True advantage?

MrMaster

Active Member
#1
Where can I find instructions on how to calculate my true (dis)advantage in any given situation, if I know exactly what and how many cards have been played etc.
Im guessing this could be in The Theory of Blackjack, am I right?
 

MrMaster

Active Member
#5
Since I can't order any books for a while (just ordered some), could anyone post the formula(s) used in the calculations?
Thank you in advance! :)
 

k_c

Well-Known Member
#7
QFIT said:
There is no formula. These calculations are done by combinatorial analysis. Further, they are only exactly correct for single deck games that deal a fixed number of rounds. The numbers in the books ignore the cut-card effect.

http://www.blackjackincolor.com/blackjackeffects1.htm
Actually the calculations can be made for any number of decks. My composition dependent program can handle 1 to 41297762 decks. (41297762 because any more overflows the long integer data type.) This program can show you theoretical EVs for any given shoe state.

My total dependent program can handle 1 to 8 decks. It computes overall EVs using full shoe total dependent basic strategy so it is more practical than theoretical. It could also handle 41297762 decks but it's more convenient to limit it to 8 decks because that's easier to display and basically that's all anyone would be interested in anyway.

k_c
 

QFIT

Well-Known Member
#8
k_c said:
Actually the calculations can be made for any number of decks. My composition dependent program can handle 1 to 41297762 decks. (41297762 because any more overflows the long integer data type.) This program can show you theoretical EVs for any given shoe state.

My total dependent program can handle 1 to 8 decks. It computes overall EVs using full shoe total dependent basic strategy so it is more practical than theoretical. It could also handle 41297762 decks but it's more convenient to limit it to 8 decks because that's easier to display and basically that's all anyone would be interested in anyway.

k_c
But if it's a CA, this still ignores the cut-card effect.
 

sagefr0g

Well-Known Member
#9
QFIT said:
But if it's a CA, this still ignores the cut-card effect.
so i guess what this means is that for most games (which have cut cards) that we really need to realize that the house advantage is really worse by around 0.02% for multiple deck and up to 0.06% for single deck than most estimates claim. and apparently this detrimental affect is realized mainly in the last round? :confused:
 

k_c

Well-Known Member
#10
QFIT said:
But if it's a CA, this still ignores the cut-card effect.
A CA calculation will always give a valid result whenever player's hand (or hands in the case of splits) as well as dealer's hand can always be completed without running out of cards. The CA doesn't care if the result is because of the so-called cut-card effect or any other effect such as floating advantage. It just goes through all of the possibilities and reports the EV for a given shoe composition.

k_c
 

QFIT

Well-Known Member
#11
k_c said:
A CA calculation will always give a valid result whenever player's hand (or hands in the case of splits) as well as dealer's hand can always be completed without running out of cards. The CA doesn't care if the result is because of the so-called cut-card effect or any other effect such as floating advantage. It just goes through all of the possibilities and reports the EV for a given shoe composition.

k_c
Exactly. Which is why overall EV cannot be correctly calculated in this manner as the frequencies of shoe compositions are ignored by and cannot be calculated by CA.
 

sagefr0g

Well-Known Member
#12
QFIT said:
Exactly. Which is why overall EV cannot be correctly calculated in this manner as the frequencies of shoe compositions are ignored by and cannot be calculated by CA.
aren't shoe compositiions essentially independent events for which over a large number of shuffles some range of true counts shall present? point being wouldn't that effect be overall a small one? sort of the differance between measuring EV for a shoe game as opposed to a CSM game? :confused:
 

QFIT

Well-Known Member
#13
sagefr0g said:
aren't shoe compositiions essentially independent events for which over a large number of shuffles some range of true counts shall present? point being wouldn't that effect be overall a small one? sort of the differance between measuring EV for a shoe game as opposed to a CSM game? :confused:
Blackjack is not really an independent event game. It would be if we shuffled after every hand. But as we don't, interesting effects take hold:

http://www.blackjackincolor.com/blackjackeffects1.htm
http://www.blackjackincolor.com/blackjackeffects2.htm
 

k_c

Well-Known Member
#14
QFIT said:
Exactly. Which is why overall EV cannot be correctly calculated in this manner as the frequencies of shoe compositions are ignored by and cannot be calculated by CA.
My ideal CA would enter data from randomly shuffled and dealt shoes into a database. This could be done as a player played a practice game. Pre-deal shoe comp would be entered and the shoe's overall EV computed and entered. Likewise card by card post-deal EVs would be entered along with the shoe comp. Since the data would be based on exact calculation a relatively small amount of data would be needed to start getting valid data. Every time the player practiced more data would be added. The database could be queried for all sorts of information.

My game and computer CA does the dealing and shows EV as the player practices but does not create a database. Player can see, though, pre-deal overall EV using basic strategy and how well his count is doing in determining bet size on each hand. He can also see post deal EVs and see how well his count index plays are doing. What is lacking is a database to aggregate the results but I still feel it is a good tool.

k_c
 

MrMaster

Active Member
#15
Hi again!
I tried a few things and figured out how to calculate EV for a hand!:)
I checked it with k_c's program and it was correct!!
Now im wondering about insurance, I understand that it's -0,5 because you get half of your bet back, but with no holecard it says that insurance with 10 has an EV of -53,6! Why is this?
Im wondering about blackjack aswell, do you multiply its EV with 1,5?
 

k_c

Well-Known Member
#16
MrMaster said:
Hi again!
I tried a few things and figured out how to calculate EV for a hand!:)
I checked it with k_c's program and it was correct!!
Now im wondering about insurance, I understand that it's -0,5 because you get half of your bet back, but with no holecard it says that insurance with 10 has an EV of -53,6! Why is this?
Im wondering about blackjack aswell, do you multiply its EV with 1,5?
I don't know what you input but you must be talking about surrender and not insurance. If ENHC (no peek) is the rule then unconditional EV assuming player's non-blackjack loses to dealer's blackjack is always displayed. This will be the case even though the option to 'Display conditional EVs where applicable' is selected because in that case it makes no sense to say dealer has checked for blackjack and doesn't have it. Unconditional EV will also always be displayed if early surrender is the rule for the up card that was input. (ES can only apply to up cards of 10 or A because dealer can never have blackjack otherwise.) Conditional EV for standing on a player blackjack versus any up card will be 150%. Unconditional EV for standing on a player blackjack versus up cards (2-9) will always be 150%. Unconditional EV for standing on a player blackjack versus up cards (10,A) will be less than 150% unless it is impossible for dealer to have blackjack given the shoe comp that was input. If you specifically input both a player and dealer blackjack then EV is displayed as 0% for both unconditional and conditional EV.

k_c
 

MrMaster

Active Member
#18
Oh yes, my mistake, I was talking about surrender...:laugh:

And QFIT, you told me earlier that there is no formula, so what i have been doing is calculating the odds for every single card that is possible to come out of a deck.
For example if the dealer is showing a 10 and player has, let's say 10,6.
So the first player card (10) has 16/52 probability of coming out, the dealers first card (10) has a probability 15/51 of coming out, and the players second card (6) has a probability of 4/50.
The first thing I tried to calculate was what are the odds of the dealer getting a bj in this situation assuming the player stands. They are obviously 4/49 since there are 4 aces and a total of 49 cards left. I checked it in k_c's program and I was correct!

Thank you for your replies!
 
Last edited:

QFIT

Well-Known Member
#19
MrMaster said:
Oh yes, my mistake, I was talking about surrender...:laugh:

And QFIT, you told me earlier that there is no formula, so what i have been doing is calculating the odds for every single card that is possible to come out of a deck.
For example if the dealer is showing a 10 and player has, let's say 10,6.
So the first player card (10) has 16/52 probability of coming out, the dealers first card (10) has a probability 15/51 of coming out, and the players second card (6) has a probability of 4/50.
The first thing I tried to calculate was what are the odds of the dealer getting a bj in this situation assuming the player stands. They are obviously 4/49 since there are 4 aces and a total of 49 cards left. I checked it in k_c's program and I was correct!

Thank you for your replies!
This will work in limited circumstances. In fixed round games, or at the first hand after a shuffle when you can only draw one card and you also take into account the different deck compositions for the dealer hands depending on the cards that you might draw.
 

k_c

Well-Known Member
#20
MrMaster said:
Oh yes, my mistake, I was talking about surrender...:laugh:

And QFIT, you told me earlier that there is no formula, so what i have been doing is calculating the odds for every single card that is possible to come out of a deck.
For example if the dealer is showing a 10 and player has, let's say 10,6.
So the first player card (10) has 16/52 probability of coming out, the dealers first card (10) has a probability 15/51 of coming out, and the players second card (6) has a probability of 4/50.
The first thing I tried to calculate was what are the odds of the dealer getting a bj in this situation assuming the player stands. They are obviously 4/49 since there are 4 aces and a total of 49 cards left. I checked it in k_c's program and I was correct!

Thank you for your replies!
That's the idea of how a combinatorial analyzer works. It's just math. The math doesn't change whether or not a cut card is encountered. The sole dependency is the shoe composition. This works for the beginning of a shoe, the end of a shoe or any point in between with the prerequisite that all player and dealer hands used in a computation must be always completed without running out of cards.

k_c
 
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