Proper bet spreading for Ace sequencing

[zengrifter]

It could be a myriad of different games. Those casino bastards haven't limited 6:5 payout on just SD.

1D-2D Ace track doesn't require mnemonic assist. zg

 

[zengrifter]

Are we remiss to not mention this controversial book, without which I
would not have become proficient with my rudimentary Ace tracking >>
http://www.blackjackaceprediction.com

 

[blackriver]

r u guys just sequencing single deck? i like grifters style of doing it single deck so u dont have to make usper gross spreads on ur aces. otherwise if u do a shoe game, how many keys will you remember? youll probably need 3 cards per key , blah

 

[zengrifter]

r u guys just sequencing single deck? i like grifters style of doing it single deck so u dont have to make usper gross spreads on ur aces.

I do it mostly at 2D, but this thread has got me thinking about 6:5 1D. zg

 

[bigplayer]

Math - 2D simple keycard approach, no mneumonic assit >

1. I have a 1 in 2 chance of spotting correct keycard 50% edge/2 = 25%

2. A 1 in 6 chance of getting (extra) Ace one of my two hands.

3. 50%/2/6 = 4% approx advantage per two hands.

Did I miss anything? zg

Yes you did miss something. If you have a 1 in 2 chance of getting the ace your edge is 50%/2 MINUS the advantage the dealer gains if he gets it instead of you. If you whiff on your ace and it goes to the dealer you are a huge underdog.

 

[aslan]

Yes you did miss something. If you have a 1 in 2 chance of getting the ace your edge is 50%/2 MINUS the advantage the dealer gains if he gets it instead of you. If you whiff on your ace and it goes to the dealer you are a huge underdog.

That would be mitigated by the fact that others in the game might get the ace besides the dealer.

 

[zengrifter]

Yes you did miss something. If you have a 1 in 2 chance of getting the ace your edge is 50%/2 MINUS the advantage the dealer gains if he gets it instead of you. If you whiff on your ace and it goes to the dealer you are a huge underdog.

So then how much % do I deduct to allow for that? zg

 

[beyondbj]

Math - 2D simple keycard approach, no mneumonic assit >

1. I have a 1 in 2 chance of spotting correct keycard 50% edge/2 = 25%

2. A 1 in 6 chance of getting (extra) Ace one of my two hands.

3. 50%/2/6 = 4% approx advantage per two hands.

Did I miss anything? zg

sorry i dont quite understand why 1 in 6 chance of getting an extra ace

will have to divide the advantage for /6

 

[beyondbj]

Yes you did miss something. If you have a 1 in 2 chance of getting the ace your edge is 50%/2 MINUS the advantage the dealer gains if he gets it instead of you. If you whiff on your ace and it goes to the dealer you are a huge underdog.

it seems dealer upcard is ace is 19% underdog for player

suppose its 1/3 chance

so the two hand have to deduct 6% advantage (each 3%) , right ?

 

[Elhombre]

1D-2D Ace track doesn't require mnemonic assist. zg

grif dont be so lazy.
A young 20 year old Chinese Student won the worldmemorychampionchips
he memorised a deck of 52 cards in 24,22 seconds.
This month in Guangzhou China www. worldmemorychampinchips. com

Ben Pridmore UK memorised 28 decks of cards in 1 hour.

the best E.H.

 

[zengrifter]

grif dont be so lazy.
A young 20 year old Chinese Student won the worldmemorychampionchips
he memorised a deck of 52 cards in 24,22 seconds.
This month in Guangzhou China www. worldmemorychampinchips. com

Ben Pridmore UK memorised 28 decks of cards in 1 hour.

I couldn't remember where my car keys were today. zg

 

[zengrifter]

it seems dealer upcard is ace is 19% underdog for player

suppose its 1/3 chance

so the two hand have to deduct 6% advantage (each 3%) , right ?

3% works. I'm glad I decided to only make my counting max bets on those. zg

 

[aslan]

grif dont be so lazy.
A young 20 year old Chinese Student won the worldmemorychampionchips
he memorised a deck of 52 cards in 24,22 seconds.
This month in Guangzhou China www. worldmemorychampinchips. com

Ben Pridmore UK memorised 28 decks of cards in 1 hour.

the best E.H.

Mnemonically, it's not that hard if you're good at making ditties.

Much easier is to write down a 500+ digit number and then recite it by heart without looking. Most anyone can do this if they can reduce their favorite song to numbers mnemonically.

 

[apex]

Math - 2D simple keycard approach, no mneumonic assit >

1. I have a 1 in 2 chance of spotting correct keycard 50% edge/2 = 25%

2. A 1 in 6 chance of getting (extra) Ace one of my two hands.

3. 50%/2/6 = 4% approx advantage per two hands.

Did I miss anything? zg

Another thing to consider when calculating your edge, is that when you don't get the ace you get a different card. If getting an Ace first gives you a 50% advantage, catching a 6 first has to be a significant disadvantage. As an example, here is a statement that can't be true:

1.) 1 in 2 correct keycard

2.) 1 in 13 catch an ace (chance if not tracking at all in a complete deck or decks)

3.) 50%/2/13 = approx 2% advantage

I might be thinking about this wrong, but it seems you need to somehow account for the other possible cards.

 

[zengrifter]

I hope I haven't been over-betting! zg

 

[blackriver]

Another thing to consider when calculating your edge, is that when you don't get the ace you get a different card. If getting an Ace first gives you a 50% advantage, catching a 6 first has to be a significant disadvantage. As an example, here is a statement that can't be true:

1.) 1 in 2 correct keycard

2.) 1 in 13 catch an ace (chance if not tracking at all in a complete deck or decks)

3.) 50%/2/13 = approx 2% advantage

I might be thinking about this wrong, but it seems you need to somehow account for the other possible cards.

i dont understand why you would devide by 13

it mightve been 50%/2 + 50%/13
25% + 4% = 29%
but i this is obv wrong because the perk of catching a random ace is almost exactly negated by the cost of all the other small cards that hurt

simplified, iits just ~50%/2 because if u dont get THAT ace then your just where you started (slightly worse for removal of an ace. w/o looking it up this is probably worth ~1%)

 

[blackriver]

if a 1 card key is asking for trouble

and a 2 card key is hard to remember try just remembering the colors whichll either be black-black or red-red half the time and the other half the time just remember the second card's suit and the other one will be narrowed to the opposite color. if you can rememebr a 2(~1.5) card key you remove having to divide your edge by (a full) 2 for the event that you observe the wrong key.

if you do a real 2 card key (rank and suits) this also helps you estimate how tight the shuffle is. if you see an ace followed by Ks and 9d, in the next show where you see a 9d/ks* followed immediately by the other then the chances of all 3 being clumped goes up. if the keys are 4 cards appart then its probably still worth at least doubling ur bet, but deff less than u would if the keys came out back to back

*depends on how they went to the discard tray

i think your effectiveness depends on how well they riffle. if it was blackchipjim's dealer who just strips the deck then no biggy, but if its a tight riffle then this may be worthless.

how many keys are you trying to remember per shoe? i would try to just get 1 or 2 keys that followed 2+ aces or an ace with faces after it. dont waste ur capacity on aces followed by a bunch of small cards. theyd be worth using but not if it costs doing 1 or 2 keys right

(sorry if any/all of this is obv. thinkin out loud for myslf too)

 

[zengrifter]

i dont understand why you would devide by 13

it mightve been 50%/2 + 50%/13
25% + 4% = 29%
but i this is obv wrong because the perk of catching a random ace is almost exactly negated by the cost of all the other small cards that hurt

simplified, iits just ~50%/2 because if u dont get THAT ace then your just where you started (slightly worse for removal of an ace. w/o looking it up this is probably worth ~1%)

Thank you! zg

 

[zengrifter]

and a 2 card key is hard to remember try just remembering the colors whichll either be black-black or red-red half the time and the other half the time just remember the second card's suit and the other one will be narrowed to the opposite color. if you can rememebr a 2(~1.5) card key you remove having to divide your edge by (a full) 2 for the event that you observe the wrong key.

if you do a real 2 card key (rank and suits) this also helps you estimate how tight the shuffle is. if you see an ace followed by Ks and 9d, in the next show where you see a 9d/ks* followed immediately by the other then the chances of all 3 being clumped goes up. if the keys are 4 cards appart then its probably still worth at least doubling ur bet, but deff less than u would if the keys came out back to back

*depends on how they went to the discard tray

i think your effectiveness depends on how well they riffle. if it was blackchipjim's dealer who just strips the deck then no biggy, but if its a tight riffle then this may be worthless.

how many keys are you trying to remember per shoe? i would try to just get 1 or 2 keys that followed 2+ aces or an ace with faces after it. dont waste ur capacity on aces followed by a bunch of small cards. theyd be worth using but not if it costs doing 1 or 2 keys right

(sorry if any/all of this is obv. thinkin out loud for myslf too)

Excellent! zg

 

[psyduck]

If aces are so important

that you want to sequence them, why not assign a more negative value to aces and count them?