Tuesday, November 4, 2008

The O/L/6 System for Baccarat

“The O/L/6”

Here’s a fresh idea for Baccarat:

Ordinarily, systems are presented with a preliminary explanation of the “why.” I am a firm believer that although all readers would benefit from an understanding of the math and logic and probabilities behind a system, most readers will skip through the explanation of the “why“ and focus their efforts on the “how to.” Unfortunately, the results are that many players will dismiss the system as ineffective without working on it or they will play the system without a thorough understanding of the possibilities and will become prematurely disenchanted when they do not see immediate results. With this in mind, I’ll start by giving you the system, that way you can get on with your gambling, or ridicule or scorn.
This is a simple system of bet selection and progression. Simply bet that the pattern exhibited by the last 6 decisions will not repeat precisely with the next 6 decisions. Hence the name, Opposite Last Six, or O/L/6.

Your first bet is 1 unit and is placed on the opposite decision as the 6th prior outcome. The previous pattern is then bet against and bets are doubled until you win or lose the 6th decision.
When you win, you begin anew. Your first bet after a win is the opposite of the 6th prior decision at that point.
Example: You look at your scorecard and the last 6 decisions are B B P B P P, your next bet will be P and if you lose, then P again and if that bet loses, the next bet is placed on B (following the previous pattern). Continuing this example, your first bet was 1 unit on P, then after a loss you bet 2 units on P, and then after a loss, you bet 4 units on B and you Won. NOW your next bet will be based on the 6 most recent decisions: B P P B B B. Your next bet is one unit on P (opposite of the sixth previous decision which was B).
The highest bet you will have to place is 32 units and a lost series will cost you 63 units.
Here’s the simple math: There are 64 possible combinations in each string of 6 decisions. Each of these 64 possibilities are equally likely. If you chose any particular pattern of 6 decisions, BBBBBB, for example, and you bet a 6-step martingale, you can expect to lose this bet (in a true 50/50 game), once in every 64 attempts (64 groups of 6 decisions). This method (in true 50/50) form would mathematically produce no gain nor loss but rather a regular return to even or zero (or zero less the house edge).
BUT by betting that the last 6 decisions will NOT repeat in precisely the same order, you are betting that an event that has a 1 in 64 likelihood of occurrence will not happen back-to-back.

More "bad math" to think about:
If a shoe gives you and average of 68 decisions, and if you sit out the first sixand if you then bet as I have outlined above you can expect to win one unit about every three decisions, when winning. This should be about 20 units per shoe.
So - in 5 shoes you should win about 100 units (without a loss). IF (and here's a big if) - IF you lose only once in five shoes, then you will be set back 64 units and your net will be 36 units for 5 shoes or 7+ units per shoe.
NOW - any of you with notes from actual shoes - please take a moment and look for a pattern of 6 decisions immediately followed by the same pattern (not a streak because we decided earlier that we would bet with a streak and not against, so 6 one way followed by 6 the other way would be a loser for us - i.e BBBBBBPPPPPP or PPPPPPBBBBBB = loser).
IF (another big if) - IF you find that a losing pattern shows up more often than one time in five shoes, THEN this system will perform less than 7 units per shoe on average and perhaps it will be a big loser.
BUT if you find that the losing pattern occurs less often than once in five shoes, THEN this system should perform at a rate of more than 7 units per shoe.
I’ve not used this system, but I have run it through some actual shoes and it performs very well in limited testing. If you have any shoes, please take a moment and look for any string of 6 decisions that is followed immediately by the same exact string of 6 decisions (excluding streaks of 12). I want to know if this occurs more often than once in 5 shoes.
A friend emailed me the data from 91 shoes with the decisions grouped into columns of 6. This is not how I envisioned checking my theory BUT it did make it easy to sort of spot check the frequency of repeating patterns of 6. I found 11 such occurences in his 91 "real" shoes. This number was closer to the mathematical expectancy than I had hoped. The basic math shows approximately 10 betting opportunities per shoe, with all producing 1 unit gains except the 11 which cost us 64 units (91 x 10 = 910; 11 x 64 = 704; 910 - 704 = 206 unit gain). The average therefore is slightly more than 2 units per shoe not considering commissions. Not nearly what I had hoped for. However, there may be a money management technique that makes better use of the fact that the event is occuring less often than expected.

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