Aquí están todos los comentarios publicados en el sitio, con los debates más recientes en primer lugar.
Para participar en cualquiera de estos debates, puede responder en la página del artículo.
Most of the time i loss when the count is plus and im winning the count is negative count.In my experience to play blackjack for many years im broke still and this game unbeatable especially the dealer they so lucky.
I’m in the process of learning the ‘Illustrious 18,’ and I recognize your situations above. My question has to do with multiple-card 16s. Let’s say the TC = -1 and I’m dealt 8, 5 v. dealer 10. I hit and receive a 3. So now I have 3 cards totaling 16 instead of a stiff 16. I’ve read elsewhere (WizardofVegas) that the odds every so slightly favor standing on a 3-card 16 rather than hit. After all, I’ve already received one little card from the deck. Does my chance of busting increase even just a little bit? Also, at how many cards what would you stand regardless of count? Every once in a while I’ll draw a 5-card 16 and I always stand, ’cause I think there’s no way I’m going to receive four consecutive little cards. Thoughts?
If you are using card counting instead of basic strategy, just ignore that “Stand with 3 or more card 16 vs ten” idea. Use the current count to decide whether to hit or stand, regardless of how many cards are in your hand.
No matter how many cards are in your 16, you should still hit it if the count is negative. Think about it this way – even though your hand already has several low cards in it, your current running count is telling you there are still more in the deck.
There is nothing special about the next card out of the shoe after you have been dealt four consecutive low cards. Your running count and true count will tell you whether the next card is more likely to be small or big, simply because it reflects the average composition of the remaining deck.
En primer lugar, ¿tiene los números para el 99,7% del tiempo para 3 desviaciones estándar, sólo por curiosidad?
Lo que realmente me interesa para esta lección es cómo gestionarías las sesiones múltiples. Como en la práctica tendremos que descansar para dormir, una sesión de 90 horas no es realista, así que utilicemos las sesiones de 3 horas como ejemplo. Si ahorro $3000 y voy al casino según las reglas de este ejemplo y después de 3 horas soy uno de los desafortunados que ha perdido $316, entonces abandono por esta noche y vuelvo mañana. ¿Recalculas tu apuesta máxima y tus spreads basándote en un bankroll de $2684?
Para ir más lejos, podría suponer que la sesión de 90 horas no tiene por qué ser continua, sino que podría pensarse en 30 sesiones de 3 horas durante las cuales consideras mentalmente que estás en una sesión larga "con descansos". A las 90 horas, ¿volverías a calcular tu apuesta máxima y tus spreads para un bankroll de $1825? Como puedes ver, esto podría expandirse hasta el infinito (si estás abajo $1175 después de 90 horas, sólo considéralo un punto bajo en una sesión de 1000 horas que podrías terminar por delante al final, así que sigue pagando en tus cálculos iniciales para un bankroll de $3000).
Sé que nunca debes limitar tus ganancias, así que subir esto no es un problema y tiene sentido calcular las ganancias de tu última sesión para optimizar cada nueva, pero cuando vas en negativo hay un fondo finito. Si estoy ahorrando $3000 para empezar y estoy bajando, ¿hay un punto de "abandono"? ¿Tengo que seguir ahorrando mis pagas para seguir subiendo a 3000 hasta que finalmente consiga algunas sesiones en positivo?
Corrí los números de la desviación estándar de 3 para usted, utilizando los mismos criterios que la lección.
Los resultados:
3 horas: +$530 a -$485
12 horas: +$1104 a -$924
48 horas: +$2388 a -$1668
90 horas: +$3452 a -$2102
Sin embargo, recientemente descubrí que los números del GameMaster se obtuvieron de una forma diferente a la que yo había supuesto, y desafortunadamente ahora necesitan una advertencia... Él calculó la varianza aquí utilizando sólo el tamaño medio de la apuesta y la fórmula para SD. Esto hace que sus resultados subestimen la volatilidad de este esquema de apuestas, porque hay más varianza en una apuesta que promedia $12 que en una apuesta directa de $12. En consecuencia, esta página necesita una revisión a fondo. Está en mi lista.
En cuanto a cómo tratar una cantidad prolongada de juego, si estás empezando con un bankroll pequeño rara vez tienes la capacidad de reducir el tamaño de tus apuestas si estás perdiendo, porque el juego no sigue siendo rentable con spreads más pequeños y es probable que ya estés en los límites de tu capacidad de spread debido al pequeño bank. Para la mayoría de los jugadores principiantes que estarían dispuestos a aumentar otro banco si pierden este, cambiar el tamaño no es una opción realista. En su lugar, su única opción real es seguir jugando mientras puedan permitirse el margen de apuesta y cubrir con seguridad cualquier doble o split que surja.
(Las cosas son muy diferentes si se trata de grandes cantidades de fondos. Entonces tienes cierta flexibilidad para redimensionar según sea necesario para reducir el riesgo de ruina).
I see that the majority of this thread is very old, and it’s been slightly hijacked, but it does help segway into a thought I’ve been having.
If you put everything else aside and look at only the order of the cards coming out of the deck, it seems there should be a point at which you should deviate from basic strategy regardless of the true count. the reason for this thought is basic probability.
Lets start with a dice example: rolling a single dice one time, the odds of getting a 6 are 1 in 6, or .1666. roll a single dice again, the odds of getting a 6 are still one in 6 cause the first roll has no effect (or no memory). that’s a basic statistic, but when you look at the odds of getting two consecutive 6’s, now it’s a PROBABILITY problem. the odds are .02777, which is a massive difference.
now translating this to blackjack, I’m thinking that at the basic level we’re looking at the card count. we’d drawing positive, negative, and neutrals. in a deck, we have 20, 20, and 12 respectively. so drawing a positive card is a 5 in 13 chance, or .38% a second positive is a .37%, then .36% and so on. Unlike the dice, there’s a memory, so each card drawn effects the odds of the second card. The tricky part is when we look at the probability of drawing 3 consecutive positives, which is a .05 chance.
So the pattern that we see is that each single card changes the numbers for the next draw by about .01% chance, which is pretty small and about inconsequential in comparison to the effect of the probability of an individual sequence. So how does this effect the game when we put everything into account and try to use this information in a game.
First off the running or true count would have an inconsequential effect at the beginning of each hand for the purposes of the probability of drawing a positive or negative card. As we saw, each single card removed will only change the probably by about .01% and we can expect that percentage to be roughly the same regardless of the number of decks. so if we use a hand as an example with 4 players where you’re on the end with a 12 against the dealers 10, then basic strategy says “hit till 17 or better” and there’s no variation on that in the I18 fab4 or otherwise. but what if the other 3 players before you all hit at least once and get a positive count card every time? To me that says that your odds of busting are extroadinarily high, since you only 3 faces that will require a 2nd hit, and if you DO draw one of those, that’s going to be the 4th positive card in a row, and makes a 5th positive card a .006% chance. on that 2nd hit, your odds of drawing a card that won’t cause a bust is even less than that cause that math doesn’t even account for getting a 4 followed by a 6 on the first and second hit respectively. I’ve probably already talked too much math to keep anybodies attention and haven’t even mentioned odds of getting a first hit card that would make you stay/bust but I think I’ve made the point that while basic strategy just says “hit till 17 or better” if you look at the flow of the cards, it would appear that a stay would be a better play.
so the point of the long story is a question: Am I wrong about something here? my thought is that this type of probability is ignored when counting cause there hasn’t been an easy way to boil it down into something easy to remember/implement at the tables. Am I anywhere close to right?
There is another explanation from Wizard of odds,FAQ , about myth of poor player made you lose money in BJ.
This author simulates 1.5 b hands of plays. One player always played basic strategy ( A), and the other player (B) always played a different strategy, different from the basic. The end result were the A player lost 0. 28% and the b player lost 11.% after 1.5 B hands. It’s doesn’t Mather how the other play, the result is the same in the long run.
that makes sense, I guess I should have pointed out that my point wasn’t that a poor player will make you lose, but that other players at the table receiving cards will give insight it to what could potentially be coming out of the deck.
I was persuing another avenue of thought from all of the poor player myths such as taking the dealers bust card and whatnot. Just simply the effect that multiple players can have on your play in terms of opportunities and insight vs one on one with the dealer.
Gracias por el sitio web y los consejos gratuitos. Me doy cuenta de que te contradices en algunos lugares y que tu tarjeta de estrategia básica no se alinea con lo que predicas en las lecciones. ¿Podría comprobarlo y volver a alinearlo?
Gracias
People like Tomi will always blame others for his poor decisions. If the person is to blame for the losses you earn, do yo thai them each time you hit a blackjack or you win? I doubt it. If he caused you to lose, then defacto he changed the card order and caused you to win as well!!
That is only true in this situation (with two 6’s and two 10’s. In any other case it would depend on the number of cards left and the values of those cards that will help or hurt you
I think you are misunderstanding how to calculate the true count. Just divide the running count by the number of decks that you haven’t seen. It does not matter whether the dealer ever uses those cards or not. They are still unseen cards that should be used in the calculation. For example, if you are one half deck in to the 2-deck shoe, there are still 1.5 decks left. A running count of +3 would translate to a true count of +2. (3/1.5 = 2).
Most of the time i loss when the count is plus and im winning the count is negative count.In my experience to play blackjack for many years im broke still and this game unbeatable especially the dealer they so lucky.
Hola Ken,
I’m in the process of learning the ‘Illustrious 18,’ and I recognize your situations above. My question has to do with multiple-card 16s. Let’s say the TC = -1 and I’m dealt 8, 5 v. dealer 10. I hit and receive a 3. So now I have 3 cards totaling 16 instead of a stiff 16. I’ve read elsewhere (WizardofVegas) that the odds every so slightly favor standing on a 3-card 16 rather than hit. After all, I’ve already received one little card from the deck. Does my chance of busting increase even just a little bit? Also, at how many cards what would you stand regardless of count? Every once in a while I’ll draw a 5-card 16 and I always stand, ’cause I think there’s no way I’m going to receive four consecutive little cards. Thoughts?
If you are using card counting instead of basic strategy, just ignore that “Stand with 3 or more card 16 vs ten” idea. Use the current count to decide whether to hit or stand, regardless of how many cards are in your hand.
No matter how many cards are in your 16, you should still hit it if the count is negative. Think about it this way – even though your hand already has several low cards in it, your current running count is telling you there are still more in the deck.
There is nothing special about the next card out of the shoe after you have been dealt four consecutive low cards. Your running count and true count will tell you whether the next card is more likely to be small or big, simply because it reflects the average composition of the remaining deck.
En primer lugar, ¿tiene los números para el 99,7% del tiempo para 3 desviaciones estándar, sólo por curiosidad?
Lo que realmente me interesa para esta lección es cómo gestionarías las sesiones múltiples. Como en la práctica tendremos que descansar para dormir, una sesión de 90 horas no es realista, así que utilicemos las sesiones de 3 horas como ejemplo. Si ahorro $3000 y voy al casino según las reglas de este ejemplo y después de 3 horas soy uno de los desafortunados que ha perdido $316, entonces abandono por esta noche y vuelvo mañana. ¿Recalculas tu apuesta máxima y tus spreads basándote en un bankroll de $2684?
Para ir más lejos, podría suponer que la sesión de 90 horas no tiene por qué ser continua, sino que podría pensarse en 30 sesiones de 3 horas durante las cuales consideras mentalmente que estás en una sesión larga "con descansos". A las 90 horas, ¿volverías a calcular tu apuesta máxima y tus spreads para un bankroll de $1825? Como puedes ver, esto podría expandirse hasta el infinito (si estás abajo $1175 después de 90 horas, sólo considéralo un punto bajo en una sesión de 1000 horas que podrías terminar por delante al final, así que sigue pagando en tus cálculos iniciales para un bankroll de $3000).
Sé que nunca debes limitar tus ganancias, así que subir esto no es un problema y tiene sentido calcular las ganancias de tu última sesión para optimizar cada nueva, pero cuando vas en negativo hay un fondo finito. Si estoy ahorrando $3000 para empezar y estoy bajando, ¿hay un punto de "abandono"? ¿Tengo que seguir ahorrando mis pagas para seguir subiendo a 3000 hasta que finalmente consiga algunas sesiones en positivo?
Corrí los números de la desviación estándar de 3 para usted, utilizando los mismos criterios que la lección.
Los resultados:
3 horas: +$530 a -$485
12 horas: +$1104 a -$924
48 horas: +$2388 a -$1668
90 horas: +$3452 a -$2102
Sin embargo, recientemente descubrí que los números del GameMaster se obtuvieron de una forma diferente a la que yo había supuesto, y desafortunadamente ahora necesitan una advertencia... Él calculó la varianza aquí utilizando sólo el tamaño medio de la apuesta y la fórmula para SD. Esto hace que sus resultados subestimen la volatilidad de este esquema de apuestas, porque hay más varianza en una apuesta que promedia $12 que en una apuesta directa de $12. En consecuencia, esta página necesita una revisión a fondo. Está en mi lista.
En cuanto a cómo tratar una cantidad prolongada de juego, si estás empezando con un bankroll pequeño rara vez tienes la capacidad de reducir el tamaño de tus apuestas si estás perdiendo, porque el juego no sigue siendo rentable con spreads más pequeños y es probable que ya estés en los límites de tu capacidad de spread debido al pequeño bank. Para la mayoría de los jugadores principiantes que estarían dispuestos a aumentar otro banco si pierden este, cambiar el tamaño no es una opción realista. En su lugar, su única opción real es seguir jugando mientras puedan permitirse el margen de apuesta y cubrir con seguridad cualquier doble o split que surja.
(Las cosas son muy diferentes si se trata de grandes cantidades de fondos. Entonces tienes cierta flexibilidad para redimensionar según sea necesario para reducir el riesgo de ruina).
I see that the majority of this thread is very old, and it’s been slightly hijacked, but it does help segway into a thought I’ve been having.
If you put everything else aside and look at only the order of the cards coming out of the deck, it seems there should be a point at which you should deviate from basic strategy regardless of the true count. the reason for this thought is basic probability.
Lets start with a dice example: rolling a single dice one time, the odds of getting a 6 are 1 in 6, or .1666. roll a single dice again, the odds of getting a 6 are still one in 6 cause the first roll has no effect (or no memory). that’s a basic statistic, but when you look at the odds of getting two consecutive 6’s, now it’s a PROBABILITY problem. the odds are .02777, which is a massive difference.
now translating this to blackjack, I’m thinking that at the basic level we’re looking at the card count. we’d drawing positive, negative, and neutrals. in a deck, we have 20, 20, and 12 respectively. so drawing a positive card is a 5 in 13 chance, or .38% a second positive is a .37%, then .36% and so on. Unlike the dice, there’s a memory, so each card drawn effects the odds of the second card. The tricky part is when we look at the probability of drawing 3 consecutive positives, which is a .05 chance.
So the pattern that we see is that each single card changes the numbers for the next draw by about .01% chance, which is pretty small and about inconsequential in comparison to the effect of the probability of an individual sequence. So how does this effect the game when we put everything into account and try to use this information in a game.
First off the running or true count would have an inconsequential effect at the beginning of each hand for the purposes of the probability of drawing a positive or negative card. As we saw, each single card removed will only change the probably by about .01% and we can expect that percentage to be roughly the same regardless of the number of decks. so if we use a hand as an example with 4 players where you’re on the end with a 12 against the dealers 10, then basic strategy says “hit till 17 or better” and there’s no variation on that in the I18 fab4 or otherwise. but what if the other 3 players before you all hit at least once and get a positive count card every time? To me that says that your odds of busting are extroadinarily high, since you only 3 faces that will require a 2nd hit, and if you DO draw one of those, that’s going to be the 4th positive card in a row, and makes a 5th positive card a .006% chance. on that 2nd hit, your odds of drawing a card that won’t cause a bust is even less than that cause that math doesn’t even account for getting a 4 followed by a 6 on the first and second hit respectively. I’ve probably already talked too much math to keep anybodies attention and haven’t even mentioned odds of getting a first hit card that would make you stay/bust but I think I’ve made the point that while basic strategy just says “hit till 17 or better” if you look at the flow of the cards, it would appear that a stay would be a better play.
so the point of the long story is a question: Am I wrong about something here? my thought is that this type of probability is ignored when counting cause there hasn’t been an easy way to boil it down into something easy to remember/implement at the tables. Am I anywhere close to right?
There is another explanation from Wizard of odds,FAQ , about myth of poor player made you lose money in BJ.
This author simulates 1.5 b hands of plays. One player always played basic strategy ( A), and the other player (B) always played a different strategy, different from the basic. The end result were the A player lost 0. 28% and the b player lost 11.% after 1.5 B hands. It’s doesn’t Mather how the other play, the result is the same in the long run.
that makes sense, I guess I should have pointed out that my point wasn’t that a poor player will make you lose, but that other players at the table receiving cards will give insight it to what could potentially be coming out of the deck.
I was persuing another avenue of thought from all of the poor player myths such as taking the dealers bust card and whatnot. Just simply the effect that multiple players can have on your play in terms of opportunities and insight vs one on one with the dealer.
so is this the way you should bet when you are counting cards?
Yes, this lesson shows a good way of calculating an appropriate bet spread for counting.
Gracias por el sitio web y los consejos gratuitos. Me doy cuenta de que te contradices en algunos lugares y que tu tarjeta de estrategia básica no se alinea con lo que predicas en las lecciones. ¿Podría comprobarlo y volver a alinearlo?
Gracias
¿Tiene algún punto concreto en el que cree que hay un problema?
how can you calculate DD BJ T/C positive or negative count is only few cards to deal.thanks
People like Tomi will always blame others for his poor decisions. If the person is to blame for the losses you earn, do yo thai them each time you hit a blackjack or you win? I doubt it. If he caused you to lose, then defacto he changed the card order and caused you to win as well!!
I agree with the person before you , I’ve seen many times when a player takes a card while the dealer has a bust card out and screws the whole table
It seems you didn’t understand any of the article. Oh well, you can lead a horse to water…
Fine. I recommend you go play your one hand and then give up the game. 🙂
That is only true in this situation (with two 6’s and two 10’s. In any other case it would depend on the number of cards left and the values of those cards that will help or hurt you
But he is not playing millions of hands..it’s ONE hand that Clueless has hurt you and that ONE hand can cost you
I think you are misunderstanding how to calculate the true count. Just divide the running count by the number of decks that you haven’t seen. It does not matter whether the dealer ever uses those cards or not. They are still unseen cards that should be used in the calculation. For example, if you are one half deck in to the 2-deck shoe, there are still 1.5 decks left. A running count of +3 would translate to a true count of +2. (3/1.5 = 2).