Card Eating

[lanbo]

One of you could help me out with a mathematical question.
So the table min is $10 and the table max is $100, and I am allowed to play all 7 spots. So I was thinking of a way to basically get as much EV as possible out of this by playing all 7 spots on low counts with minimum bets (to eat up cards), and $100 on 7 spots at TC >= 2.

Would I generate the same EV spreading from 7X$10 to 7X$100 as spreading from 1X$70 to 1X$700?

How much lower would the ROR be betting 7 hands instead of one?

Is this an efficient method in terms of generating the most EV since I will be eating up cards on low counts.. instead of just playing 1X$10 on low counts?


Any insight would be appreciated.

Thanks!

[Three]

You should reduce the number of hands at advantage counts to either 1 or 2. You don't want card eating at those counts. To maintain the same risk with multiple hands you have to bet less on each hand compared to fewer hands. There is little difference between playing 1 or 2 hands heads up at the same risk profile but any more than that you have to choose between higher risk or lower EV. With 1 hand you get your $100 bet out and eat 5.4 cards/round or $18.519/card. With 2 bets you should lower your bets to $75 ($73 is the exact amount but not a practical bet) to maintain the same risk. You eat 8.1 cards/round on average and get $150 out on the table each round so you are betting $18.519/card (exactly the same). At 3 hands the optimal bet to maintain the same risk is 57% of the original wager per spot. So without rounding for a practical bet you get $171 and eat 10.8 cards and are now getting $15.83/card on the table. Even rounding the longer distance to get a practical bet of $60 instead of $57 you would only get 16.67/card on the table. As you keep adding hands the return per card at the same risk will continue to reap diminishing returns.

How much lower would the ROR be betting 7 hands instead of one (using an 85k BR)?

The ROR would go up immensely if you bet the same/spot on 7 spots as you would on 1 spot.

Is this an efficient method in terms of generating the most EV since I will be eating up cards on low counts.. instead of just playing 1X$10 on low counts?

NO

Would I generate the same EV spreading from 7X$10 to 7X$100 as spreading from 1X$70 to 1X$700?

[Three]

Is this an efficient method in terms of generating the most EV since I will be eating up cards on low counts.. instead of just playing 1X$10 on low counts?

In low counts you want to eat the most cards at the least cost. If you bet 7x10 you would eat 21.6 cards at a cost of $70 bet. That's $3.24/card. At 1x10 you eat 5.4 cards for $1.85/card. You are not better off at 7x10. If you could lower your bet to 7x5 you would be eating cards at a lower cost ($35 at risk to eat 21.6 cards, a cost of $1.62 bet/card eaten). You would still have a higher ROR for those lower counts. The same ROR would be arond 7x$3. I am not sure the conversion factor for 7 hands as it is not something that is practical or useful in most cases. My qick conversion of 150% of the bet divided by the number of hands increasingly losses accuracy as hands are added but is pretty good for 2 hands. That would say bet $2.14 and be on the low side so I went with 2x$3. It may be higher.

[lanbo]

You should reduce the number of hands at advantage counts to either 1 or 2. You don't want card eating at those counts. To maintain the same risk with multiple hands you have to bet less on each hand compared to fewer hands. There is little difference between playing 1 or 2 hands heads up at the same risk profile but any more than that you have to choose between higher risk or lower EV. With 1 hand you get your $100 bet out and eat 5.4 cards/round or $18.519/card. With 2 bets you should lower your bets to $75 ($73 is the exact amount but not a practical bet) to maintain the same risk. You eat 8.1 cards/round on average and get $150 out on the table each round so you are betting $18.519/card (exactly the same). At 3 hands the optimal bet to maintain the same risk is 57% of the original wager per spot. So without rounding for a practical bet you get $171 and eat 10.8 cards and are now getting $15.83/card on the table. Even rounding the longer distance to get a practical bet of $60 instead of $57 you would only get 16.67/card on the table. As you keep adding hands the return per card at the same risk will continue to reap diminishing returns.


The ROR would go up immensely if you bet the same/spot on 7 spots as you would on 1 spot.

NO

Thanks for the reply - the issue is that my max bet is $400 X 2, however the table max is only $100.. so wouldn't I be better off getting as much on the table at high counts, since I wouldn't be betting less than $100 / spot?

In low counts you want to eat the most cards at the least cost. If you bet 7x10 you would eat 21.6 cards at a cost of $70 bet. That's $3.24/card. At 1x10 you eat 5.4 cards for $1.85/card. You are not better off at 7x10. If you could lower your bet to 7x5 you would be eating cards at a lower cost ($35 at risk to eat 21.6 cards, a cost of $1.62 bet/card eaten).

This makes a lot of sense! The issue is that the table minimum is $10, and if I do not play the extra spots, usually another person comes by and fills up the spot.. depleting me of the opportunity to play that hand at a high count.

What would the threshold have to be to pay the extra $ per round to eat up cards at $10 X 7 spots to speed up the game and reserve the spot?

[Three]

I will leave this to others as 7 hands I would be guessing. That is a good way to give bad info and lose credibility. If you won't risk losing a spot you are stuck with playing 7 spots min to max. The question then becomes is this worth it? That depends on the rules of the game and other conditions.

[seriousplayer]

In low counts you want to eat the most cards at the least cost. If you bet 7x10 you would eat 21.6 cards at a cost of $70 bet. That's $3.24/card. At 1x10 you eat 5.4 cards for $1.85/card. You are not better off at 7x10. If you could lower your bet to 7x5 you would be eating cards at a lower cost ($35 at risk to eat 21.6 cards, a cost of $1.62 bet/card eaten). You would still have a higher ROR for those lower counts. The same ROR would be arond 7x$3. I am not sure the conversion factor for 7 hands as it is not something that is practical or useful in most cases. My qick conversion of 150% of the bet divided by the number of hands increasingly losses accuracy as hands are added but is pretty good for 2 hands. That would say bet $2.14 and be on the low side so I went with 2x$3. It may be higher.

Card eating requires a big bankroll. Suppose your bet is $20 and spread to two hands with $10 on each hand when the casino has an advantage. You play 80 hands per spot in an hour. By playing two hands your loss decrease to $8 an hour instead of losing $10 an hour.

[Skull]

Hi! Very interesting thread. I know nothing of the mathematical aspect of it, but here goes a few simulations (Hi-Lo Sweet 16 and Fab 4, 6 deck H17 DAS RSA LS 75% pen)

Would I generate the same EV spreading from 7X$10 to 7X$100 as spreading from 1X$70 to 1X$700?

One hand:

1hand.jpg

Seven hands:

7hands.jpg


Conclusion: EV pretty much the same, but more than double the SCORE for the 7-hand approach!

How much lower would the ROR be betting 7 hands instead of one (using an 85k BR)?

One hand:

1handROR.jpg

Seven hands:

7handsROR.jpg


Conclusion: MUCH SMALLER RoR for the 7-hand approach!


I might have done something wrong, but the odds are the 7-hand approach is way better, notwithstanding cover issues, casino tolerance, etc..., of course.


I hope it helped.

Best!

[SURFER]

Play the table all 7 spots going from $10 to $70 IE 10 20 30 / once the count hits tc 3 or above play them all at $100.00 JMO

[SURFER]

And its right, LOL