting the house edge here, but the numbers seem to work.

If using this on a single zero wheel with a smaller house edge, especially one that follows EURO style playing, the house odds are reduced by 75%.

I plan on trying this in Vegas $25 min single zero tables next month.

Thoughts?

Quote: jjcrocco

To start out, I know all systems are flawed. Even so it seems you can improve your odds by following some simple rules.

That's the "secret" behind Martingale and other such systems - you increase the chance of winning, at a cost of losing much, much more if you do happen to lose. A 10-step Martingale (where you can double your bet 9 times - e.g. if the minimum is 5 and the maximum is 5000, your bets are 5, 10, 20, 40, 80, 160, 320, 640, 1280, and 2560) on a single-zero wheel "wins" about 784 times out of 785, but every loss costs you 1023 times what every win gains.

Quote: jjcrocco

I have used this system and won more than lost.

Not surprising, as the probability of winning > 1/2. If the bet was, "Roll a six-sided die; if it is 1-5, you win $1, but if it is 6, you lose $100", you win that bet 5/6 of the time, but that doesn't make it a good bet to make.

Quote: jjcrocco

I have lost, but usually because of the human element and not following the rules, and on the rare run when 10-13 times in a row happens – just bad timing.

It's this mysterious "bad timing" that helps to pay for pretty much everything in the casino in the first place. Well, that, and the price of Cirque du Soleil tickets, but that's another story.

Quote: jjcrocco

So, I play all 3 outside even bets, red-black, odd-even, and 1:18-19:36.
I watch each spin until any of them come in 3 times in a row and that’s when I start to bet 5 then 10 then 20. After the 20 loses, I concede and go back to 5 then 10 to 20 again. With this method, if it comes in on the 4th, 5th, or 6th spin I win 1 unit ($5 in most cases). If it comes in on the 7th, 8th, or 9th spin I lose $30. If it comes in on the 10th, 11th, or 12th spin, I lose $65.
Now with that said, most times it comes in on the 4th or 5th or even 6th in which case I win 1 unit. I only need to win 6 of these to cover the loss of 6 units when the occasional 7th – 9th run happens.
Now the way I see it, the odds of it coming in on the 4th 5th or 6th is 8 times more likely than coming in on the 7th 8th or 9th.
So in theory, if 4-5-6 comes in 8 times for every 1 time of 7-8-9 then I should be up 2 units.
I know I am not counting the house edge here, but the numbers seem to work.

If using this on a single zero wheel with a smaller house edge, especially one that follows EURO style playing, the house odds are reduced by 75%.

I plan on trying this in Vegas $25 min single zero tables next month.

Thoughts?

Here are my thoughts:
4 happens 18/37 of the time; you gain $5
Remember, the probability of 4 reds in a row is the same as 3 reds followed by a black (and less than the probability of 3 reds followed by either a black or 0)
5 happens 19/37 x 18/37 of the time; you gain $5
6 happens (19/37)² x 18/37 of the time; you gain $5
A run of 6 happens (19/37)³ of the time; you lose $35 and start your bet progression over again.
In any "group" of 1-3 bets, the probability of gaining $5 is 18/37 x (1 + 19/37 + (19/37)²) = 43794/50653, and the probability of losing $35 is 6859/50653. You are 43794/6859 = about 6.385 times as likely to win as you are to lose, but your losses are 7 times as large as your wins, so you end up losing overall.

Can someone answer me this . . .
What are the odds of 3 coming in a row?
What are the odds of 6 coming in a row?
This is assuming the next spin is different.
In simple math 3 in a row happens 1 in 8 (1-2, 1-4, 1-8)
In simple math 6 in a row happens 1 in 64 (1-2, 1-4, 1-8, 1-16, 1-32, 1-64)
In theory 3 in a row will happen 8 times before 6 in a row does once.
In my betting system, if I bust after 6, I lose 7 units, but if I win I the next 3 spins I get 1 unit back, so I only lose 6 units.
If I win 8 units before losing 6 units, I am ahead 2 units.
thoughts?

Quote: jjcrocco

Thoughts?

With the Marty you are wallowing in your losing sessions. You are pressing when you are losing.

I personally like to press when I am winning. I never see anybody having any fun wallowing in this or any other miserable situation.

Quote: jjcrocco

In my betting system, if I bust after 6, I lose 7 units, but if I win I the next 3 spins I get 1 unit back, so I only lose 6 units.

If you win.
If, on the other hand, you lose all three of the next three spins, you lose another 7 units.
In "simple math" (which I assume means each bet has a 50% chance of winning), you are 7 times as likely to win 1 unit back as you are to lose 7 more. On an actual roulette wheel, it is less than 7 times likely, so your wins do not cover your losses.

Quote: jjcrocco

. . . Even so it seems you can improve your odds by following some simple rules. I have used this system and won more than lost. . .
. . . the house odds are reduced by 75%.

I plan on trying this in Vegas $25 min single zero tables next month.

Thoughts?

No No No No NOOOOOOOOooooooooo.

I don't know what you mean by 'house odds' but your system does nothing, zero, zilch to affect or reduce the house edge.

If you want to win more often than you lose, simply put 35 chips down on 35 different numbers each spin. You will win far more often than you lose.... Then you will lose.

Oh, and if you think that waiting for an outcome to occur 8 times or so before wagering will help you, then yes. It will. Because and only because it will mean that you will wager less often.

Your various posts are almost perfect examples of how the gamblers fallacy and Martingale systems are often interpreted by fools.

I can barely believe that anyone would be so clueless as to advocate such systems here, except maybe with the intent of generating traffic to the site.

Quote: odiousgambit

With the Marty you are wallowing in your losing sessions. . . . wallowing in this or any other miserable situation.

Actually, I disagree.
With Marty, you are gloating in your frequent insignificant wins while waiting patiently for inevitable oblivion.