CyrusV
CyrusV
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November 13th, 2018 at 6:23:27 AM permalink
Hi

I am after some rationale on the following observation regarding Baccarat shoes.

Often I notice a shoe running "dead even" regarding Banker vs Player outcomes after 20, 30, 50, 'the entire shoe (70 hands)'. I accept it is by no means a unique event.

However when you consider the maths, for a shoe to be equal in terms of Banker v's Player results say after 50 hands;

The number of possible combinations 50^2, to be in Equilibrium 126,410,606,437,752, to be Imbalanced 999,489,300,404,872.

In basic terms it has a 88.8% chance of it not being in Equilibrium and a 11.2% chance of it being equal, Yet it can be quite a common event (after 20/30 hands).

I want to lay this to rest and realise it is not remarkable, however I struggling to understand why I see a shoes within the same session defy the statistical expectation 89% v's 11% with a degree of regularity.

Help me understand why this isn't remarkable.
LuckyPhow
LuckyPhow
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November 13th, 2018 at 2:15:20 PM permalink
Quote: CyrusV

Often I notice a shoe running "dead even" regarding Banker vs Player outcomes after 20, 30, 50, 'the entire shoe (70 hands)'.



The Wizard has a large number of sample 8-deck baccarat hands. Over an average shoe of 82 hands, the Banker on average wins one more hand than does the Player. Because the probability of either Banker or Player winning is almost equal, one should expect the total number of hands won to remain pretty close to one another.

I'm not quite sure how you are mixing in "combinations" in your analysis. If a shoe begins with 3 Banker wins, it could still have 20 wins in 40 hands. Another shoe might begin with 3 Player wins and a total of 20 wins in 40 hands [ignoring Ties for simplicity], even though the "combination" of Ps and Bs at any intermediate point might differ from that of the first shoe. I'm unsure how this defies your "statistical expectation." Apologies if I sound confused, because I am.

Quote: CyrusV

Help me understand why this isn't remarkable.


It isn't remarkable because the game was designed that way. The Banker and the Player each have an almost identical probability of winning any random hand. That results in them sometimes [sometimes too frequently?] finishing with identical win totals for a shoe.
CyrusV
CyrusV
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November 14th, 2018 at 11:52:28 PM permalink
Thank you for your response, sure I understand that for all intense purpose a Baccarat outcome is akin to a 50/50 coin toss.

However if you choose a smaller sample of say 4 hands (ignoring ties), we can have any of the 16 possibilities.

B B B B
B B B P
B B P B
B B P P
B P B B
B P B P
B P P B
B P P P
P B B B
P B B P
P B P B
P B P P
P P B B
P P B P
P P P B
P P P P

Of which 6 are in equilibrium and 10 are imbalanced. The larger the sample the greater the ratio swing to imbalanced over equilibrium.

If you toss a coin 50 times, I would not expect 25 heads and 25 tails, indeed the maths points to a 11% probability of this occurring and a 88% of it not occurring. This doesn't fit with what I'm seeing at the tables or as you point out what the Wiz's shoes indicate. So what am I not grasping here??

The reason I pose the question, is sometimes I notice (as an example) a shoe sitting on 25 Banks and 23 Players. Two more outcomes needed to reach the 50 mark. The maths point to a 88% chance of not being in equilibrium, therefore bet Bank for 2 hands maximum, skipping the second bet if the you win the first. Only to see the shoe sit at 25 each side after 50 hands.
OnceDear
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OnceDear
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November 15th, 2018 at 2:08:32 AM permalink
Quote: CyrusV


The reason I pose the question, is sometimes I notice (as an example) a shoe sitting on 25 Banks and 23 Players. Two more outcomes needed to reach the 50 mark. The maths point to a 88% chance of not being in equilibrium, therefore bet Bank for 2 hands maximum, skipping the second bet if the you win the first. Only to see the shoe sit at 25 each side after 50 hands.

The maths might have pointed to 88% chance of non-equilibrium before any rounds were dealt. If 48 rounds are in the past with only 2 rounds to play, then those 48 outcomes HAVE been decided. your 88% has moved on and you are looking at a new future sample size of about 2 rounds. You know how those two rounds could go.
What you propose is similar to tallying the hands up to the penultimate one and (wrongly) expecting to know with almost absolute certainty which way the last hand will go.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
unJon
unJon
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November 15th, 2018 at 6:10:02 AM permalink
The question isnít the probability that the shoe is even at exactly 20 or exactly 30 hands. Itís the probability that it is even at least once during hands 20-40. Or am I misunderstanding? Thatís 11 shots at evenness.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
CyrusV
CyrusV
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November 15th, 2018 at 10:30:20 AM permalink
Quote: unJon

The question isnít the probability that the shoe is even at exactly 20 or exactly 30 hands. Itís the probability that it is even at least once during hands 20-40. Or am I misunderstanding? Thatís 11 shots at evenness.



Nope, it was the shoe being even at a exact point within a shoe.

Quote: OnceDear

The maths might have pointed to 88% chance of non-equilibrium before any rounds were dealt. If 48 rounds are in the past with only 2 rounds to play, then those 48 outcomes HAVE been decided. your 88% has moved on and you are looking at a new future sample size of about 2 rounds. You know how those two rounds could go.
What you propose is similar to tallying the hands up to the penultimate one and (wrongly) expecting to know with almost absolute certainty which way the last hand will go.



Thanks for that, I do understand that odds\expectation only relate to the number of bets actually placed. I would avoid a situation such as 25 v's 24, as it only offers a single shot (or just bet the table minimum). Rather would look for 25 v's 21 or 22 scenarios. Now I understand spotting something like this and when it gets down to the final bet it is just another 50-50 outcome. Personally I prefer to treat bets as within a series and not individually.

I understand from your reply, that this 88% simply never enters the equation, implied probability perhaps (Barstow)? Does such a thing exist!!

Thanks for your response.
unJon
unJon
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November 15th, 2018 at 10:37:22 AM permalink
Now Iím confused. The chance of being even at exactly that point is low (call it 11%). The OP says even at 20, 30, 40 or even 70 hands. The chance of being even at any one of those points (which would trigger a memory that itís ANOTHER even shoe) is much higher. Back of envelope (not correct but close for the lazy) would be 1-(1-0.11)^4. Or 37%. Add in 50 and 60 points to check and you get to 50/50.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
gordonm888
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gordonm888
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November 15th, 2018 at 1:41:53 PM permalink
Quote: OnceDear

The maths might have pointed to 88% chance of non-equilibrium before any rounds were dealt. If 48 rounds are in the past with only 2 rounds to play, then those 48 outcomes HAVE been decided. your 88% has moved on and you are looking at a new future sample size of about 2 rounds. You know how those two rounds could go.
What you propose is similar to tallying the hands up to the penultimate one and (wrongly) expecting to know with almost absolute certainty which way the last hand will go.



Sorry, I think you're confused on this one.

Problem: in 50 hands (or 50 coin tosses) what is the frequency distribution for how often the dealer wins m hands, m =0-50. When you do that analysis, you will find that the chance of winning exactly 25 of 50 hands is low. I think the OP is citing standard statistics.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
OnceDear
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OnceDear
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November 15th, 2018 at 3:40:02 PM permalink
Quote: gordonm888

Sorry, I think you're confused on this one.

Problem: in 50 hands (or 50 coin tosses) what is the frequency distribution for how often the dealer wins m hands, m =0-50. When you do that analysis, you will find that the chance of winning exactly 25 of 50 hands is low. I think the OP is citing standard statistics.

Thanks Gordon,
I was taking the OP's word for it that the chance of it not being a 25:25 split was of the order of 88%. I.e. 12% probability that it would be. I did think that was on the high side but couldn't be bothered to check. Actually that was why I used the words 'might have'.
I was just trying to allude to the fact that after 48 known outcomes (24 of each), that the 88% / 12% was ancient history with no predictive value for those last 2 outcomes. I.e. if we were facing just two coin tosses in the future, the probability of exactly 1 head and 1 tails would be 50% and certainly not 12%
I stand by that point, even if the originally quoted stats were in error.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
TomG
TomG
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November 15th, 2018 at 7:31:53 PM permalink
Quote: CyrusV

Help me understand why this isn't remarkable.



In craps, there is an 11% chance of rolling a nine. Would you find that remarkable? In blackjack, there is less than an 11% chance of getting dealt 20. I don't find those all that remarkable. It isn't remarkable, because when something happens even with an 89% chance of it not happening, it generally isn't considered remarkable.

Now if you have records of it happening much more often than 11% of time, that would be different

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