9 Sequencing Baccarat?

[Gronbog]

But that is moving the goal posts. Knowing what the next card will be is completely different from knowing what one of the next two cards is going to be.

Yes, but even knowing the probability that the 9 will occur within the next 2 cards gives you enough information to increase your edge. Think long term. Think "Ace Sequencing" in blackjack. It's not necessary to get the Ace every time in order to have an edge over time.

[Meistro123]

Yes, but even knowing the probability that the 9 will occur within the next 2 cards gives you enough information to increase your edge.

Oh, what is your edge in that case with a player bet?

[Gronbog]

Not at all. If you get it you win, if you don't get it you lose. There is no advantage if there is an equal chance of the card going to the player or banker.

Really? You think that if the nine is first and you bet on player you are guaranteed to win the hand and that if the nine goes to banker you are guaranteed to lose the hand? Of course not. You just have an increased advantage on that hand if you know where the 9 will land. If you know that the 9 is one of the first or second cards, you will still have an advantage, but it will not be as high as with exact knowledge. [edit: the following sentence not entirely true. See my later posts in this thread.] The advantage further degrades the less precision you have with respect to the position of the 9.

This is exactly how card counting works, for what it's worth. Your count gives you an increased probability of receiving favourable cards on the next hand. You have an advantage so you bet more, but there is no guarantee that you will get those cards on that next hand or win your hand even of you do. The increased probability alone along with repetition is what gives you your overall edge. The same applies to card sequencing in any game. You don't get the card every time, nor do you need to. You don't win every time when you do get it. But over time, with repetition, you get the edge.

[Gronbog]

Oh, what is your edge in that case with a player bet?

I gave the formula in an earlier post. The edges needed to do the calculation are in the Eliot Jacobson article, also posted earlier in this thread.

[ZenMaster_Flash]

In the old-fashioned "Big Baccarat" game the
cards are destroyed after each shoe is complete.
Those games have become very rare. I have not
seen one in use in many years.

What exists is two versions of Mini-Baccarat.

In general, the low stakes game uses ASM
and the players never get to touch the cards.
For higher stakes the "Macau Style" ASM game
is commonplace. The sole difference being that
the players do get to touch the cards.


If I am reading Bushie correctly he appears to
be attempting to 'track' the 9's.

Without precision, I cannot see the value in this.
The e.v. for the two sides is almost identical for
Banker/Player receiving a 9 or an 8 as their first
card.

One places a bet before the cards are dealt of course.
Now ~ can someone explain to me how even if I knew
for certain that either the first or second card is a nine
(9) that could conceivably aid the bettor.

I cannot see it. I find myself wondering if this notion has
been developed by someone who imagines that the game
is trackable. IF one "counts nines" and finds a shoe with an
unusually rich surplus in nines (9's) it fails to help anyone
as the E.O.R. for the 9 only slightly favors the Player bet.
There is no way to know which hand will get a particular card.

[Gronbog]

With the numbers from https://www.888casino.com/blog/edge-...the-known-card ...

You know that one of the first two cards is a 9, but not which one. You bet on player.

• If the card is first out, it goes to player and you have a 21.528% advantage. This happens 50% of the time.
• If the card is second out, it goes to banker and you have a 20.641% disadvantage. This also happens 50% of the time.

So your expectation is 0.5 x 21.528% - 0.5 x 20.641% = 0.4435%

Furthermore, since banker and player each get two of the first 4 cards, you have the same advantage even if you only know that one of the first 4 cards is a 9 (left as an exercise). Also left as an exercise, do you still have an advantage if you only know that one of the first 5 cards is a 9? How about one of the first 6 cards?

[Gronbog]

Interesting addendum to my previous post. It can be shown that your advantage increases if you only know that one of the first 3 cards is a 9 (vs the first 2).

[Iwantmoney]

This is was in Macau MGM when they had manual shufflers yes or no

John Chang's team with a partner team, and some other teams, won more than 40,000,000 HKD in Macau, in two years, by "catching" the first card in some baccarat games.

[ZenMaster_Flash]

I guess that my mathematical chops are weak,
but someone needs to tell me how ANY knowledge
of the card(s) to be dealt is possible if the cards were
not manually shuffled.

[peterlee]

This is was in Macau MGM when they had manual shufflers yes or no

MGM did has hand shuffle tables, two passes single riffle each.
But JC's winning was not from them.

[Iwantmoney]

Wow, you still remember their two passes this was back in 2012. Then how else could you catch the first card and where

MGM did has hand shuffle tables, two passes single riffle each.
But JC's winning was not from them.

[Hackjack]

This is my first post here, so take it for what it is. But I see in ZenMaster's response the standard AP response...nothing could ever be new in the world, no techniques outside the known ones could ever work, etc. Work backwards: knowing that the first card is a 9 gives the player an advantage - that is an undeniable fact. So now work through various ways to find out what the first card is.

I asked a high profile AP recently about what he thought of an idea I had to take advantage of the fact that most hand shuffles are not anywhere near random, and he basically told me that I was an idiot for even thinking it. But the simulations I've now run are proving quite promising, at least against certain shuffles. So don't let responses like Zenmaster's totally discourage you from at least investigating new approaches to these things. People get set in their ways and will criticize anything new.

[James989]

With the numbers from https://www.888casino.com/blog/edge-...the-known-card ...

You know that one of the first two cards is a 9, but not which one. You bet on player.

• If the card is first out, it goes to player and you have a 21.528% advantage. This happens 50% of the time.
• If the card is second out, it goes to banker and you have a 20.641% disadvantage. This also happens 50% of the time.

So your expectation is 0.5 x 21.528% - 0.5 x 20.641% = 0.4435%

Furthermore, since banker and player each get two of the first 4 cards, you have the same advantage even if you only know that one of the first 4 cards is a 9 (left as an exercise). Also left as an exercise, do you still have an advantage if you only know that one of the first 5 cards is a 9? How about one of the first 6 cards?

If the card is second out, it goes to banker and you have a 20.641% disadvantage

Above statement is incorrect, if card "9" goes to banker and you bet on PLAYER , you have a 23.489% disadvantage , so your expectation = 0.5x21.528% - 0.5 x 23.489% = -0.9805% !
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