A modified martingale. Why won't it work?

#1
First I will say that if I have thought of it, someone else has. Since it isn't being touted as the wonder of BJ, I am 99% sure this doesn't work. On the other hand, I can't help but ask.

I am good at math but do not have any experience with running simulations and finding probabilities etc. which some of you may have at your disposal. Now of course the Martingale system blows up (into the thousands/millions very quickly) at losing streaks that will absolutely happen, completely devastating winnings that accumulate relatively slowly. My proposal is this:

Do a modified Martingale that begins with a $5 and ends at $20 (or 40?). If you lose your 20, start the system over again. Do this only at a positive count to reduce losses.

I am playing 6 deck, H17, DAS, resplit aces, surrender.

Now guessing that loss percentage will be around 49% though I am positive that some will know better and that it should be lower especially with counting involved.

(.49)^3=.118 or 1:8.5
Win = 7.5x5 - 1x(20+10+5) = 37.5 -35 = $2.5

There would also be the addition of 3:2 blackjack which would increase winning and not affect losses. Even several bad runs will not result in the projected losses of a true martingale or hopefully the wins associated with several good runs.

Please tell me what you think. I would love to see the math that would show if this works or not. Thank you for your time.
 
#2
Hey wolf,

I recently asked the same thing. It works for me and mathematically it makes sense to me, though I don't remember enough from statistics to be able to articulate it with a formula (this is a part of statistics, not math in general).

How it works for me, I generally win more than lose, so I have a little "puddle" of winning chips. When I lose, my "puddle" goes down. Maybe when I start at a table, I'll go down but almost always go back up. I've been using this for years and it only fails when I break my own rules (double when I shouldn't or double one too many times, split on a double...ect).

College left a bad taste in my mouth regarding this statistical probability but if you remember the Bell Shaped Curve, lets just say that my experience has been, statistically, almost always positive. I've tripled my minimum bet with success.

There's regulars here who argue against it, but their main arguments tend to involve either the fact that "martingales" is usually thought in terms of eternal doubling ups (which I dont practice and you explained dont want to do) or they think in terms of, in the last doubling hand, you still run a xx% chance of losing. Somewhere between the two is the gray area of "you're going to lose more than win because of statistical probability" but no one has made the mathematical formula for that yet...and I'm sure it works, mainly because it does.

The hard part is sticking to your rules, especially if drinking or tired/sleepy.
 

Sucker

Well-Known Member
#4
lilawolf said:
I would love to see the math that would show if this works or not.
OK - here's the math:Any time the house has a one-half of one percent advantage, for every $100 you bet, you will lose an average of 50 cents.

This is the ONLY math you need to know about this, or ANY strategy that's based on betting. PERIOD. END OF DISCUSSION. No matter HOW much some people will try to pretend, and tell you otherwise.
 

Machinist

Well-Known Member
#5
Sucker said:
OK - here's the math:Any time the house has a one-half of one percent advantage, for every $100 you bet, you will lose an average of 50 cents.

This is the ONLY math you need to know about this, or ANY strategy that's based on betting. PERIOD. END OF DISCUSSION. No matter HOW much some people will try to pretend, and tell you otherwise.
ABSOLUTELY NOTHING BUT THE FACTS HERE!!!!!!!!!!! The truth the whole truth and nuttin but the truth.......But some people can't handle the truth.:rolleyes:


Machinist
 

21gunsalute

Well-Known Member
#6
lilawolf said:
(.49)^3=.118 or 1:8.5
Win = 7.5x5 - 1x(20+10+5) = 37.5 -35 = $2.5
Please explain how you derived these numbers. I can't make head or tail out of this.

Martingale works in theory because you keep doubling your losses. If you don't keep doubling your losses you'll never get your money back! You can stop your progression @ $20 (or anywhere you want to) and start over, but that won't stop the losing streak from continuing. You'll lose less during such a losing streak, but the loss will still be fairly significant with no forseeable way to recover such a loss.

Under the method you outlined above, if you lose 8 hands in a row you'll be out $85. What's your next move? Are you going to go back to your $5, $10, $20 progression? Are you going to bet $85 or $90 on one hand to try to get back to even or slightly ahead? What if you lose again? Then you'll be out $170-175 and approaching table limits at most places.

What's the upside to this progression? I don't see one. Are you going to use the same progression if you're winning? You won't ever win much with this system because we all know we're going to lose more hands than we win and you're only spreading 1-4, and probably not spreading at optimal times. You mentioned using counting with this system, but if you're going to count then why bother using a progression at all? And now that I think about it, the term progression is nothing but a misnomer because they're all digressions not progressions.
 
#7
You will lose approximately 49% of the time (though if you have a more accurate percentage to put here go ahead). Since we are looking at losing three hands in a row you do
(.49)^3 = .118
1/.118 gives you a 1 in 8.5 chance of losing three in a row.
So for the 7.5 times that you win your bet ($5) for one loss: 7.5x5=37.5
When you lose, it will be a loss of $35, giving you a net win of $2.50 + anything you have won on black jacks.

After a loss you would begin back with one unit. You would NOT put in a bigger bet (as someone asked) to try and regain everything because that would be a true martingale which I clearly said was not what I was doing. Here you are minimizing the max loss though limiting it's recovery benefits. I think at a certain length of the chain your benefits will outweigh the losses (in the long run).

Would this work for someone who is not counting (my fiance?). I see what what commenter was talking about with this system being at odds with the bets you should make with the count.
 

21gunsalute

Well-Known Member
#8
lilawolf said:
You will lose approximately 49% of the time (though if you have a more accurate percentage to put here go ahead). Since we are looking at losing three hands in a row you do
(.49)^3 = .118
1/.118 gives you a 1 in 8.5 chance of losing three in a row.
So for the 7.5 times that you win your bet ($5) for one loss: 7.5x5=37.5
When you lose, it will be a loss of $35, giving you a net win of $2.50 + anything you have won on black jacks.

After a loss you would begin back with one unit. You would NOT put in a bigger bet (as someone asked) to try and regain everything because that would be a true martingale which I clearly said was not what I was doing. Here you are minimizing the max loss though limiting it's recovery benefits. I think at a certain length of the chain your benefits will outweigh the losses (in the long run).

Would this work for someone who is not counting (my fiance?). I see what what commenter was talking about with this system being at odds with the bets you should make with the count.
You still aren't making any sense.
 

Sucker

Well-Known Member
#9
Although I DID say "End of discussion"; I guess it's only fair of me to point out the error in your math:

You're handling pushes as though they were wins. It's true that you'll lose 49% of the hands, but that doesn't mean you're going to WIN 51%. The correct figure for WINS is something like 43%. This means that you'll lose 49 out of 92 DECISIONS, or 52%.

You tell me that you're good with math; I believe that if you go back and refigure, using the proper input; you'll understand the fallacy of this and any OTHER system that relies on betting patterns only.
 

JJR

New Member
#11
Wolf, the math you're doing is for winning 3 in a row. Losing 3 in a row would be (.51)^3 = .132651

1/.132651 is approximately a 1 in 7.5 chance of losing 3 in a row.
6.5 x 5 = 32.5 vs. $35 loss is a net loss of $2.5

Sorry, but you would need an advantage over the house for the martingale to work, even when you cut it off at 3 or 4 losses, even with a straight 50/50 game it would only break even.
 
#12
Thank you sucker. I now see what I was missing. As I said at the beginning, I knew something had to be wrong with it.... Counting it is then. Working on learning the rule variations now.
 
#13
Unit for unit, this math makes sense. Its a good argument against my system and when I get home ill do some math of my own. Its possible that the 2:1 payoff from blackjacks puts the math in my favor.


Allthough my result are very marginally positive, I always walk away from a table after an up swing in momentum, or a few blackjacks in a row.

I am not disagreeing with those who are against this strategy, I would like to ask the same to try it...either a casino or electronically. I've been doing it for years live and recently got Hoyle on the laptop and aces blackjack on my phone.

I sincerely appreciate and respect those who explained their reasoning. I hope to be an advantage player by your paradigm eventually. If I can keep my unit bet more stav
Ble ill enjoy the game more I'm sure.
 

dacium

Well-Known Member
#14
This is no logic in this.

The mathematical optional bet is the bet at the kelly criterion that matches your EV from a positive count. Doing any modified martingale, even only at high counts, will simply increase your variance and risk bombing out.
 

beating vegas

Well-Known Member
#15
JJR said:
Wolf, the math you're doing is for winning 3 in a row. Losing 3 in a row would be (.51)^3 = .132651

1/.132651 is approximately a 1 in 7.5 chance of losing 3 in a row.
6.5 x 5 = 32.5 vs. $35 loss is a net loss of $2.5

Sorry, but you would need an advantage over the house for the martingale to work, even when you cut it off at 3 or 4 losses, even with a straight 50/50 game it would only break even.
martingales dont work in the long run . i see alot of people at casinos do it and they leave screaming
 

NightStalker

Well-Known Member
#16
Your system is a winning system

lilawolf said:
First I will say that if I have thought of it, someone else has. Since it isn't being touted as the wonder of BJ, I am 99% sure this doesn't work. On the other hand, I can't help but ask.

I am good at math but do not have any experience with running simulations and finding probabilities etc. which some of you may have at your disposal. Now of course the Martingale system blows up (into the thousands/millions very quickly) at losing streaks that will absolutely happen, completely devastating winnings that accumulate relatively slowly. My proposal is this:

Do a modified Martingale that begins with a $5 and ends at $20 (or 40?). If you lose your 20, start the system over again. Do this only at a positive count to reduce losses.

I am playing 6 deck, H17, DAS, resplit aces, surrender.

Now guessing that loss percentage will be around 49% though I am positive that some will know better and that it should be lower especially with counting involved.

(.49)^3=.118 or 1:8.5
Win = 7.5x5 - 1x(20+10+5) = 37.5 -35 = $2.5

There would also be the addition of 3:2 blackjack which would increase winning and not affect losses. Even several bad runs will not result in the projected losses of a true martingale or hopefully the wins associated with several good runs.

Please tell me what you think. I would love to see the math that would show if this works or not. Thank you for your time.
Not because it's some random martingale. Only because your losing probability is 49% <50. So you can flat bet or do whatever bet in +ive counts, will get you the money..

The truth is: Card's have no memory, It doesn't matter whatever your bet in the previous hand, what so ever was the result... Your expectation per round is based on the composition of remainder shoe always...
 
#17
Lost 21 of 22 today at casino !!!

(1.0)
(1.0)
(1.0)
(1.0)
(2.0)
(1.0)
(1.0)
(1.0)
1.0
(1.0)
(0.5)
(1.0)
(1.0)
(0.5)
(0.5)
(1.0)
(0.5)
(0.5)
(1.0)
(1.0)
(2.0)
(1.0)

Today I lost 21 of 22 hands. The count was always either neutral or slightly positive. I always stop playing in negative counts. The game was 8d s17 DAS
LS. I lost 19.5 units in this 22 hand sequence. Luckily LS option saved me a few. All hands were played strickly at BS. No indices were used because the count did not drift far enough to either range.
My point of this post is to detail progressions can not and will not work with actual results like these unless you have a tremendous BR.
 
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