- #1
judsonjames
- 1
- 0
Hi all, thanks for looking. My first post. I am not a scientist, so forgive me for using the space here, however I have some math questions, easy, I am sure:
Given the standard deck of cards, with values 1 2 3 4 5 6 7 8 9 10 10 10 10 11,
Q1) How do we find the centre average? To be exact, do we divide by 13 cards, or 14 numbers? Given that the ace 1/11 counts as two numbers, I am not sure how to manage this 'singularity'.
Q2) As there are a number of 10's, the centre average may be thrown to the right. Where on a bell curve may this be positioned?
If anyone is interested in, this is not card counting at all, but keeping probability to whole numbers, please write.
Also,
Q3) The origins of blackjack are shrouded in mystery. 2 french mathematicians are approached in the 17th century by an unknown gambler, wanting to understand the probablity of a certain game he devised. How could they have computed thus? was 17thy century maths totally stellar? And what's more, given the endless tomes of math out there on the subject, making it all look impossibly impossible, how did they devise the formula, and not leave a shred of evidence?
And,
Q4) Our nearby casino rated there table rules to give a house advantage of 0.5% approx. How is that arrived at?
Just because one lose more games than you win (101/200 hands, if played accurately), doesn't mean you don't make more $ by doubling up when the odds are better, or better payouts for blackjacks. (Assuming one betted with a consistent percentage of ones purse, to observe rate of play, such as 5-10% of purse). One plays the same game regardless of the stakes.
And further, some of the 'standard play' charts out there seems quite silly, like, being suggested to hit on 17, with a 4/13 of holding (9/13 throwing the game away), while the dealer on 10, if he hits with 2 3 4 5 6 7, either has to hit again, or draws the hand, thus having 6/13 odds at that point, better than 4/13, then further odds may make the dealer go over. Any ideas please.
As one can see this is not card counting, however feel free to contribute any or all information, even if based from card counting or whatever. I have seen the miles of math out there...I can count to 13, and divide by 100, that's about it.
I am assuming we are using a continuous shuffler with all cards inplay at all times. (6 decks). (Shouldn't make much difference but by my reckoning more decks are better).
Thansk so much for your interest
Jud
Given the standard deck of cards, with values 1 2 3 4 5 6 7 8 9 10 10 10 10 11,
Q1) How do we find the centre average? To be exact, do we divide by 13 cards, or 14 numbers? Given that the ace 1/11 counts as two numbers, I am not sure how to manage this 'singularity'.
Q2) As there are a number of 10's, the centre average may be thrown to the right. Where on a bell curve may this be positioned?
If anyone is interested in, this is not card counting at all, but keeping probability to whole numbers, please write.
Also,
Q3) The origins of blackjack are shrouded in mystery. 2 french mathematicians are approached in the 17th century by an unknown gambler, wanting to understand the probablity of a certain game he devised. How could they have computed thus? was 17thy century maths totally stellar? And what's more, given the endless tomes of math out there on the subject, making it all look impossibly impossible, how did they devise the formula, and not leave a shred of evidence?
And,
Q4) Our nearby casino rated there table rules to give a house advantage of 0.5% approx. How is that arrived at?
Just because one lose more games than you win (101/200 hands, if played accurately), doesn't mean you don't make more $ by doubling up when the odds are better, or better payouts for blackjacks. (Assuming one betted with a consistent percentage of ones purse, to observe rate of play, such as 5-10% of purse). One plays the same game regardless of the stakes.
And further, some of the 'standard play' charts out there seems quite silly, like, being suggested to hit on 17, with a 4/13 of holding (9/13 throwing the game away), while the dealer on 10, if he hits with 2 3 4 5 6 7, either has to hit again, or draws the hand, thus having 6/13 odds at that point, better than 4/13, then further odds may make the dealer go over. Any ideas please.
As one can see this is not card counting, however feel free to contribute any or all information, even if based from card counting or whatever. I have seen the miles of math out there...I can count to 13, and divide by 100, that's about it.
I am assuming we are using a continuous shuffler with all cards inplay at all times. (6 decks). (Shouldn't make much difference but by my reckoning more decks are better).
Thansk so much for your interest
Jud
Last edited: