Betting Multiple Hands

[zengrifter]

This is not true. If you play 2 hands at high counts one vs one at 150% of your normal betting spread than you will be making more money. Play 2 hands heads up.

Not true - UNLESS the two hands buys you increased pene%. zg

 

[rollem411]

Not true - UNLESS the two hands buys you increased pene%. zg

I still disagree. Your putting more money out on the table which means more $ per hour.

 

[rukus]

I still disagree. Your putting more money out on the table which means more $ per hour.

the more $ per hour concept builds off how many rounds per hour you can get at an advantage. by playing more than one hand heads-up you will reduce the number of rounds you get at those high counts. when playing with others at the table, they will cut your # rounds at the high counts anyway, so you might as well play 2 or 3 hands each round.

 

[rollem411]

the more $ per hour concept builds off how many rounds per hour you can get at an advantage. by playing more than one hand heads-up you will reduce the number of rounds you get at those high counts. when playing with others at the table, they will cut your # rounds at the high counts anyway, so you might as well play 2 or 3 hands each round.

I agree completely with what you are saying, however, I think your still missing what I am saying.

You will make more money because lets say a max bet...a 1x$100 bet that you have out won't yield you as much as 2x$75, which is the optimum bet for spreading to two hands.

 

[rukus]

I agree completely with what you are saying, however, I think your still missing what I am saying.

You will make more money because lets say a max bet...a 1x$100 bet that you have out won't yield you as much as 2x$75, which is the optimum bet for spreading to two hands.

i agree on a given round you will make more betting multiple hands. but you are missing what we are saying :devil:. we are saying that overall, you will get more rounds at a given advantage when you play less hands per round. here is how i think of it in my head --

lets take an example and say there are X cards remaining before the cut card in a high TC segment. Each hand (yours and dealer's) use Y cards. what is the optimal number of hands to play?

play 1 hand per round at max bet of $100: you will get X/(2Y) rounds at $100 each for a total of X/(2Y)*100 total money in action.
play 2 hands per round at $75 each: you will get X/(3Y) rounds at $150 each for a total of X/(3Y)*150 total money in action.

for the sake of this example, assume your advantage across rounds remaining does not change.

then in an ideal world, where you can use up fractions of a card (and Y is a fraction, somewhere slightly less than 2.7 in a positive TC, but let's call it 2.7), the results for the two situations above would be exactly the same. however, in the real world, you will not use 2.7 cards, you will use 3, or an extra 0.3 cards per hand than you ideally would. you will more quickly use up these slight 0.3 fractions the more hands you play, such that after just one round of playing two hands, you've already used about one extra card worth of those fractions than in the ideal world. playing one hand per round, it would take you two rounds to use up that extra card. so basically you will be able to get more rounds in playing just one hand. does that make sense? maybe this is a bad/incorrect way to look at it?

also note the results in the two cases above are exactly the same because we assume the optimal amount on 2 hands is to play 75% each. what if it's only 70% on each hand? then you see for sure that playing one hand is the optimal choice.

there is a discussion of the optimal number of hands to play in BJA3. i dont recall Don's arguments off the top of my head.

 

[FLASH1296]

My take on this issue

IF the situation is advantageous and you are betting 2 hands as opposed to one hand - you put your $$ to work and get 2 hands while depleting less than twice the amount of cards.

Both you and the dealer will, on average, consume 2.7 cards per hand.

If you play ONE HAND heads-up for TWO rounds, you will use up 2.7 x 4 = 10.8 cards.

If you play heads-up TWO hands on ONE round you use up 2.7 x 3 = 8.1 cards.

f you play ONE HAND heads-up for THREE rounds, you will use up 2.7 x 6 = 16.2 cards.

If you play heads-up THREE hands on ONE round you use up 2.7 x 4 - 10.8 cards.

Clearly multiple hands under advantageous conditions is substantially better as it will deplete less cards to play the same number of hands. This is superior as you'll get more hands to play under advantageous True Counts.

With others at the table the situation re: playing multiple hands improves even more.

Look at the most extreme case. There are 6 spots and the table is full:

In situation A.you play 1 spot.
In situation B you play 2 spots.
In situation C you play 3 spots.

Ergo, you are playing 1/6, 1/3, or 1/2 of all spots.

I prefer to seize control of as many hands as possible under advantageous betting conditions.

"Do the math"

 

[rukus]

IF the situation is advantageous and you are betting 2 hands as opposed to one hand - you put your $$ to work and get 2 hands while depleting less than twice the amount of cards.

Both you and the dealer will, on average, consume 2.7 cards per hand.

If you play ONE HAND heads-up for TWO rounds, you will use up 2.7 x 4 = 10.8 cards.

If you play heads-up TWO hands on ONE round you use up 2.7 x 3 = 8.1 cards.

f you play ONE HAND heads-up for THREE rounds, you will use up 2.7 x 6 = 16.2 cards.

If you play heads-up THREE hands on ONE round you use up 2.7 x 4 - 10.8 cards.

Clearly multiple hands under advantageous conditions is substantially better as it will deplete less cards to play the same number of hands. This is superior as you'll get more hands to play under advantageous True Counts.

With others at the table the situation re: playing multiple hands improves even more.

Look at the most extreme case. There are 6 spots and the table is full:

In situation A.you play 1 spot.
In situation B you play 2 spots.
In situation C you play 3 spots.

Ergo, you are playing 1/6, 1/3, or 1/2 of all spots.

I prefer to seize control of as many hands as possible under advantageous betting conditions.

"Do the math"

you will play more hands, less rounds. what we care about is total money in action and that is equal to money/round*#rounds. my thoughts above try and shoe that #rounds decreases faster than money/round increases when you play more hands each round.

 

[rollem411]

i agree on a given round you will make more betting multiple hands. but you are missing what we are saying :devil:. we are saying that overall, you will get more rounds at a given advantage when you play less hands per round. here is how i think of it in my head --

lets take an example and say there are X cards remaining before the cut card in a high TC segment. Each hand (yours and dealer's) use Y cards. what is the optimal number of hands to play?

play 1 hand per round at max bet of $100: you will get X/(2Y) rounds at $100 each for a total of X/(2Y)*100 total money in action.
play 2 hands per round at $75 each: you will get X/(3Y) rounds at $150 each for a total of X/(3Y)*150 total money in action.

for the sake of this example, assume your advantage across rounds remaining does not change.

then in an ideal world, where you can use up fractions of a card (and Y is a fraction, somewhere slightly less than 2.7 in a positive TC, but let's call it 2.7), the results for the two situations above would be exactly the same. however, in the real world, you will not use 2.7 cards, you will use 3, or an extra 0.3 cards per hand than you ideally would. you will more quickly use up these slight 0.3 fractions the more hands you play, such that after just one round of playing two hands, you've already used about one extra card worth of those fractions than in the ideal world. playing one hand per round, it would take you two rounds to use up that extra card. so basically you will be able to get more rounds in playing just one hand. does that make sense? maybe this is a bad/incorrect way to look at it?

also note the results in the two cases above are exactly the same because we assume the optimal amount on 2 hands is to play 75% each. what if it's only 70% on each hand? then you see for sure that playing one hand is the optimal choice.

there is a discussion of the optimal number of hands to play in BJA3. i dont recall Don's arguments off the top of my head.

I understand that you decrease the amount of rounds you play when playing 2 hands vs 1 hand. Also, like you said, if you are not betting the optimum when spreading to 2 hands, you may as well stick to 1.

Quick question to solve the matter...You are playing a heads up game and the TC is +5. Would you rather play 1 hand or would you move to 2?

Like I said, I have run sims with the 2 scenarios and the 2 handed play has given me better results.

 

[Kasi]

Quick question to solve the matter...You are playing a heads up game and the TC is +5. Would you rather play 1 hand or would you move to 2?

I agree with both of you lol.

But since the optimal bet would be 2*$73, rather than 2*$75, one would be slightly better off playing just 1 hand heads-up vs dealer. That is more EV with same risk. 2*$75 is same EV but slightly higher risk - nothing gained lol.

After 100 top-bet opportunities 1 hand =$10K wagered. Spreading to 2 hands at every top bet opportunity would only yield 66.7 top bet opportunities in the same amount of time. So 66.7*2*$73=$9738 wagered.