Im guessing this could be in The Theory of Blackjack, am I right?
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?
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
But if it's a CA, this still ignores the cut-card effect.
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
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?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?![]()
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.
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?
Hi again!
I tried a few things and figured out how to calculate EV for a hand!![]()
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!
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!