My theory on multi hands while playing 6D

For instance, let's assume the game is 6D with 80% pen, we're 4 decks into the pack, the TC is +15 and only 12 aces have fallen from the pack. We have about 45 cards left 'till the cut card falls. Let's also assume in this situation that there is on average 16 play positions (including the D) worth of cards still left in the pack before the the cut card falls. Also, let's assume we are in a perfect world and the pit crew doesn't have any care at all about us being AP's or how many hands we play. We also have plenty of $ for a bk-roll. Now let's also assume we will be playing 1 on 1 with the D in any of the following manners while finishing out the shoe:

1) we can play one position with a total of eight hands to the D and eight hands to the player -or-

2) we can play three positions with a total of four hands to the D and twelve to the player -or-

3) we can play seven positions with a total of two hands to the D and fourteen to the player.

Which option would mathematically provide the highest return to the player? I assume that option #3 is best. More BJ's to the player???

you must play the table max bet on all positions. When you get a natural, you get paid 3/2 while the dealer gets paid 7 to one because he beats all positions.

It contains a discussion of card-eating while playing multiple hands. The results will surprise you. Once you've learned to count, that book contains the most useful blackjack information you'll find anywhere.

I think you'll get more money in action during the hot part of the shoe by playing one spot. I don't remember the co-variance for 3 hands and have never bothered to figure the co-variance on 7 hands, but off the top of my head, once you create bets of equivilent risk, you're better off, in this heads up configuration, to play 1 spot.

This is all very puzzling when I keep hearing things like this from veteran AP's:

I think you'll get more money in action during the hot part of the shoe by playing one spot.

This is why I want to bring this up for discussion.

I don't see how it can be this way as BP is saying. It doesn't make sense to me to allow the dealer to have any more hands than I can keep him from getting during a hot shoe situation.

I have read books that say to spread to two hands when we get to 2TC or better. If that is true, why not play as many hands as is allowed in order to get more of the "big card" hands such as shown in option 2 and 3 below:

1) we can play one position with a total of eight hands to the D and eight hands to the player -or-

2) we can play three positions with a total of four hands to the D and twelve to the player -or-

3) we can play seven positions with a total of only two hands to the D and fourteen to the player.

Do we really want to share so many aces and faces with the dealer while playing 1 on 1, one hand at a time? I can't make sense of that.

In option 3 above, we are cutting the dealer out of the loop as much as is possible. Makes sense to me. Comments please.

Stealth Bomber,

Assuming you know exactly when the cut card will appear, if the count is still high as you place your bets for the final round, you should bet all seven spots, regardless of how many spots you've been playing.

Here, the card-eating effect doesn't apply (because this is the last round), and as per your posing of the question, I'm assuming in the follwing discussion that your bankroll is so large that you needn't be concerned with the covariance of your hands.

In fact, you would probably be best served by varying the number of spots you're playing on the "penultimate round" (the round before the final round) to try to ensure that the cut card will fall early in the final round rather than late in the penultimate round.

For example, say you're choosing your bets at a time when the shoe contains 10 cards before the cut card. Now you have three options.

(1) You can make this round the final round by playing all 7 spots, which will get you 7 more high-count hands. If you play 4 or more spots, this will also be the last round (and will probably be the last round if you play 3), so if you're planning to play that many, you should just play all 7.

(2) You can play just 1 hand on this penultimate round, and then play all 7 spots on the final round. This will almost certainly get you a total of 8 high-count hands... "almost certainly" because the chance of two hands consuming 10 cards is rather small, particularly with a high count.

(3) Finally, you can play 2 spots on this penultimate round, so that between your two hands and the dealer's one hand, you'll use an average of 8.1 cards (2.7 per hand). Then, play all 7 spots for the final round. If it works, this option will get you 9 more high-count hands. However, you run a high risk of hitting the cut card on the penultimate round, in which case you'll have gotten only 2 high-count hands, rather than the certain 7 of option #1 or the very likely 8 of option #2.

I would lean towards option #2 in this case, because I believe (but have not checked!) that the odds of running out on the penultimate round are less than the breakeven odds of 1 in 8.

Here's a formula you can use to calculate "S", the number of spots to play on the penultimate round:

S = INT((n-min)/2.7) - 1

Here, "n" is the number of cards remaining in front of the cut card, "min" is the target (no, not Patterson's TARGET!) number of cards that you are planning to have in front of the cut card at the start of the final round, and "2.7" is the average number of cards used by each hand... note that at high counts, this value may differ, but I have not run any sims to see what it may be. Also, the "INT" function means to chop off the decimal value of the argument, so if ((n-min)/2.7) is equal to 4.7, when you take the INTEGER, you get 4.

If S is between 1 and 7 (inclusive), then play S spots on this round and play all 7 on the final round. If S is zero or negative, then this round will in fact be the final round, so bet all 7 spots. If S is greater than 7, then play all 7 spots, then recalculate S at the start of the next round to see whether it will be the final round or the penultimate round.

"min" is "how close do you want to cut it?". If the number of cards actually used in the round is equal to 2.7 times the number of hands, then you'll have "min" cards (with a little leeway for the rounding introduced by the INT function) before the cut card for the final round. "min" has to be a positive number, but the smaller you make it, the better the chance of running out of cards on what you had hoped would be the penultimate round.

Let's walk through an example to see how this all works. Say you have 22 cards before the cut card, and you're shooting for having 5 left there to start the next round, which means "min" = 5. Then on this penultimate round you should play

S = INT((22-5)/2.7) - 1 = INT(6.3) - 1 = 6 - 1 = 5

So play 5 spots this round, then play 7 on the final round.

As for how to select "min", this seems to be an optimization problem. My gut feeling (and all A.P.'s know the value of a gut feeling!) is that the larger the value of "n", the larger the value you should select for "min". Thus, for the 10 card example, choose "min" = 3 and play one hand. However, in the 22 card case, I chose a higher "min", since the player has more opportunities to split.

Dog Hand

is a way to get more money in action for a given risk criterion. If the assumption is made that the risk is nill as the bank is infinite for the purposes of such calculations, then it would be best to pile all of the money on a single spot. What possible argument could there be to the contrary?

In the real world two hands is almost always best, but you are talking about an imaginary world. Two hands gets a better DI and more money in action relative to risk than does one hand or three hands. More than two hands is only optimal when one is being constrained by table maxes and wants to get more money in action. With infinite table limit and infinite bank then one should bet an infinite amount on one hand when ever he has an edge.

Assuming you know exactly when the cut card will appear, if the count is still high as you place your bets for the final round, you should bet all seven spots, regardless of how many spots you've been playing.

Here, the card-eating effect doesn't apply (because this is the last round), and as per your posing of the question, I'm assuming in the follwing discussion that your bankroll is so large that you needn't be concerned with the covariance of your hands.

In fact, you would probably be best served by varying the number of spots you're playing on the "penultimate round" (the round before the final round) to try to ensure that the cut card will fall early in the final round rather than late in the penultimate round.

This is all very good and true. But it is aside from the exact point I'm trying to make.

I will try again in the form of a question: What is the benefit in allowing the D to have any hands more than the absolute minimum necessary during a very high count which includes extra aces more than the average amount?

why aren't more hands even better? Remember, the more cards we get as the player, the more aces we get and thus the less aces the dealer will eat during the high count.

Another way to look at it is: Let's just say there are 16 cards left to put into play. Of those 16 there are 12 faces and 4 aces. Which of the following 3 options would you choose for your playing proceedure?:

1) you can play one position with a total of eight hands to the D and eight hands to the player -or-

2) you can play three positions with a total of four hands to the D and twelve to the player -or-

3) you can play seven positions with a total of only two hands to the D and fourteen to the player.

Let's also assume we are in a very well banked real world with a 100K BK roll. The money must be managed and is not limitless.

why aren't more hands even better? Remember, the more cards we get as the player, the more aces we get and thus the less aces the dealer will eat during the high count.

The aces we get don't matter. Getting one bj for $200 is just as good as getting two of them for $100 each. It is the same return. Remember, you said that the bank is big and it doesn't matter. So I assume by this you mean that what ever amount you choose to play that you can pile it on one hand or spread it among as many as you wish. This being the case you'd like to get as many rounds as possible during the high count. The more hands you play the fewer rounds it takes you to deplete the rich count and thus the less action you get in.

Two hands is better than one when bankroll IS a consideration. You can put 50% more money in action within the same risk criterion. The desirability index is better. You also make better use of your time. But in terms of total return for the shoe, one hand or two make no difference when heads up. You put 50% more money in action but use up 50% more cards per round at the same time, so it is a wash.

Now, what I'm saying is that if neither table limit nor bankroll are a concern then pile it all up on one hand. Multiple hands only helps you when you increase your action.

If you bet as much as they allow you to bet on one (or two) hands and your bank allows you to bet more within your risk criterion then it is good to play as many hands as you can afford. But this is the special case of being constrained by table limits.

it goes without saying, if you know it is the last round you would of course play all the spots at the table if allowed. Playing one hand is for the purpose of getting additional rounds.

The aces we get don't matter. Getting one bj for $200 is just as good as getting two of them for $100 each.

because, we can lesson the variance by playing multiple hands while getting possibly even more money on the table. Therefore, if we are going to play one hand of $200 we might be mathematically better served by playing two hands of $150 or even maybe three hands of $125 or four hands of $110 and so on. Therefore, the aces are more valuable to us rather than letting the dealer have more than is necessary.

Another thought:

50% of all dealer BJ's will be with the ace as the whole card. That may present an additional factor to consider.

because, we can lesson the variance by playing multiple hands while getting possibly even more money on the table. Therefore, if we are going to play one hand of $200 we might be mathematically better served by playing two hands of $150 or even maybe three hands of $125 or four hands of $110 and so on. Therefore, the aces are more valuable to us rather than letting the dealer have more than is necessary.

To get a clear answer you have to ask a clear question. I was under the impression that bank was of no consideration and you could bet as much as you wanted on any or all hands. Lay down the ground rules and the optimal method for those specifications can be found. I'm saying that if you want to bet a total of X dollars then you can make more by betting that entire amount on one hand rather than spreading it over more than one.

Now, specify what you are talking about. Is there a table limit? Is the limit less than you'd like to bet? Is there some finite amount that you allow yourself to bet?

Again, forget the aces and whether you will get them vs the dealer. What you are looking at is the ev on each round, the dollars you can get in action on that round, and how many rounds you will get to play. These are the only critical factors, not how you and the dealer will share the cards.

It doesn't matter how many boxes you play, your advantage will remain the same (percentage-wise). Even if you get Blackjacks on all the seven spots against the dealer ten, you won't get a penny if he has an ace under.On the other hand, a Blackjack on the single hand you play might get a maximum reward.So, the number of cards you leave to the dealer is totally irrelevant.
Having said that, it's still a good thing to play two or even three spots in order to get more money on the table without alerting the pit.I would play seven spots only if I'm sure that the cut card will come out anyway next round.

Francis Salmon

But based on what you say, if it is true, then why ever play more than one hand. Why not just stack it all on one hand? Maybe I can answer my own question: By playing more hands, we can lessen variance and lessen the possibility of heat, assuming that table max is not an issue. That said; what is the perfect number of hands to maximize the lessening of variance?

But based on what you say, if it is true, then why ever play more than one hand. Why not just stack it all on one hand? Maybe I can answer my own question: By playing more hands, we can lessen variance and lessen the possibility of heat, assuming that table max is not an issue. That said; what is the perfect number of hands to maximize the lessening of variance?

I'll first say one more thing on probabilities. You can't reduce the chance of a dealer catching an ace. If there are 30 cards remaining and 6 aces in the pack, then the dealer has 2 chances in 30, or 1/15, of catching an ace in his first two cards. The chances that the dealer will catch an ace will be exactly the same whether you play one hand or six hands.

Now, on the probability that YOU will catch an ace: without increasing your action over multiple hands you can only increase the chance of catching a diluted ace. With two hands you have twice the chance of catching an ace, each ace being worth half as much because you have only bet half as much. What you are doing with multiple hands is not eating the dealer's cards but perhaps eating your own cards that you may like to get on future rounds.

Playing multiple hands is a good thing to do but you simply have the wrong objectives in mind. Multiple hands can get around table limits, it can get you more cards dealt per hour, it can increase the amount of action you can put on the table within the same risk criterion, and sometimes there is even cover value. But these reasons are not specific to the issue you raised.

If getting around table limits is not a consideration, there is generally not a good mathematical reason to play a third or fourth hand unless you know it is the last round before the shuffle. The increase in action for the same risk when going to the second hand is substantial. Going to the third hand the gain is minimal and often negated by the card eating effect.

About stealing cards: the more players that are at the table the more the value of playing multiple hands. But you are not stealing the dealer's cards, you hope to steal them from your fellow players. You are hoping that during the rich count that you will get more than your fair share of the big cards... the same as a backcounter does.

I'll first say one more thing on probabilities. You can't reduce the chance of a dealer catching an ace. If there are 30 cards remaining and 6 aces in the pack, then the dealer has 2 chances in 30, or 1/15, of catching an ace in his first two cards.

DD',

Shouldn't the dealer's odds of catching at least one ace in his first two cards under these conditions be given by the following:

1 - (24/30)*(23/29) = 0.3655 ?

Dog Hand

Let's just say there are 16 cards left to put into play. Of those 16 there are 12 faces and 4 aces. Which of the following 3 options would you choose for your playing proceedure?:

1) you can play one position with a total of eight hands to the D and eight hands to the player -or-

2) you can play three positions with a total of four hands to the D and twelve to the player -or-

3) you can play seven positions with a total of only two hands to the D and fourteen to the player.

Stealth Bomber,

If this situation ever actually arises, choose option 1 and bet the table MAX, but split EVERYTHING!

Think about it... if the dealer shows an X, and does not have BJ, then he has a pat 20. If you have a BJ, smile and take the $. If you have X-X, split them, and keep on splitting, since every hand you make will either push (20) or win (21). If you have A-A, split them as well, since the only net loss for you is if you draw the two remaining A's, and if RSA is allowed, even that wins for you!

Even better for you is if the dealer shows an A. Of course you'll take insurance... hell, insure for MORE, if allowed (some casinos do). If the dealer does not have a BJ, then he must have A-A, and given the deck configuration you specified, he'll bust... thus, if you have a BJ, DOUBLE DOWN! If you have either pair, split 'em 'til the cows come home... or at least until you are sure the remaining cards contain at least two X's, to bust the dealer.

Sometimes you have to think outside the box!

Dog Hand

I posted quickly to make my point and actually gave the probability that would exist if there were only one ace left among the 30 cards rather than six.

What you give:

1 - (24/30)*(23/29) = 0.3655

1- (24/30) is certainly the probability of the first card being an ace. That's 20%. I won't comment on the combined probability of the two cards as there are others here with greater expertise on such calculations. My intent was only to show that the number of player hands does not change the dealer's probability of catching an ace. I'm sure someone else will comment as to whether your calculation is correct.

What I do in real life is make my best guess and then email MathProf:)