2 hands during low counts

because often people will post any little slight exception and cause others to lose sight of the principle that is attempting to be illustrated.

More money in action means simply that more money, overall, will be bet. The post to which I was responding claimed that playing extra hands in negative counts would cause you to need to spread bigger. That is absolutely untrue. For example, suppose your big bet is 16 units (all on one or spread over two, doesn't matter). Your small bet is one hand of 1 unit. You have some specific win rate and you have a 1:16 spread. Now let's suppose that you decide to play a second hand of 1 unit during negative counts. Even if there were absolutely no measurable gain from card eating, you would still not need to increase your bet spread to attain the same win rate. At the point you spread to the second hand in - counts you have decreased your spread from 1:16 to 1:8. If there is no card eating gain you would have to increase positive ev bet sizes to regain your prior win rate. But, you would not have to increase them to the spread size you had before. If you now spread 1:16, meaning a top bet of 32 units, your win rate would be double what it was initially. So what I'm saying when addressing the comment of this particular poster is simply that multiple hands in negative counts require smaller spreads "in all cases" to achieve the same win rate.

Now, the comments I made previously in the thread concern increasing your win rate via multiple hands in negative counts without increasing positive ev action. This is often possible but depends on some variables, mainly the size of your spread and the number of other players at the table.

are always a barrier to optimal play, which is based on the assumption that you can bet what you want. If your unit size if equal to the table minimum, and they require you to double the minimum for two hands, obviously you are not going to gain by increasing to 4X action in negative counts. Playing at tables where the minimum is equal to your unit size is sometimes desirable so that you can enjoy less of a crowd by playing the highest table you can afford. But it takes away lots of options. Playing $100 units I am often able to play a second hand of only $25. A $25 player may often be able to play a second hand of only $10. The rules are not the same everywhere. The original poster was asking for cases when you could gain from such strategies, not for cases where it would not be possible.

an advanced counter ?
"For another, spreading to two hands only in +ev situations just looks too obvious to me. It is what nearly all beginner counters do"
just kidding.

"And two handed play offers the opportunity for a great many more low cost cover options. "
what kind of cover do you refer to ? different bet sizes ?

I've seen this advocated before and I am curious about this approach. Could you give a little detail on how you'd do it? Feel free to email me if you prefer.

Cheers,
sargon

It is well known that the table minimum is usually valid for every spot of this table. So playing $25 on one spot and $10 on an other would not be possible.
But if you have to play 25$ on both spots, the additional card consumption will never compensate for the doubling of your ev-loss per round even if you're playing head on.With one box 5.6 cards and with two players 8.4 cards which is only 50% more.
Let's face it,DD, you put a lot of energy in this thread trying to prove that playing two spots is advisable on mathematical grounds but this is not possible. I get the impression that the real reason why you play two spots is because you want them to be available in high counts. This reason is good enough, why are you looking for something else?

Francis Salmon

Is that table limit constraints can sometimes intefere with optimal strategy.

It is well known that the table minimum is usually valid for every spot of this table. So

playing $25 on one spot and $10 on an other would not be possible.

Impossible at a $25 minimum table, perhaps. Quite often possible at a $10 table. I very frequently make bets of unequal sizes on two hands. You are making the assumption that table minimum always equals unit size. It may or may not be. I could be at a $25 table playing one hand of $100 and another of $50.

I have no desire to give the answers here. I have a desire to show people the right path and I prefer to do this in the most vague way possible. I am glad to help with specific answers when it comes to vague theoretical issues, but not actual casino tactics. I can assure you that anyone who has done a good amount of research on their own already knows what I'm talking about and wishes that I would say no more about it. I'll respect their wishes.

I don't see a reason why you should bet $25 on one of the spots in negative situations.
Don't get me wrong: I'm not criticizing your tactics. I just say that it has no mathematical foundation.

Francis Salmon

You might note that this is NOT a $10 bettor playing a second hand of $25. This is a $25 bettor who is playing a second hand of $10.

I don't see a reason why you should bet $25 on one of the spots in negative situations.

You mean you don't understand why one might have a $25 minimum rather than a $10 minimum? Hmmm, that is a tough one. Might his bankroll have anything to do with it? Could it be because 10+10=20 but 10+25=35? Might one's intended spread have anything to do with his minimum bet? Could it be that $20 constitutes a smaller fraction of your big bet than you intended to spread?. If your max is $500 and you decide on a 1:20 you will not be making $20 bets (2@$10). Pretty simple. But $35 falls within your spread criteria.

I just say that it has no mathematical foundation.

Of course not. Let's try to figure this out using some math. We'll just make sure we don't include a foundation.

Max is $500.
Spread is 1:20.
$500/20=$25.
10+10=20.
20<25

I mean, why would you ever bet $25 in a neg. count when you could bet $10?

Lets take a closer look at simple math as you posted.

Max is $500.
Spread is 1:20.
$500/20=$25.
10+10=20.
20<25

I assume you mean you are playing 2 hands of $10 in all negative counts but you usually play 1 green. If this is the case then of course simple math says 2 hands are better. However the problem is you have actually increased your spread to a 1 to 50 ratio as you go to playing one hand on plus counts. I would not recomend this type of play unless you are prepared for some heat and scrutiny. This ploy will not work at many casinos anyway due to rules of doubling your bet to play 2 hands. In this case 40>25.
There are some unique times (if conditions allow) we can benefit from card eating if table minimum and max bets allow us and no heat is invovled. At any rate...thru my eyes of action, most of the time it is impractical.

Mott

You told people who disagreed with you to just run some Sims.

None of your critics has run a single sim. I guess some people are just too smart to need computers.

I don't think you should be giving away advanced tactics for people who do not appreciate them.

I assume you mean you are playing 2 hands of $10 in all negative counts but you usually play 1 green.

In the origional example I said that the $25 bettor could actually increase his win rate by playing a second hand of $10. To all who miss the point, this was intended to illustrate that you could increase your action during negative counts, and essentially lower your total spread, yet increase you win rate. In a heads up game, with virtually any number of decks, this will be true. You will go through the negative segment at a lower net cost and will increase the frequency of positive counts per 100 dealt. Your $ in action will be $70 per 16.2 cards dealt, rather than $75 per 16.2 cards dealt, which will be the cost to the one hand of $25 player. You might even go through those cards faster.

unless you want Shiftmgr Fat Nick to whisper in your ears: "You can't play here anymore. Beat it!" BTW, counselor, I'm glad you didn't get hurt near WTC. Where were you when the twin towers collapsed?