after 1 million shoes testing the results for single chops are 50%, double chops are 25% and others' chops are 25%. My system is basically combining the counts of single and double to bet against others' counts. Since single is 50% and double is 25% we get 75%. and others is 25%. So basically is a 3 to 1 ratio, every three single or double we get a others and that's the norm. My goal here is to bet against double when we get a 5 to 1 ratio because i believe eventually the shoe will be balanced. If we lose we will wait for another double to bet against it until we win using negative progression 1,3,6,12,24,48. we will stop betting until others' counts reached 5 because the minimum for others count is around 5 per shoe.

Example:

BBpBpBpBpp ---> bet Player on next bet

1234567

0

The count here for single and double are 7 and others is 0, this meet our 5 to 1 ratio requirement. So we will bet player on the 11th hand and we WON. We still stop until the others counts reached 5

I already tested 800 shoes manually and the total return is 900 units about 1 units per shoes. I know one unit per shoe is not alot but at least i m not loosing.

please Let me know if you have any questions or advice, looking forward to improve this system.

Quote:SkyOddsMy goal here is to bet against double when we get a 5 to 1 ratio because i believe eventually the shoe will be balanced.

One problem:

"Eventually" can be a very, very, very, veryveryvery, very, did I mention "very", very long time.

Your system runs the same risk as every other progression system where any one bet covers all previous losses; you run the risk of having to exceed the table's maximum bet. Your wins will be plentiful but small; your losses may be few and far-between, but they will be large.

Another thing, what happens if you start betting when the ratio is 5 to 1, but after you start betting it just goes back to 3 to 1 like normal. You're breaking even at this point (ignoring the vig) and off by just 2. If you started when the ratio was 100 to 1, and it went back to 3 to 1 after that, you'd be off by just 97. So, you're right that you are "due", but you just can't tell when you're going to get those back. Or, at least, you can't prove mathematically when you will get those back.

You can however show that your chances of winning 5 before losing 10 are more than 50%. For that matter your chances of winning 6 before 12, or 7 before 14, 8 before 16, are each continuously much better than 50%. But, of course, you're doing a Marty so you just need 1 win. And, the problem here is, that being "owed" a thousand wins doesn't change the odds of losing 12 in a row and destroying your bankroll.

By the way, did you happen to notice when you do these chops, that 50% of the time you get 1 in a row (a one chop) and 50% of the time you get 2 in a row (a 2 chop) or more? So, on a coin flip the chances of getting a streak of 1 heads in a row (a one chop) and the chances of getting a streak of 2 heads in a row (a 2 chop) are EXACTLY THE SAME. Could make for a good prop bet.

So now it's 110 to 32, or a 3.4375 to 1 ratio. And you're now losing to the Vig.

If you were to do a D'alembert, you'd have the benefit as I stated before, of continuously having a greater than 50% chance of getting half of your losses back. If you D'alemert and the betting is at 10, the chances of getting to 5 before 20 is above 50%. Then if the betting goes to 12, the chances of getting to 6 before 24 is above 50%. Betting goes to 14, the chances of getting to 7 before 28 is above 50%. So, you're continuously way above 50% to get half back the entire time you're playing. If my math is correct there.

Anyway, basically, what you're doing doesn't change anything, mathematically anyway. It's the same as if you bet every hand, only it is slower, which might be a good thing if you're getting free drinks :)

Have you been doing this in a casino? If not go try it every day for 6 months Then come back and report.Quote:SkyOddsYes you are right i m doing martingale with a stop loss once progression 48 is reached. pBpBpBpBpBpBpBpBpBpb 12 times does not trigger the bet, it only triggers when we see a 5 to 1 ratio with a double like for example PP, then we bet player. The only way i will loose is 7 doubles in a row while ignoring the singles. All my bets must be at 5 to 1 ratio. This system is based of stats, i know that you guys want to emphasize that doing marty is same as flat betting and it will never reduce the house edge but how you guys explain my profits after 800 shoes. If Marty progression is that bad my profit should be negative by now?

http://screencast.com/t/XbSGEUo0aIdQ

Quote:SkyOddsYes you are right i m doing martingale with a stop loss once progression 48 is reached. pBpBpBpBpBpBpBpBpBpb 12 times does not trigger the bet, it only triggers when we see a 5 to 1 ratio with a double like for example PP, then we bet player. The only way i will loose is 7 doubles in a row while ignoring the singles. All my bets must be at 5 to 1 ratio. This system is based of stats, i know that you guys want to emphasize that doing marty is same as flat betting and it will never reduce the house edge but how you guys explain my profits after 800 shoes. If Marty progression is that bad my profit should be negative by now?

http://screencast.com/t/XbSGEUo0aIdQ

Well there it is, after 500+ years in existence and countless players gone broke, the game of baccarat has finally been beaten.

Obviously the French mathematician who invented it neglected to account tor the 12 times trigger 5 to 1 ratio PP 7 doubles in a row stats based martingale system. And furthermore, shame on the game analyzers at the casinos who allowed this on their floors

Someone should update the games Wikipedia page. Credit SkyOdds with the breakthrough.

+1Quote:michael99000Well there it is, after 500+ years in existence and countless players gone broke, the game of baccarat has finally been beaten.

Obviously the French mathematician who invented it neglected to account tor the 12 times trigger 5 to 1 ratio PP 7 doubles in a row stats based martingale system. And furthermore, shame on the game analyzers at the casinos who allowed this on their floors

Someone should update the games Wikipedia page. Credit SkyOdds with the breakthrough.

Quote:SkyOddsI already tested 800 shoes manually and the total return is 900 units about 1 units per shoes. I know one unit per shoe is not alot but at least i m not loosing.

please Let me know if you have any questions or advice, looking forward to improve this system.

Is your 'total return' of about one unit per shoe after commissions?

Either way, consider how much you are risking to win one measly unit.

I've read that for the 600 shoes in the Zumma tester, there are only 643 more Banker wins than Player wins.

This averages out to 1.07 more Banker wins than Player wins per shoe.

These results are comparable to yours and achieved by flat betting Banker.

Still a loss after commissions, but no 48 unit bets were necessary.

Quote:SkyOddsAccording to this site

https://imspirit.wordpress.com/tag/live-baccarat-shoes/

Quote:that siteOf course, assumptions like reversion to the mean is based on gambler’s fallacy (ref. Fallacies and Illusions), and from mathematical grounds, has absolutely no hope of winning a long term simulation.

Actually as soon as the aliens realize you have a winning system you get abducted.Quote:Baccaratfrom79There are casinos with baccarat in virtually every state. Spend your time making millions at the tables, you won't have time to post or try selling anything. Don't let the secret be known, the baccarat tables will all be pulled.

Quote:AxelWolfActually as soon as the aliens realize you have a winning system you get abducted.

LOL! Once again, Axel makes my coworkers think I'm nuts - just sitting here in the corner laughing to myself...

Random means just that. In fact you can take it one further, even if you knew every card left in a shoe, it still would not help you. Unlike every other game like BJ and poker, you can win a decision with any value of cards. Meaning you can prevail with low value, mid value or high value, you can actually prevail with no value cards. The only decision that would be guaranteed would be knowing there are 6 or 4 cards left of the same value.

When you play this game for many decades you [might] realize what I am saying.

Quote:SkyOddsYes you are right i m doing martingale with a stop loss once progression 48 is reached. pBpBpBpBpBpBpBpBpBpb 12 times does not trigger the bet, it only triggers when we see a 5 to 1 ratio with a double like for example PP, then we bet player. The only way i will loose is 7 doubles in a row while ignoring the singles.

One problem with long-term statistics is, it tends to use percentages or fractions rather than counts.

Let's look at it this way:

Suppose that, in the first 1,000,000 situations where you have B followed by PP (the start of a double chop), you have 530,000 Ps and 470,000 Bs; the 3+ chop happens 53% of the time. (Remember, according to your first post, a 3+ chop should be just as likely as a double chop.)

Now, in the next 1,000,000 situations, you have 510,000 Ps and 490,000 Bs. In the 2,000,000 double-chop situations combined, you got a 3+ chop 52% of the time, so "the double chops are catching up to the expected 50%" - but there were still more 3+ chops than there were double chops in the second 1,000,000, so if you bet against every 3+ chop, you lost money.

See why you can't depend on long-term statistics for short-term trends?

As for why you are winning, never mind how many shoes you have played - how many times have you started your progression? Assuming a 0.4932 chance of a particular bet winning (which is the probability of a player win before a bank win in an 8-deck shoe), your progression is expected to lose six times in a row once in every 59 starts. If you play it 100 times, there is a 49.3% chance of either winning all 100 times, or winning 99 and losing 1, but 99 wins is a gain of at least 99 while the one loss is a loss of 94, so you still come out ahead.

Quote:Baccaratfrom79It is a true 50/50 chance of either side, but there is no reasoning as to which side will prevail on the next decision or the next 10 or the following 30. .

Actually, in the next 10 decisions you do get an idea of what kind of results you'll get. For 1 decision it's 50/50. But, I think MustangSally has posted the stats for these before, but the chances of winning 1 in 10, 2 out of 10, 3 out of 10, 4 out of 10, etc. The odds of those happening are not 50/50. Getting 1 out of 10 is probably 90%, 5 out of 10 probably around 50%, 9 out of 10, probably like 10%.

But, for a marty the problem still exists that none of that changes the odds of losing 6 in a row.

Now, for a D'Alembert, and ThatDonGuy can probably check the math for this, but you're chances of winning 5 before losing 10 is probably something around 75%. So, basically the entire time you play the D'Alembert your chances of getting half of the losses back are well in your favor. And, if you can say that you have a 75% chance at 10, a 75% chance at 12, a 75% chance at 14 etc. etc. then all that would add up pretty quickly.

But, again for a marty, you just need to get the 1 win, so I don't see how anything would change the probabilities for you.

Quote:JyBrd0403Now, for a D'Alembert, and ThatDonGuy can probably check the math for this, but you're chances of winning 5 before losing 10 is probably something around 75%. So, basically the entire time you play the D'Alembert your chances of getting half of the losses back are well in your favor. And, if you can say that you have a 75% chance at 10, a 75% chance at 12, a 75% chance at 14 etc. etc. then all that would add up pretty quickly.

Assuming "winning 5 before losing 10" means winning a total of 5 bets before losing a total of 10 since the same starting point, the probability for an even-money bet on a double-zero roulette wheel is 87.37%.

But how many "win 5 before lose 10" runs make money in D'Alembert? Four wins in a row (+1 each) followed by four losses in a row (-1, -2, -3, -4) followed by a win (+5) is -1 for the run.

Quote:ThatDonGuyAssuming "winning 5 before losing 10" means winning a total of 5 bets before losing a total of 10 since the same starting point, the probability for an even-money bet on a double-zero roulette wheel is 87.37%.

But how many "win 5 before lose 10" runs make money in D'Alembert? Four wins in a row (+1 each) followed by four losses in a row (-1, -2, -3, -4) followed by a win (+5) is -1 for the run.

I'm not sure if we are saying the same thing. What I meant is if flat betting what is the probability of reaching a profit of +5 before a loss of -10. Or on the D'Alembert if the starting point is 10, the probability of reaching a betting point of 5 (+5) before reaching a betting point of 20 (-10). I'm not sure if that's the same as what you are saying.

If the 87.27% is accurate, that would be just for one instance. For instance, in our example the starting point on the D'Alembert game was 10, and you have an 87.27% chance of getting back to 5. The interesting thing is when the betting is at 12 for the D'Alembert, you have another 87.37% chance of getting to 6 before -24. That's already two 87.37% chances of getting half of the losses back. That's 87.37% x 87.37% . At 14 you have another 87.37% chance (that's 87.37% x 87.37% x 87.37%) of getting to 7 before 28. At 16, you have another 87.37% chance. etc. etc.

In other words, the odds are in your favor of getting half back, the entire time you are playing. At least mathematically.

Quote:JyBrd0403Actually, in the next 10 decisions you do get an idea of what kind of results you'll get. .

Sorry, you are wrong. Wrong with the aspect of might be true sometimes and might be wrong a lot of other times. Again, you can not watch/track/plot or whatever for any amount of hands and then base the next future decision off of what has happened. It will not work. What is confusing to the new-bee or the six month or one year experienced player is that, they have not played enough to realize this.

It is true at baccarat you can track or plot (whatever you call it) past decisions and base a wager on those. You might win doing that for the next time or the next 7 times or the next 25 times. But likewise, you can lose the same amount of times or greater. What seems to happen in reality is your 'calculated' tracking/plotting/figuring, etc., gets converted to a lucky guess on the random event that just happened.

.

Guys, I have played this game for decades in numerous jurisdictions and at all types of properties, with all types of players using all types of systems, plans, stats and beliefs. Some work sometimes and some don't work at times. The next day or the next shoe the one that didn't work, works. The one that worked suddenly doesn't work any longer.

Study what random is and how random works and what it really means. It will be time well spent, actually the best investment of time in baccarat gambling.

And not to confuse matters, there are only two wagers in baccarat. Going with the trend or cut. Period, there are no others. Looking at the past is just something to give you a boost, motivation, and confidence as to what wager to place. Likewise, it can hurt you immensely beyond a person's understanding. All the stats have nothing to do with the short term, few shoes on any given casino play.

Back to your statement that is quoted above. The only idea you will get is what happened in those previous 10 hands, which is not in any way indictive as to what is going to happen or not happened in the next 1 or 10 decisions.

Quote:JyBrd0403I'm not sure if we are saying the same thing.

No, we're not.

Using simulation, I get a 62% chance of reaching +5 before -10 betting player in baccarat (assuming a win probability of 0.4932).

Quote:ThatDonGuyNo, we're not.

Using simulation, I get a 62% chance of reaching +5 before -10 betting player in baccarat (assuming a win probability of 0.4932).

Okay, that actually makes more sense. I said 75% off the top of my head, but I figured it would be lower. Probably 66% for a 50/50. But, of course, my point is that the odds are in your favor of getting half back. I wasn't really going for a particular number.

Just to make sure I'm not misleading anyone. This 62% chance of getting half back would be for each instance or try, correct. If the betting is at 10 you have a 62% chance of getting to 5 before getting to 20. At 12 a 62% chance of getting to 6 before 24. At 14 a 62% chance of getting to 7 before 28.

So, if the betting goes to 20 you lost one 62% attempt to get half back(the one started at 10), at 24 you lost another 62% attempt to get half back (the one started at 12), and at 28 you lost another 62% attempt to get half back (the one started at 14).

Does that all make sense?

Quote:JyBrd0403Okay, that actually makes more sense. I said 75% off the top of my head, but I figured it would be lower. Probably 66% for a 50/50. But, of course, my point is that the odds are in your favor of getting half back. I wasn't really going for a particular number.

Just to make sure I'm not misleading anyone. This 62% chance of getting half back would be for each instance or try, correct. If the betting is at 10 you have a 62% chance of getting to 5 before getting to 20. At 12 a 62% chance of getting to 6 before 24. At 14 a 62% chance of getting to 7 before 28.

Not exactly. Starting at 12, you have a 62% chance of getting to 7 before getting to 17, and starting at 14, a 62% chance of getting to 9 before getting to 19.

However, starting at 12, you have about a 61% chance of getting to 6 before 24, and starting at 14, a 60% of getting to 7 before 28.

Every win reduces the bet by 1 (down to a minimum of 1) and every loss increases it by 1, correct? The 62% is for having five more wins than losses before having ten more losses than wins, regardless of the bet sizes.

Quote:ThatDonGuyHowever, starting at 12, you have about a 61% chance of getting to 6 before 24, and starting at 14, a 60% of getting to 7 before 28.

Right, so in this case, you'd have a .39 x .40 chances of NOT getting half the losses back, correct? I mean the chances of both failing (or reaching 24 and 28 respectiveley) would be .39 x .40, correct? These are 2 different events since 1 you exit at either 6 or 24, and the other (if it even gets to 14) would exit at 7 or 28. So the chances of both failing would be .39 x .40, correct?

Quote:JyBrd0403Right, so in this case, you'd have a .39 x .40 chances of NOT getting half the losses back, correct? I mean the chances of both failing (or reaching 24 and 28 respectiveley) would be .39 x .40, correct? These are 2 different events since 1 you exit at either 6 or 24, and the other (if it even gets to 14) would exit at 7 or 28. So the chances of both failing would be .39 x .40, correct?

It sounds as if you are talking about playing two seats at the same table simultaneously, in which case, the answer is no - the probability of both failing is much greater than .39 x .40.

If your 12 reaches 24, then your 14, is at 26; the probability is .39 x whatever the probability is of starting at 26 and reaching 7 before reaching 28.

Being "different events" is insufficient; they need to be independent events. Clearly, two betting runs are not independent if they depend on the same hand results at some point.

House Light Bulb Moment

_{TM}

I just realized...since the actual bet size is irrelevant, but only the number of wins vs. losses matters, this is a Risk of Ruin problem with starting point 2N and success point 3N (since you want N more wins than losses before having 2N more losses than wins).

Here are the calculated probabilities of winning for starting points from 2 to 50:

Start | Success % |
---|---|

2 | 65.75594593 |

4 | 64.83768474 |

6 | 63.91257864 |

8 | 62.98133693 |

10 | 62.04468035 |

12 | 61.10333868 |

14 | 60.15804834 |

16 | 59.20955001 |

18 | 58.25858612 |

20 | 57.30589854 |

22 | 56.35222617 |

24 | 55.39830268 |

26 | 54.44485422 |

28 | 53.49259729 |

30 | 52.5422367 |

32 | 51.59446358 |

34 | 50.64995353 |

36 | 49.70936494 |

38 | 48.77333739 |

40 | 47.84249016 |

42 | 46.91742097 |

44 | 45.9987048 |

46 | 45.08689285 |

48 | 44.18251167 |

50 | 43.28606245 |

behave or..... https://www.youtube.com/watch?v=ZomwVcGt0LEQuote:rdw4potusLOL! Once again, Axel makes my coworkers think I'm nuts - just sitting here in the corner laughing to myself...

Quote:ThatDonGuyIt sounds as if you are talking about playing two seats at the same table simultaneously, in which case, the answer is no - the probability of both failing is much greater than .39 x .40.

If your 12 reaches 24, then your 14, is at 26; the probability is .39 x whatever the probability is of starting at 26 and reaching 7 before reaching 28.

Being "different events" is insufficient; they need to be independent events. Clearly, two betting runs are not independent if they depend on the same hand results at some point.

Right, I see where it would not be .39 x .40. But the problem is .39 is for 12 to reach 24 before reaching 6, but during that same run you could get out at 7, when the betting is 14 half is 7, or at 8, when the betting is 16 half would be 8. etc. The exit point keeps changing during the run. So, the ".39 x whatever the prob. is starting at 26 and reaching 7 before 28", wouldn't be right either.

Quote:JyBrd0403Right, I see where it would not be .39 x .40. But the problem is .39 is for 12 to reach 24 before reaching 6, but during that same run you could get out at 7, when the betting is 14 half is 7, or at 8, when the betting is 16 half would be 8. etc. The exit point keeps changing during the run. So, the ".39 x whatever the prob. is starting at 26 and reaching 7 before 28", wouldn't be right either.

I am confused as to what you are trying to determine. It looks like you are trying to chase a moving target - i.e. if you start at 12 and then get down to 14 but then get back up to 7 before reaching 28, this is a "success" of some sort; if you get to 16 before reaching 7, but then get up to 8 before reaching 32, this is also a "success", and so on, even if it means, say, reaching 100 and then getting back up to 50 before reaching 200.

Is this correct?

Assuming it is, then the probability of, say, starting at 6, then reaching 8 before 3 (and "resetting the target"), then 10 before 4 (and resetting again), then 12 before 5, and so on until you reach 40, is only about 1 in 22.6 - but that doesn't mean you're going to be ahead the other 95.6% of the time. To get from 6 to 38 means you lost bets at 6, 7, ..., 38, which is a total loss of 688; to get from 38 back up to 19 means you won 38, 37, ..., 20, which is a total gain of 551. Since you gain 1 for every win (and offsetting loss) on your way down, and 1 for every loss (and offsetting win) on your way back up, you would need 137 "extra" win/loss pairs to break even if you reached 38.

Yes, I realize you can also, say, get from 6 to 8 and then back up to 4, and your losses are only 13 while your gains are 26, but once you get from 6 to 16, you reach the "no guaranteed profit" point. Okay, there's ony about a 20% chance of even getting to the 16 point from 6, but the higher your bet gets, the more the losses add up. "A miniscule chance of losing a lot against a huge chance of winning a little" is what ruins Martingale and D'Alembert.

Also, I did make one mistake; the probability of failure should be ".39 x the probability of starting at 26 and then reaching 28 before reaching 7", not "7 before 28".