CVCX and using two hands

[shorty]

UPDATE since posting the paragragh under this one ,I fugured out how to simply run the numbers properly .with one hand being played at 100
and then being split into two hands of $75. my ror stayed almost the same but my hourly win rate went up 33% that's HUGE
so what your going to read under this was my OLD finding and they were flawed because I split the hands in half instead of 75%
THE NEW QUESTION BROUGHT TO LITE BY ROLLING STONED is that possibly the 33% extra $ isn't really that easy to get because ide be eating up good cards, I disagree at this time! comments are muchly appreciated. bring it baby!!!

CVCX would have you believing that using two hands is pointless extra trouble.
after customizing the bets in the sim so that each of the two hands are approx half one regular bet, my ROR is just as bad and hourly rate dropped.
does this mean I should give up playing two hands? I seriously don't know
all this time I thought that you want to use as many spots as possible when TC is up, this has me seriously wondering if its worth it.

[RS]

You bet 75% of a regular 1-hand bet on 2 different hands.

ie: One hand of $100 -> 2 hands of $75 [each]. Or One hand of $500 -> 2 hands of $375 [each].


If you play the way you described, then of course your win rate will go down, because you have the same amount of money in action (at the same advantage), BUT you are eating up more cards. Imagine 1 deck (52 cards) is remaining to be played and you're sitting at a 1% advantage. If you bet 1 hand of $100, you expect to play about 18 hands, assuming 2.7 cards per spot played(including dealer). That's 18*$100*0.01 = $18 in EV. If you decide to play two hands of $75 each, then you expect to play 12 hands. 12*$150*0.01 = $18. But if you were playing 2 hands of $50, then your EV would be 12*$100*0.01 = $12. From an EV perspective, it's best to bet all your money on ONE hand when you have an advantage. Of course, you want to take heat and risk/variance into consideration. You can probably get away with betting 2 hands of $500 much easier than playing one hand of $1000. Also, by playing 2 spots, you reduce the variance a bit, because one hand can win and the other lose....even though they are related to each other and frequently win or lose at the same time [due to the fact they are likely to both win when the dealer busts and both lose when dealer pulls a 20 or 21. also both more likely to lose when dealer has a T or A up because you'll be more likely to hit and bust on both hands...or for the dealer to pull a T from under the A for a dealer BJ).

[ZeeBabar]

Rollingstone, I am not much for math but I have read that playing two hands in the manner you described is always better than playing one hand. I have also read where a few argue that, when heads up, you play one hand in positive situations and two minimum hands on negative situations to eat the cards. Another view that I sometimes practice is starting out with 2 hands, dropping to one in negative situations and playing two hands in positive situations.

comments are appreciated

[Cougfan]

I think you are doing something wrong. Perhaps if you can post the exact bet ramps and game rules / pen someone can help you narrow it down. Or just try solving for the same ROR and you should see win rate go up. Ditto if you solve for the same win rate, then your ROR should go down. If you want to keep ROR the same, then the optimal 2 hand bet is approximately 75% of the one hand bet. This should increase your win rate. But if you bet 1/2 on each hand like you said, then you should see win rate stay roughly the same and ROR go down.

I don't have access to CVCX at the moment as its on another computer, but BJRM shows the following for a random game (DD, H17, DAS, NSR, 62/104 pen with a 1-8 optimal ramp and a $90K BR) betting 1/2 Kelly for a 1.84% ROR:

One hand: Win rate = $178 per 100 hands ($75 unit size)
Two hands: Win rate = $258 per 100 hands ($53 unit size)

If I change the unit size to $50, then my win rate drops to $241 per 100, and my ROR drops to 1.37%.

This assumes you are playing 2 hands at all counts.

[shorty]

You bet 75% of a regular 1-hand bet on 2 different hands.

ie: One hand of $100 -> 2 hands of $75 [each]. Or One hand of $500 -> 2 hands of $375 [each].


If you play the way you described, then of course your win rate will go down, because you have the same amount of money in action (at the same advantage), BUT you are eating up more cards. Imagine 1 deck (52 cards) is remaining to be played and you're sitting at a 1% advantage. If you bet 1 hand of $100, you expect to play about 18 hands, assuming 2.7 cards per spot played(including dealer). That's 18*$100*0.01 = $18 in EV. If you decide to play two hands of $75 each, then you expect to play 12 hands. 12*$150*0.01 = $18. But if you were playing 2 hands of $50, then your EV would be 12*$100*0.01 = $12. From an EV perspective, it's best to bet all your money on ONE hand when you have an advantage. Of course, you want to take heat and risk/variance into consideration. You can probably get away with betting 2 hands of $500 much easier than playing one hand of $1000. Also, by playing 2 spots, you reduce the variance a bit, because one hand can win and the other lose....even though they are related to each other and frequently win or lose at the same time [due to the fact they are likely to both win when the dealer busts and both lose when dealer pulls a 20 or 21. also both more likely to lose when dealer has a T or A up because you'll be more likely to hit and bust on both hands...or for the dealer to pull a T from under the A for a dealer BJ).

I find this to be suuper interesting for some reason and thanks for the simple breakdown. that's math I could of probably figured out but never thought to do. Kudos and thanks for that info as ill probably refer back to it in the future.
this is a big relief. you see ,ive been fighting for two seats at the table when it gets busy... but what's the point? would you agree that its not worth wasting your breath over (to keep proclaiming your two hand status to every walker...)
walker= crazy half drunk casino zombie eager to lose some cash at a minus C

[shorty]

I think you are doing something wrong. Perhaps if you can post the exact bet ramps and game rules / pen someone can help you narrow it down. Or just try solving for the same ROR and you should see win rate go up. Ditto if you solve for the same win rate, then your ROR should go down. If you want to keep ROR the same, then the optimal 2 hand bet is approximately 75% of the one hand bet. This should increase your win rate. But if you bet 1/2 on each hand like you said, then you should see win rate stay roughly the same and ROR go down.

I don't have access to CVCX at the moment as its on another computer, but BJRM shows the following for a random game (DD, H17, DAS, NSR, 62/104 pen with a 1-8 optimal ramp and a $90K BR) betting 1/2 Kelly for a 1.84% ROR:

One hand: Win rate = $178 per 100 hands ($75 unit size)
Two hands: Win rate = $258 per 100 hands ($53 unit size)

If I change the unit size to $50, then my win rate drops to $241 per 100, and my ROR drops to 1.37%.

This assumes you are playing 2 hands at all counts.

hmmm.I noticed small differences with win rate and RoR... but but barely a noticeable amount.
your way is quite diff

[RS]

What happens if you're in the same situation (let's say 3 decks or 156 cards with 1% advantage), but instead of playing heads up, there are 4 other players + 1 dealer + your 1 or 2 hands.

1 hand: 6*2.7 = 16.2 cards/round : 156/16.2 = 9.63 rounds
2 hands: 7*2.7 = 18.9 cards/round : 156/18.9 = 8.25 rounds

1 hand: 9.63 * $100 * 0.01 = $9.63 in EV
2 hands: $150 * 8.25 * 0.01 = $12.37 in EV



The game is confusing.

[shorty]

[QUOTE=RollingStoned;174844]What happens if you're in the same situation (let's say 3 decks or 156 cards with 1% advantage), but instead of playing heads up, there are 4 other players + 1 dealer + your 1 or 2 hands.

1 hand: 6*2.7 = 16.2 cards/round : 156/16.2 = 9.63 rounds
2 hands: 7*2.7 = 18.9 cards/round : 156/18.9 = 8.25 rounds

1 hand: 9.63 * $100 * 0.01 = $9.63 in EV
2 hands: $150 * 8.25 * 0.01 = $12.37 in EV


so I figured out how to run the numbers in cvcx

TC 1 and up receives $100 bet on one hand .while anything under gets no bet
doing the same thing with two hands but with 75% or $75 made lots more money.just look

1 hand ,any count above TC +1 receives $100. RoR 19.3% Hourly $29.99
2 hands ,any count above TC +1 receives $75. RoR 20.3% Hourly $44.99

[shorty]

You bet 75% of a regular 1-hand bet on 2 different hands.

ie: One hand of $100 -> 2 hands of $75 [each]. Or One hand of $500 -> 2 hands of $375 [each].


If you play the way you described, then of course your win rate will go down, because you have the same amount of money in action (at the same advantage), BUT you are eating up more cards. Imagine 1 deck (52 cards) is remaining to be played and you're sitting at a 1% advantage. If you bet 1 hand of $100, you expect to play about 18 hands, assuming 2.7 cards per spot played(including dealer). That's 18*$100*0.01 = $18 in EV. If you decide to play two hands of $75 each, then you expect to play 12 hands. 12*$150*0.01 = $18. But if you were playing 2 hands of $50, then your EV would be 12*$100*0.01 = $12. From an EV perspective, it's best to bet all your money on ONE hand when you have an advantage. Of course, you want to take heat and risk/variance into consideration. You can probably get away with betting 2 hands of $500 much easier than playing one hand of $1000. Also, by playing 2 spots, you reduce the variance a bit, because one hand can win and the other lose....even though they are related to each other and frequently win or lose at the same time [due to the fact they are likely to both win when the dealer busts and both lose when dealer pulls a 20 or 21. also both more likely to lose when dealer has a T or A up because you'll be more likely to hit and bust on both hands...or for the dealer to pull a T from under the A for a dealer BJ).

wait wait , I changed my mind
ROLLING STONEd the thing is , yes were chewing up cards but so is everybody elses and whoes to say the count is gonna stay up or go down.we cant assume its gonna stay the same and well be "chewing" up good cards
the bottom line is. playing two hands is much better than one. I ran the numbers properly finally
am I right? tink about the eating up cards thing because that theory could be bull shhhhh donyt you think?

[shorty]

so I figured out how to run the numbers in cvcx

TC 1 and up receives $100 bet on one hand .while anything under gets no bet
doing the same thing with two hands but with 75% or $75 made lots more money.just look

1 hand ,any count above TC +1 receives $100. RoR 19.3% Hourly $29.99
2 hands ,any count above TC +1 receives $75. RoR 20.3% Hourly $44.99

[RS]

wait wait , I changed my mind
ROLLING STONEd the thing is , yes were chewing up cards but so is everybody elses and whoes to say the count is gonna stay up or go down.we cant assume its gonna stay the same and well be "chewing" up good cards
the bottom line is. playing two hands is much better than one. I ran the numbers properly finally
am I right? tink about the eating up cards thing because that theory could be bull shhhhh donyt you think?

Actually, we can assume the TC will stay constant, because the TC tends* to stay constant. If you're at a positive count, as you go through the shoe and take cards out, the RUNNING COUNT will decrease, but it tends* to decrease at a rate comparable to the cards being taken out, leaving the TC the same.

* When I say "tends", I mean in terms of expectation.

Imagine a fraction, 30/45 (equals 2/3). For every 2 you remove from the number (the 30), you can take 3 from the denomintor (the 45). The result is unchanged. 30/45 = 28/42 = 26/39 = 24/36 = ... 2/3


Say you have 3 decks remaining. The RC is +9. Thus the TC is +9/3 = +3. There are 3 extra little cards per deck. If you play through ONE deck, you expect 3 fewer little cards to come out than big cards, yes? So after that one deck, the RC should be at +6. But the decks remaining is now 2. TC is +6/2 = +3. After another 1 deck of play, 3 fewer little cards coming out than big cards, the RC drops from +6 to +3. But in doing so, the remaining decks goes from 2 to 1. TC is now +3/1 = +3.

[shorty]

Actually, we can assume the TC will stay constant, because the TC tends* to stay constant. If you're at a positive count, as you go through the shoe and take cards out, the RUNNING COUNT will decrease, but it tends* to decrease at a rate comparable to the cards being taken out, leaving the TC the same.

* When I say "tends", I mean in terms of expectation.

Imagine a fraction, 30/45 (equals 2/3). For every 2 you remove from the number (the 30), you can take 3 from the denomintor (the 45). The result is unchanged. 30/45 = 28/42 =
26/39 = 24/36 = ... 2/3


Say you have 3 decks remaining. The RC is +9. Thus the TC is +9/3 = +3. There are 3 extra little cards per deck. If you play through ONE deck, you expect 3 fewer little cards to come out than big cards, yes? So after that one deck, the RC should be at +6. But the decks remaining is now 2. TC is +6/2 = +3. After another 1 deck of play, 3 fewer little cards coming out than big cards, the RC drops from +6 to +3. But in doing so, the remaining decks goes from 2 to 1. TC is now +3/1 = +3.

that math you got makes sence but it doesn't mean that's what will happen.
if the count is slightly up then its just waiting to correct itself and the further in you go the more likely it will happen.correcting itself back to a nuetrel true count that is.
ok think about it this way then. forget about that shoe ,moving on to more good TC counts because lets just say we don't even bet on bad Tc counts right.
anyways your making more money with two hands on the table at any given moment when the count is up compaired to using just one hand.
I don't see how that fractions above prove anything about the future outcome

[RS]

that math you got makes sence but it doesn't mean that's what will happen.
if the count is slightly up then its just waiting to correct itself and the further in you go the more likely it will happen.correcting itself back to a nuetrel true count that is.
ok think about it this way then. forget about that shoe ,moving on to more good TC counts because lets just say we don't even bet on bad Tc counts right.
anyways your making more money with two hands on the table at any given moment when the count is up compaired to using just one hand.

I gave two examples before.

I don't see how that fractions above prove anything about the future outcome

Then re-read it. Or better yet -- show me how the true count (when it's at say, +4), will "correct itself back to neutral".