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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.
First, do you have the numbers for the 99.7% of the time for 3 standard deviations, just out of curiosity?
What I’m really curious about for this lesson is how you would manage multiple sessions. Since in practice we’ll have to break for sleep, a 90 hour session isn’t realistic, so let’s use the 3 hours sessions as an example. If I save up $3000 and hit the casino per the rules in this example and after 3 hours I’m one of the unlucky ones that’s down $316, then I quit for the night and come back tomorrow. Do you recalculate your max bet and spreads based on a bankroll of $2684?
To go further, I could surmise that the 90 hour session doesn’t have to be continuous but could be thought of as 30, 3 hour sittings during which you mentally consider it that you’re on one long session “with breaks.” At 90 hours would you then recalculate your max bet and spreads for a bankroll of $1825? As you can see this could be able to expand to infinity (if you’re down $1175 after 90 hours, just consider it a low point on a 1000 hour session that you could end up ahead at the end so just keep paying in your initial calculations for a $3000 bankroll).
I know you should never limit your wins, so going up this isn’t a problem and it makes sense to figure in winnings from your last session to optimize each new one, but when you’re going negative there’s a finite bottom. If I’m saving $3000 to start out and I’m going down, is there a “walk away” point? Do I have to keep saving my paychecks to keep coming back up to 3000 until I finally get some sessions in the positive?
I ran the numbers of the 3 standard deviation for you, using the same criteria as the lesson. The results: 3 hours: +$530 to -$485 12 hours: +$1104 to -$924 48 hours: +$2388 to -$1668 90 hours: +$3452 to -$2102
However, I recently discovered that the GameMaster’s numbers were derived in a different way than I had assumed, and unfortunately they now need a caveat… He calculated the variance here by just using the average bet size and the formula for SD. This causes his results to underestimate the volatility of this betting scheme, because there is more variance in a bet spread that averages $12 than in a straight flat bet of $12. This page really needs a thorough reworking as a result. It’s on my list.
As for how to treat an extended amount of play, if you are starting with a small bankroll you rarely have the ability to scale down your bet sizes if you are losing, because the game doesn’t stay profitable with smaller spreads and you are likely already at the limits of your ability to spread because of the small bank. For most beginning players who would be willing to raise another bank if they lose this one, resizing isn’t a realistic option. Instead there only real option is to keep playing as long as they can afford the bet spread and safely cover any doubles and splits that arise.
(Things look quite a bit different if you are dealing with larger bankroll numbers. Then you have some flexibility to resize as needed to reduce the risk of ruin.)
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.
Thanks for the website and the free advise. I do notice that you contradict yourself in some places and that your basic strategy card doesn’t align with what you are preaching in the lessons. Can you double check, and realign? Thanks
Just like in any number of decks. You divide the running count by the number of unseen decks. Let’s say you are playing a deeply dealt double deck game, and 1.5 decks have been used already.
If your running count is +3, you divide that by the number of unseen decks, which is 0.5. +3 / 0.5 = +6. Your true count is +6.
Sucks that this is the trend in Vegas. I stopped playing at the Venetian in LV about a year ago because of all the bs 6:5 games. Recently, I saw the NY NY just started doing this on their shoe games. I hope all blackjack players boycott these games so the greedy casinos are forced to go back to 3:2!
Is it true you guys are still getting shoe boxes dealt? We are from Asia and there is none to be found. It is all Continious Shuffling machines we are playing against. Can never be the same and cannot be counted anymore. Drop me a line if anyone knows how to beat these CSM. poiandrew at yahoo dot com
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
Hi 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.
First, do you have the numbers for the 99.7% of the time for 3 standard deviations, just out of curiosity?
What I’m really curious about for this lesson is how you would manage multiple sessions. Since in practice we’ll have to break for sleep, a 90 hour session isn’t realistic, so let’s use the 3 hours sessions as an example. If I save up $3000 and hit the casino per the rules in this example and after 3 hours I’m one of the unlucky ones that’s down $316, then I quit for the night and come back tomorrow. Do you recalculate your max bet and spreads based on a bankroll of $2684?
To go further, I could surmise that the 90 hour session doesn’t have to be continuous but could be thought of as 30, 3 hour sittings during which you mentally consider it that you’re on one long session “with breaks.” At 90 hours would you then recalculate your max bet and spreads for a bankroll of $1825? As you can see this could be able to expand to infinity (if you’re down $1175 after 90 hours, just consider it a low point on a 1000 hour session that you could end up ahead at the end so just keep paying in your initial calculations for a $3000 bankroll).
I know you should never limit your wins, so going up this isn’t a problem and it makes sense to figure in winnings from your last session to optimize each new one, but when you’re going negative there’s a finite bottom. If I’m saving $3000 to start out and I’m going down, is there a “walk away” point? Do I have to keep saving my paychecks to keep coming back up to 3000 until I finally get some sessions in the positive?
I ran the numbers of the 3 standard deviation for you, using the same criteria as the lesson.
The results:
3 hours: +$530 to -$485
12 hours: +$1104 to -$924
48 hours: +$2388 to -$1668
90 hours: +$3452 to -$2102
However, I recently discovered that the GameMaster’s numbers were derived in a different way than I had assumed, and unfortunately they now need a caveat… He calculated the variance here by just using the average bet size and the formula for SD. This causes his results to underestimate the volatility of this betting scheme, because there is more variance in a bet spread that averages $12 than in a straight flat bet of $12. This page really needs a thorough reworking as a result. It’s on my list.
As for how to treat an extended amount of play, if you are starting with a small bankroll you rarely have the ability to scale down your bet sizes if you are losing, because the game doesn’t stay profitable with smaller spreads and you are likely already at the limits of your ability to spread because of the small bank. For most beginning players who would be willing to raise another bank if they lose this one, resizing isn’t a realistic option. Instead there only real option is to keep playing as long as they can afford the bet spread and safely cover any doubles and splits that arise.
(Things look quite a bit different if you are dealing with larger bankroll numbers. Then you have some flexibility to resize as needed to reduce the risk of ruin.)
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.
Thanks for the website and the free advise. I do notice that you contradict yourself in some places and that your basic strategy card doesn’t align with what you are preaching in the lessons. Can you double check, and realign?
Thanks
Do you have a specific spot where you think there is an issue?
how can you calculate DD BJ T/C positive or negative count is only few cards to deal.thanks
Just like in any number of decks. You divide the running count by the number of unseen decks.
Let’s say you are playing a deeply dealt double deck game, and 1.5 decks have been used already.
If your running count is +3, you divide that by the number of unseen decks, which is 0.5.
+3 / 0.5 = +6.
Your true count is +6.
Sucks that this is the trend in Vegas. I stopped playing at the Venetian in LV about a year ago because of all the bs 6:5 games. Recently, I saw the NY NY just started doing this on their shoe games. I hope all blackjack players boycott these games so the greedy casinos are forced to go back to 3:2!
Is it true you guys are still getting shoe boxes dealt? We are from Asia and there is none to be found. It is all Continious Shuffling machines we are playing against. Can never be the same and cannot be counted anymore. Drop me a line if anyone knows how to beat these CSM. poiandrew at yahoo dot com
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