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**AsymBacGuy / Re: Why bac could be beatable itlr**

« **on:**January 10, 2020, 10:46:16 pm »

Since I can't touch the SM machines topic, let's compare baccarat with roulette.

At roulette every spin will provide symmetrical probabilities, since the probability of each number or groups of numbers remains the same (1/37, 2/37, etc).

Say the whole model we are playing into is symmetrical by any means.

At baccarat every BP hand will be formed by two distinct and very different probabilities: 50%/50% and 57.93%/42.07%. Those different probabilities alone makes baccarat an asymmetrical game.

Of course every fkng shoe dealt will present different values of such asymmetricity, either in terms of numbers and, more importantly, in term of distributions.

Everybody reading my pages (btw, thanks to you) knows that the asymmetrical 57.93/42.07% value should come out on average 8.4% of the total hands dealt.

A probability value very similar to betting 3 numbers at a single zero roulette (8.1%).

Every player having a decent familiarity of both roulette and baccarat would expect that a similar probability (3 numbers vs asym hand) will produce similar dispersion values taken on the same 75 hands sample.

It seems this is not the case.

Easy to argue that a shoe formed by a finite number of cards burnt hand after hand is quite different from a so called perfect symmetrical world happening at roulette.

More importantly is to notice that when a 3 numbers group hit at roulette the winning probability is 100%, whereas at baccarat we are still fighting with a well lower 15.86% edge.

On the other hand, every other spin not hitting our 3 numbers provides a 100% losing event whereas at baccarat we still get a "fair" 50% (taxed) probability to win.

Itlr, a perfect math plan should be oriented either to bet P trying to escape the 42.07% unfavourable winning probability or, it's way better, to catch the 57.93% winning probability when betting B.

In truth a wonderful virtual player capable to always bet P without crossing one time a single asym hand will get very tiny profits (p=50%, yet certain card distributions happening on symmetrical situations help the Player side thus enlarging a bit the P probability). But there's a more excellent player, that is whoever is capable to bet B as he/she assessed that an asym hand will come out more likely within a more restricted range than what math dictates.

Some very experienced players (Alrelax and Sputnik surely belong to this list) have raised the ability to catch or abandon the situations where B or P winning probability ranges are more or less restricted than what the old 50.68/49.32 ratio dictates.

But the common denominator we have to put in first place is that shoes are not randomly shuffled (say it's

There are many ways to detect this, I prefer to choose a strict objective betting placement following the best "randomness" definition ever made by some statistics experts.

as.

At roulette every spin will provide symmetrical probabilities, since the probability of each number or groups of numbers remains the same (1/37, 2/37, etc).

Say the whole model we are playing into is symmetrical by any means.

At baccarat every BP hand will be formed by two distinct and very different probabilities: 50%/50% and 57.93%/42.07%. Those different probabilities alone makes baccarat an asymmetrical game.

Of course every fkng shoe dealt will present different values of such asymmetricity, either in terms of numbers and, more importantly, in term of distributions.

Everybody reading my pages (btw, thanks to you) knows that the asymmetrical 57.93/42.07% value should come out on average 8.4% of the total hands dealt.

A probability value very similar to betting 3 numbers at a single zero roulette (8.1%).

Every player having a decent familiarity of both roulette and baccarat would expect that a similar probability (3 numbers vs asym hand) will produce similar dispersion values taken on the same 75 hands sample.

It seems this is not the case.

Easy to argue that a shoe formed by a finite number of cards burnt hand after hand is quite different from a so called perfect symmetrical world happening at roulette.

More importantly is to notice that when a 3 numbers group hit at roulette the winning probability is 100%, whereas at baccarat we are still fighting with a well lower 15.86% edge.

On the other hand, every other spin not hitting our 3 numbers provides a 100% losing event whereas at baccarat we still get a "fair" 50% (taxed) probability to win.

Itlr, a perfect math plan should be oriented either to bet P trying to escape the 42.07% unfavourable winning probability or, it's way better, to catch the 57.93% winning probability when betting B.

In truth a wonderful virtual player capable to always bet P without crossing one time a single asym hand will get very tiny profits (p=50%, yet certain card distributions happening on symmetrical situations help the Player side thus enlarging a bit the P probability). But there's a more excellent player, that is whoever is capable to bet B as he/she assessed that an asym hand will come out more likely within a more restricted range than what math dictates.

Some very experienced players (Alrelax and Sputnik surely belong to this list) have raised the ability to catch or abandon the situations where B or P winning probability ranges are more or less restricted than what the old 50.68/49.32 ratio dictates.

But the common denominator we have to put in first place is that shoes are not randomly shuffled (say it's

**physically**impossible to arrange cards by so called perfect random models).There are many ways to detect this, I prefer to choose a strict objective betting placement following the best "randomness" definition ever made by some statistics experts.

as.