Proper bet spreading for Ace sequencing
- Thread starter Jack_Black
- Start date
No, I do my best sequenzing and bet only one spot, first place.
The problem is with 2 or 3 spots, if there sits an other person
you have no rights of decision. (spelling ?)
If there sits a counter or a perfect BS player you could bet 2 spots.
I like to bet only one spot because the small flux.
Sure the hourly win rate of betting 2 spots is bigger, I would never bet
3 spots, IMHO its nonsens or mental weakness.
There is no greater frequency of aces in percent to the bets
if you bet more spots,
example: betting 2 spots 13 times, you get usually 2 aces
when i am sequenzing I would get 6 aces in 26 big bets at 2 spots =8%
or 3 aces at 1 spot = 8%
Ok the hourly gain is bigger.
But I found out, when I bet only first base I get a bit more aces per spot, as if I bet 2 spots.
That means the first spot is the best in the long run.
The good news the flux are very very small.
If I mustn't care about money and would have a very big bankroll ,
I would bet the first 2 spots.
And there are the costs of the waiting bets, often I only sit at the table
and make no waiting bets, tell the dealer I have a bad streak and want to
wait.
If you bet only 1 spot, it often happens that you get no ace maybe
in 7 big bets, so you have a better camo than betting 2 spots.
Unfortunately for camo(fortunately for you) you hit the aces very often
and the old lady next to you cries "every time he bets big, he gets an ace"
you have a problem to explain , "pure luck, I had a good feeling,that's because I tip, 7th sense.
E.H.
I bet
The problem is with 2 or 3 spots, if there sits an other person
you have no rights of decision. (spelling ?)
If there sits a counter or a perfect BS player you could bet 2 spots.
I like to bet only one spot because the small flux.
Sure the hourly win rate of betting 2 spots is bigger, I would never bet
3 spots, IMHO its nonsens or mental weakness.
There is no greater frequency of aces in percent to the bets
if you bet more spots,
example: betting 2 spots 13 times, you get usually 2 aces
when i am sequenzing I would get 6 aces in 26 big bets at 2 spots =8%
or 3 aces at 1 spot = 8%
Ok the hourly gain is bigger.
But I found out, when I bet only first base I get a bit more aces per spot, as if I bet 2 spots.
That means the first spot is the best in the long run.
The good news the flux are very very small.
If I mustn't care about money and would have a very big bankroll ,
I would bet the first 2 spots.
And there are the costs of the waiting bets, often I only sit at the table
and make no waiting bets, tell the dealer I have a bad streak and want to
wait.
If you bet only 1 spot, it often happens that you get no ace maybe
in 7 big bets, so you have a better camo than betting 2 spots.
Unfortunately for camo(fortunately for you) you hit the aces very often
and the old lady next to you cries "every time he bets big, he gets an ace"
you have a problem to explain , "pure luck, I had a good feeling,that's because I tip, 7th sense.
E.H.
I bet
What is having a hard time sinking in is how do you increase ace hits purely by knowing they are coming but not betting more than one hand? Do you mean ace hits at max bet, or just ace hits? If just ace hits, how does knowing they are coming help them land on your one spot? Forgive me for being so dense. Thanks.
Ok, I'll quit jerking you around now. :laugh:
if one of the keycards gets shuffled behind the ace, that usually means that that particular shuffle procedure is not trackable. Move onto another joint.
or, when tracking aces in 2d and up games, you are using 2 keycards before the ace as to confirm that you have the correct keycards to begin with. that's why I disagree with ZG's method of using 1 keycard. maybe it is the right one, maybe it isn't. but seeing the 2nd one come up after the first one will alleviate that.
Or, you're just screwed for that round, until the next group of keycards/ace or the next shuffle. hey, no one said this was going to work 100% of the time.
if one of the keycards gets shuffled behind the ace, that usually means that that particular shuffle procedure is not trackable. Move onto another joint.
or, when tracking aces in 2d and up games, you are using 2 keycards before the ace as to confirm that you have the correct keycards to begin with. that's why I disagree with ZG's method of using 1 keycard. maybe it is the right one, maybe it isn't. but seeing the 2nd one come up after the first one will alleviate that.
Or, you're just screwed for that round, until the next group of keycards/ace or the next shuffle. hey, no one said this was going to work 100% of the time.
zengrifter said:
So then how much % do I deduct to allow for that? z
g
g
Assuming that you are using a single key card. You said that you have a 1 in 6 chance of getting an extra ace on your hand. I'm assuming this because the ace will be one of the six cards following your key-card.
Situation A: You are playing heads-up
In this situation the likelihood of the ace following on the spots is as follows:
Hand 1, first card: 1/6
hand 2, first card: 1/6
Dealer hole card: 1/6
Hand 1, second card: 1/6
Hand 2, second card: 1/6
Dealer up card: 1/6
the 52% advantage (or 40% something in 6:5) doesn't change regardless whether the ace is your first or second card. According to Exhibit CAA, it seems there is slight difference in player disadvantage depending on whether ace is dealer's hole card (-38%) or dealer's face card (-36%) - let's just average it out to be -37%. Assume further that HE is 0.45% - and we are not adjusting this for the fact that there is one less ace is the game.
Assume bet size, B, is $100.
In this case I would calculate your edge as follows:
1/3 of the time the Ace is either card on first hand:
Hand 1: 52% x B ($52)
Hand 2: -0.45% x B (-$0.45)
Overall: 51.55% x B ($51.55)
1/3 of the time the Ace is either card on second hand:
Hand 1: -0.45% x B (-$0.45)
Hand 2: 52% x B ($52)
Overall: 51.5% x B ($51.55)
1/3 of the time the Ace is either of dealer's cards
Hand 1: -37% x B (-$37)
Hand 2: -37% x B (-$37)
Overall: -74% x b (-$74)
Thus your EV in this situation would be 1/3 x 51.55 + 1/3 x 51.55 + 1/3 x -74 = $29.10
Your wager in this round is $200 (two bets of $100) - thus your advantage is $29.1/$200 = 14.55%
Now let's consider the situation where you are not playing heads-up but are instead sharing the table with other players who between them take up two spots.
In this situation the likelihood of the ace following on the spots is as follows:
A) ZG 1, first card: 1/6
B) ZG 2, first card: 1/6
C) Player 1, first card: 1/6
D) Player 2, first card: 1/6
E) Dealer hole card: 1/6
F) ZG 1, second card: 1/6
A) 52%xB + -0.45% B = 51.45% ($51.45)
B) -0.45%B + 52%B = 51.45% ($51.45)
C) -0.45%B + -.45%B = -0.9% (-$0.90)
D) -0.45%B + -.45%B = -0.9% (-$0.90)
E) -37%B + -37%B = -74% (-$74)
F) 52%B + -0.45% = 51.45% ($51.45)
These numbers add up to $78.55 and give you an advantage of 78.55/200 = 39.28% - much higher because there is a 3/6 chance of you getting the ace versus dealer's chance of 1/6 (a 3:1 ratio) compared to a 4/6 chance to 2/6 chance in heads-up situation A (a 2:1 ratio).
The calculations above show how I view this mathematical question - not sure whether others agree with it - but if the percentages are right then you certainly wouldn't be better overbetting if you are using your standard max bet.
In this situation the likelihood of the ace following on the spots is as follows:
A) ZG 1, first card: 1/6
B) ZG 2, first card: 1/6
C) Player 1, first card: 1/6
D) Player 2, first card: 1/6
E) Dealer hole card: 1/6
F) ZG 1, second card: 1/6
A) 52%xB + -0.45% B = 51.45% ($51.45)
B) -0.45%B + 52%B = 51.45% ($51.45)
C) -0.45%B + -.45%B = -0.9% (-$0.90)
D) -0.45%B + -.45%B = -0.9% (-$0.90)
E) -37%B + -37%B = -74% (-$74)
F) 52%B + -0.45% = 51.45% ($51.45)
These numbers add up to $78.55 and give you an advantage of 78.55/200 = 39.28% - much higher because there is a 3/6 chance of you getting the ace versus dealer's chance of 1/6 (a 3:1 ratio) compared to a 4/6 chance to 2/6 chance in heads-up situation A (a 2:1 ratio).
The calculations above show how I view this mathematical question - not sure whether others agree with it - but if the percentages are right then you certainly wouldn't be better overbetting if you are using your standard max bet.
I posted these calculations nearly over two weeks ago. Nobody actually ever commented on them - it would be great to hear whether people agree or disagree with the maths. I had been hoping to receive some kind of feedback on them
Solo Player, I would think that sequencing does WAY BETTER than straight-up counting, regardless whether 2d or 6d.
Solo Player, I would think that sequencing does WAY BETTER than straight-up counting, regardless whether 2d or 6d.
No, your math could not be much further off. Let's say you do have a 1/6 chance of getting an ace on any one hand.
Your edge when your first card is an ace: 51%.
Your edge when your first card is some random card other than an ace: -4.3% approximately.
Your edge: (1/6)*51% + (5/6)*-4.3% = 4.9%
So many people have been tricked by this problem, don't be one of them!
Your edge when your first card is an ace: 51%.
Your edge when your first card is some random card other than an ace: -4.3% approximately.
Your edge: (1/6)*51% + (5/6)*-4.3% = 4.9%
So many people have been tricked by this problem, don't be one of them!