Now let's put the craps system ideas into baccarat.

That craps system relies upon the distant probability to get four distinct consecutive players in a row to make each 4 or more passes.

Our progressive betting sounds as

$10-20-40-80

$20-40-80-160

$30-60-120-240

$40-80-160-320

Total $1500, that is 150 units.

Whenever we win we restart the $10 betting, whenever we lose we'll go toward the next betting step.

At craps this system is so solid that you'll need a lot of sessions to lose your entire 150 units bankroll. Odds are that in the process you'll be in the positive field in the vast majority of the times.

Say we want to assign at any single baccarat column a kind of new shooter, thus whenever a new column starts it's like this column impersonates a new shooter.

For example a BBBPBBBPPPBBPBPPPB sequence will endorse the action of 8 distinct shooters getting each 2 passes (as the first hand of the shoe is a neutral indicator), zero passes, 2 passes, 2 passes, 1 pass, zero passes, zero passes and 2 passes.

In this "fortunate" example we didn't get forward the first step betting line, thus we'll get all winnings.

Of course any 5+ streak will make us a first-step loser, thus thereafter we need a proper cumulative amount of not 5+-hands to get an overall win.

Now we'll get singles, doubles, triples and 4-streaks to get a winning situation, the only situation we'll lose is whenever a 4+ situation will come out.

In a word, we'll lose our entire bankroll when a shoe will produce four or more 5+ consecutive streaks, a thing that it'll surely happen but by which degree of probability?

Now say we do want to put in action just the players getting two wins in a row. After all doubles are the more likely results at baccarat, aren't they?

Then our new betting patterns are doubles, triples, 4-streaks and 5-streaks. At the price of missing singles opportunities, now we know that the probability to lose our entire bankroll is not existent at all other than from a theorical point of view.

Show me how many times you had crossed shoes producing four or more consecutive 6+ streaks. Answer: zero.

But we can make a further adjustment, that is to classify how many times different classes of winning/losing patterns had acted consecutively along the way.

We can't prevent shoes to produce consecutive 5+-streaks, but this happening is a perfect negation either of the general asymmetrical card distribution and of the whinsical asym strenght favoring B side.

That's now that so called math experts must put their knowledge in their a.sses, even though they can easily opine that no matter what, our bets are getting a money return lower than 1.

Yep, but for their misfortune, when properly assessed the statistical advantage will be higher than what a math edge can do.

Is this mathematical big.hornsh.it?

Probably, but we're eager to get people facing our bets.

as.