More min bets to surpass a negative count?

[Johnston]

Let's say a casino does not allow mid entries and sitting out while playing, which is common in more casinos than many think. If a shoe with 6 decks or greater has a TC of -4 (random number I thought of nothing special) or lower would it make sense to place out more min bets with hopes of displacing the negative true count to a positive? I am asking this because with double deck games, true counts fluctuate at a quicker rate compared to 6 deck games. I understand that an AP would typically just leave the shoe at a high negative count, but out of curiosity would it make sense to just play more hands, as it could potentially burn through more low cards and increase the likelihood of a higher true count later in the shoe?

[KrishnaSentMe]

I am not sure if you can think of any option other than playing out -ve count if you can’t sit out and watch someone else eating them for you.

[Freightman]

Assuming you can’t Wong out. Now, assum8ng heads up, min bet is the same for 2 squares as it is for one, then by all means play 2 which will get you to the cut card faster. If 2 squares is 2x min, then don’t. If more than you at table, just play 1.

The fewer the decks, the greater the volatility. This scenario is one which encourages one to learn more negative indices.

I recall playing a table with another ap who wonged out for a whiz break. When he got back, we were close to the end of the shoe and I was putting out max bets. If you’re a winger, have a Wong out strategy which takes into acco7nt number of players, tc, deck pen and where you are in the shoe. I tend to play all.

[UNCBear4SJ]

The interesting thing is, at negative counts the "streaks" tend to become longer. So, if you slowly feed into any weird streak while negative, it can be effective cover play by making you look like more of a ploppy. You can of course adjust split and double decisions to account for this and not put too much at risk when a streak has gotten you to a larger than warranted wager for the count. At that point it becomes more art than math/science and you're basically preserving your playing longevity.
I'd also agree with Freightman's guidance -- if others at table, let them eat the cards, but if the casino allows table min at 2x when you're heads up, go head and spread to two to get through the rough patch. If you "stack and rack" (slowly), you'll also sometimes get some nice results despite the negative counts due to that weird streak effect.

[DSchles]

Sorry, guys, the above advice is simply not correct. The "race" to the cut card doesn't tell the whole story; you're all forgetting that there's actual money on the table for those extra hands per round you're playing!

Suppose it takes you or the dealer approximately the same amount of time to play your hand. And suppose you bet $1 on each hand you play. So, playing heads up, you bet one dollar each round, and two hands are dealt. Now, spread to two hands of $1 each. You now are betting $2 per round, and three hands are dealt. Time-wise, you aren't going to get to the cut card any faster playing two hands than you are playing one! And yes, I've heard ad nauseam how you can play your two hands faster than one hand twice, because you can "wave" your hand over both and try to speed up the game when you stand, etc. But the counter-argument to that is that, when you play alone, you receive 50% of the hands, but when you play two spots, you receive two-thirds of the hands, and each or both might require a double or a split, which takes more time. Bottom line: If you want to "race" to the cut card, you aren't going to get there much faster playing two hands than you are playing one, BUT, ... during that time, playing two hands, you're going to have bet TWICE as much money on two-thirds as many rounds! DUH! Do you really want to do that? I think not!

The principle of eating cards doesn't work unless the total amount bet on the two hands can be LESS than 1.5 times what you bet on the one hand. For example, if you were betting $10 per hand but then decided to bet two hands of $5, in negative counts, that would be a good idea. But betting two hands of $10 each would surely be a bad idea. And betting one hand of $5 would be the best idea of all!

Don

[Freightman]

Yeah, you’re right! I was thinking along the lines of getting to the next shoe faster.

[DSchles]

Yeah, you’re right! I was thinking along the lines of getting to the next shoe faster.

And, you might even do that. But, ... it will cost you more money in the process.

Don

[Freightman]

And, you might even do that. But, ... it will cost you more money in the process.

Don

Possibly yes, possibly no. Depends on quality of the new shoe. The idea is to lose some waiting min bets on the crappy shoe making up for it on the hopefully positive new shoe.

[JohnGalt007]

Possibly yes, possibly no. Depends on quality of the new shoe. The idea is to lose some waiting min bets on the crappy shoe making up for it on the hopefully positive new shoe.

Then to make the decision to stick around and wait for the shoe to stay positive preferable to the decision to simply leave and find a new table, you'd require the probability for the shoe to swing back to a positive count to outweigh the probability of a new shoe becoming positive at all within a finite number of steps (min bets). Modeling the running count as a random walk on the integers, the former probability is roughly inversely proportional to the square root of n+1, where -n is the current running count, whereas the latter probability is 0.5. The former probability is less than the latter after just a single min bet - a very rough approximation to be sure, but even with that back-of-the-envelope reasoning, it's also cheaper and less time-consuming to simply find a new table.

[Freightman]

Then to make the decision to stick around and wait for the shoe to stay positive preferable to the decision to simply leave and find a new table, you'd require the probability for the shoe to swing back to a positive count to outweigh the probability of a new shoe becoming positive at all within a finite number of steps (min bets). Modeling the running count as a random walk on the integers, the former probability is roughly inversely proportional to the square root of n+1, where -n is the current running count, whereas the latter probability is 0.5. The former probability is less than the latter after just a single min bet - a very rough approximation to be sure, but even with that back-of-the-envelope reasoning, it's also cheaper and less time-consuming to simply find a new table.

I seldom have a choice of tables - ergo - I play the hand that I’m dealt.

[DSchles]

See the optimal departure study in BJA3!

Don

[Frankie]

Additionally, many casinos won't allow you to bet the minimum on multiple hands (this might be a relatively new practice, as I don't recall seeing this 15+ years ago). So if you play two hands, you have to bet twice the minimum. Three or more, three times the minimum.

[Freightman]

Additionally, many casinos won't allow you to bet the minimum on multiple hands (this might be a relatively new practice, as I don't recall seeing this 15+ years ago). So if you play two hands, you have to bet twice the minimum. Three or more, three times the minimum.

Different places, different rules. I’ve seen both 1x and 2x min for 2 spots and both 5x and table max for 3 spots.