math about aces sequencers

if i play shoe blackjack with 8 decks with a cut of 50 cards and i remember 20 key cards .i know the shuffle . of this 20 we lost maybe 5 and the others 15 key card and ace are in a range of 1 to 4 card distance. what is the ecuation to do and with how many players. i can play stand only the hand i see the key card and bet to 4 players. can anyone help me with the ecuation and edge? thanks i will be waiting the answer. thanks again.

Quote: nicoramone

if i play shoe blackjack with 8 decks with a cut of 50 cards and i remember 20 key cards .i know the shuffle . of this 20 we lost maybe 5 and the others 15 key card and ace are in a range of 1 to 4 card distance.

Obviously, English is not your first language. Just curious, what do you mean when you say "key card". Does that mean a Ten?

When you say "15 key card and ace are in a range of 1 to 4 card distance," how many aces are you referring to?

Quote: gordonm888

Obviously, English is not your first language. Just curious, what do you mean when you say "key card". Does that mean a Ten?

When you say "15 key card and ace are in a range of 1 to 4 card distance," how many aces are you referring to?

Hi gordonm888,

I don't think ("key card" mean a Ten).

"key card" is use for ACE-Tracking/ACE-Location, in AP Blackjack.

I use to do a lot of ACE-Tracking some 30 Years ago and make a lot of money from it, before I become a Games Inventor.

P.S. Back then there are only a few people know about it.

sorry for my english spanish is my first languague.
key card is the card that will be up the ace in the dischard . i am refering that 15 aces are in difference with it key card in a range of 1 to 4 cards distance after the shuffle . key card can be a ten a 3 a 4 . is the card up the ace to remember when is comming the ace in the next shoe after the shuffle.

so can you help me with the question please. or send me pm . thanks and yes key card is that and is for tracking aces .

No comprendo nada.
Preguntalo en español en PM y voy a probar traducirlo, si comprendo.

ok. te escribo en castellaño por privado a vos? gracias

ahi lo puse en español.
la idea es secuenciar aces . se juega con zapato . penetracion zcla manual que conozco. 8 mazos . 32 ases. se ve una partida entera y se memorizan la carta que quedo sobre cada as en el descarte. luego hacen la mezcla y el corte. en el corte se pierden entre 4 y 7 ases. luego empieza la nueva partida quedando en juego 25/28 ases. con la mezcla se pierden otros 5 /8 ases . quedan en juego 20 . de esos 20 la distancia entre la carta llave q se memorizo y el as es entre 1 carta y 4/5 cartas de distancia.
uno puede jugar parado fuera de la mesa y apostar solo al ver las cartas llave a todos los jugadores que desee. el obtener un as como primera carta significa un 54% de ventaja. cual es la ecuacion para saber como es mejor jugarlo y de que ventaja hablamos . saludos.

Lo comprendo. Gracias.

bueno sigo a la espera de una ecuacion para ajustar el juego y ver q es lo mas conveniente y la ventaja de cuanto es en cada caso. saludos.

Quote: nicoramone

ahi lo puse en español.
la idea es secuenciar aces . se juega con zapato . penetracion zcla manual que conozco. 8 mazos . 32 ases. se ve una partida entera y se memorizan la carta que quedo sobre cada as en el descarte. luego hacen la mezcla y el corte. en el corte se pierden entre 4 y 7 ases. luego empieza la nueva partida quedando en juego 25/28 ases. con la mezcla se pierden otros 5 /8 ases . quedan en juego 20 . de esos 20 la distancia entre la carta llave q se memorizo y el as es entre 1 carta y 4/5 cartas de distancia.
uno puede jugar parado fuera de la mesa y apostar solo al ver las cartas llave a todos los jugadores que desee. el obtener un as como primera carta significa un 54% de ventaja. cual es la ecuacion para saber como es mejor jugarlo y de que ventaja hablamos . saludos.

Tentative translation.

« Thé idea is to follow aces. Based on the shoe, whose penetration ans manual shuffle are known. 8 decks, 32 aces. You observe a full game (shoe) and memorize which card ended up on top of each ace in the discard. After the shuffle, the cut will put away 4 to 7 aces. The new game begins with 25/28 aces active. With the shuffle some further 5 to 8 aces got lost. 20 remain in play.
For these 20, the distance between the ‘key’ memorized card and the corresponding ace will be between 1 and 4/5 cards dealt.
One can prepare aside of the table and wager only when a key card is dealt to any player you determine. Getting an ace as first card is a 54% advantage. What is the equation for knowing the best way to play that, and what is the expected advantage? Thank you. »

someone who can help me with the math ecuations for that?

Quote: nicoramone

so can you help me with the question please. or send me pm . thanks and yes key card is that and is for tracking aces .

Hi nicoramone,

Sorry, I don't give out my Trade Secrets.

P.S. I don't think there is many AP Players give out their Trade-Secrets here or any where that I know of.

BTW. ACE-Tracking is the most Powerful technique AP Blackjack I know of (House-Edge 51+%, if you Track it with High-cards), the second is Section-Tracking, and some others, and the last is Card-counting.

I know this, because I use them all, 30+ years ago (it lasted over 10+ years), and finally be told not to Play Blackjack All over the UK, but can still go in the Casinos.

After my AP Blackjack career. That is when I started my FIRST New Table Games (Casino Hold'em®) in 2000, The Number-1 Casino-poker play Online in 2,000+ Casinos.

The rest, shall we say is history.

hi i am nicolas from argentina and i am too an advantage player sometimes on teams sometimes alone.i play rulette ds and vb and rc . fortunes roulettes. black jack. craps.
i dont want any secret i count and shuffle track but i know too that ace sequencer is the best. i know the technique i only want a math ecuation to make it best only that .
i am not telling you to tell me how to play i only want help to put that on math ecuations to know with how many players to play or to play stand only betting to others. thanks for your answer. if you know aboit that you know that all before i write is correct . thanks

waiting for an answer or is not a math thread?

Quote: nicoramone

waiting for an answer or is not a math thread?

Most of the Math guys here don't know what ACE-Tracking is, and the ones that know ACE-Tracking (if any) are not Math-guys and properly don't want to share what they know.

ok i think the best is to eliminate the post. nobody will answer . i tink that was a advantage player forum but maybe i was wrong. if somwbody ask me about ballistic i answer to help but here i think is different. thanks anyway.

It is best to figure it out by yourself. No one will share secrets on open forum. You expecting something for nothing, that does not work.

Quote: DeMango

It is best to figure it out by yourself. No one will share secrets on open forum. You expecting something for nothing, that does not work.

He just wanted the math formula. I think the answer is you have to sim it though.

i dont want any secret. i want a math equation . that does not work? jajajajjajajajjajanjanajabanjajajajajjajajja

How do you want to establish an equation when we don’t have any parameters on the shuffle effect?

Quote: nicoramone

waiting for an answer or is not a math thread?

Welcome to the forum. It is a math based forum. But you are asking a complicated question. There may be someone who has the requisite math skills (not me!) who will do the work for you. But I think your exact question is not clear enough.
It is easy enough to understand that you are saying that after a certain card is seen there is an increased chance the next card is an ace. For anyone to answer, based on your experience, what is the chance the ace will follow the tracked card? 1/13 would of course be without tracking. What is it with your tracking skills?

By the way, it is possible that there are members here that make their living by betting this way. And they would be unlikely to let anyone know this. They will probably be upset that this is even being discussed on this forum. Some perceive that any advantage play openly discussed here immediately makes it directly to casino management.

sorry if this is not an advantage forum for advantage players i will eliminate all now. if this thread is in the wrong place please delete it . not problem. the shuffle is not the problem i want an ecuation for the values i give before . only that. for moderators if the post is incorrect or is not the place to talk about advantage delete it.

the equation is without the shuffle . that i indicate are final numbers aproximated. i want a math ecuation for play it better and to know the max and the min of edge. and the better numbers of players on table to play.

Let's assume the question is an interesting mathematical puzzle and the answer depends on the number of decks in use. For simplicity let's assume there are 4 decks, so there are 16 Aces in the shoe. As the cards are dealt you remember the card that goes next to each Ace and, if unshuffled, would come out just before the Ace next time. Let's assume one of these "key" cards is Kh. The shuffle isn't perfect so there's an enhanced chance when you see the same Kh coming out the next card will be an Ace. Similarly for 15 other cards (and if Aces come out together one of the key cards will be an Ace).

I assume the proposition is that if the last card of the previous hand was, in this example, Kh and you sit at the first hand, then you should increase your bet.

There are four Kh, so firstly three of these aren't the one you saw and the chance of an Ace is (presumably) about 1/13. The fourth one is and the chance of an Ace is larger than 1/13, but how likely is it the cards have stuck together? Let's say you're twice as likely to get an Ace. Then the total chance is 5/52 (=9.6%).

If the last card wasn't one of the "key cards" then chances of an Ace is less (about 6.8%) (the maths is assume on average it's 1/13 with no info, you've used up 80/2704 so have 128/2074 left among 36 cards).

So 16 times out of 52 you'd bet a bit bigger and 36 times you'd bet a bit smaller. I'll leave it to those who know whether that gives you an advantage (I'm guessing it does from the comments).

Note I've used 1/13 rather than using 16/207 for simplicity.

thank you. there are 8 decks . with a cut card at 75%. i write all the other. i can play in four places. and the shuffles in a high percentaje maintenance the key and ace not more of 4 cards between. obviously aces are not 32. with cut maybe 25/27. maybe 5/7 are lost more tahn 4 cards between . if i see last or lasts players key cards what is the best scenario to play . or play all the places with team? what is the minimum advantage and maximum . if we know that an ace at first card is 54% advantage. what edge we are talking about ? thank you very much for your answer.

The problem with your scenario is in an ideal case the key card is the last card used on the previous hand when, according to your assumptions, an Ace has 1/4 chance of being the next card out. If there are 25 key cards then about half the time you'll be assuming the next card is more likely to be an Ace. However only 1 time in 8 (since there are 8 Kh and only one of them is the real one that's next to an Ace) you'll have the advantage. (The logic is similar if you have duplicate key cards but you're better off as the chances of a real one is higher.) If you have a real key card, using your number, the advantage is 1/4*54% - 3/4*5% = 9.75%; if you don't then it's -0.5% (I'm making a simple assumption but technically the figure should be worse). 1 time out of 8 it's +9.75%, the other 7 it's -0.5% : so average = +0.8%.

Half the time you have a key card +0.8%, half the time you know nothing -0.5%. So it really depends on how much variance the casino will let you have and the values seem to be similar to counting (except in your case you get a small advantage 50% of the time).

It probably works with fewer decks, but 8 makes it difficult to know whether you have a real key card or not (there's 7 chances in 8 you're wrong).

i will delete the post . maybe the language maybe other thing . thanks to all .