When the Count is High should I play 2 hands of 100 or 1 hand for 200?

[Bendtackle]

My thinking is it will decrease the variability, but slightly decrease the expected value, because it most likely will decrease the count faster and allow less hands to be played at that count.

(Note: I usually play at a six deck table with a shuffle point of a 5 deck penetration. The casino allows for 6 people to be at a table, and most likely the table will have 5 or 6 people at it. I use High Low.)

 

[Deathclutch]

Two hand of 75 is the for a max bet of 100.

Two hands of 150 for a max bet of 200.

 

[Bendtackle]

Two hand of 75 is the for a max bet of 100.

Two hands of 150 for a max bet of 200.

What are you talking about?

 

[Lonesome Gambler]

When you bet two hands, you can bet 150% of your single-spot max bet on each hand. So if your max bet is $200, you can play two hands of $150. It decreases variance and the win rate is similar.

 

[Bendtackle]

When you bet two hands, you can bet 150% of your single-spot max bet on each hand. So if your max bet is $200, you can play two hands of $150. It decreases variance and the win rate is similar.

So what you are saying is my guess was correct in that it decreases the variance while at the same time decreasing the expected value? In order to make the expected value about the same I would have to bet 150% of what I put on 1 spot?

Unfortunately I can not increase my wager. I currently play at a $10 minimum table with my max bet of $50. My stack size is not large enough for me to bet higher. Would it be better to bet two hands of 25 or one of 50? What is the difference in variation and expected value by playing two hands of $25 compared to one at $50?

 

[Deathclutch]

What are you talking about?

I'm talking about the answer to your question in your thread title.

 

[Bendtackle]

I'm talking about the answer to your question in your thread title.

I understand what you are getting at now. I need to compare same sum bet amounts, as the example in my topic question $200 is the max amount I can afford to bet. Therefore your answer is invalid, because I can not afford to bet 2 $150 hands (sum of $300). If I could afford to bet $300 I would have been betting it on a single hand.

 

[Deathclutch]

I understand what you are getting at now. I need to compare same sum bet amounts, as the example in my topic question $200 is the max amount I can afford to bet. Therefore your answer is invalid, because I can not afford to bet 2 $150 hands (sum of $300). If I could afford to bet $300 I would have been betting it on a single hand.

But if your max bet is usually $200 you can handle betting two hands of $150. I'm going to let you do your comparisons but I think you'll soon figure out what I mean.

 

[Bendtackle]

But if your max bet is usually $200 you can handle betting two hands of $150. I'm going to let you do your comparisons but I think you'll soon figure out what I mean.

The most money I can put on the table is $200. I simply don't have the stack size to put out more. Maybe I am missing something, but I don't see the relevance of comparing $150 at 2 hands (sum of $300) and 1 hand at $200 (sum of $200). I am researching the topic as we post, so maybe I will figure out what you mean.

 

[assume_R]

Maybe I am missing something, but I don't see the relevance of comparing $150 at 2 hands (sum of $300) and 1 hand at $200 (sum of $200). I am researching the topic as we post, so maybe I will figure out what you mean.

What he was saying is that your variance will stay the same if you bet 2x$150 as if you bet 1x$200. I'm pretty sure the correct way to think about it is that your EV is your advantage times the total amount of money you have on the table. If you spread that $$ over several hands, your variance will decrease (but the EV stays the same).

Now, if you are comfortable with the variance and RoR associated with a 1x$200 bet at a certain count, then you can keep the same RoR but get more money out on the table (2x$150, or $300 total). The EV in terms of % is the same, but the total expected $$ is greater.

Does that clarify it? And for other posters, is my advice 100% correct in all situations?

 

[fwb]

Therefore your answer is invalid, because I can not afford to bet 2 $150 hands (sum of $300). If I could afford to bet $300 I would have been betting it on a single hand.

If your max bet is the size of your entire bankroll, you're doing it wrong.

 

[Bendtackle]

What he was saying is that your variance will stay the same if you bet 2x$150 as if you bet 1x$200. I'm pretty sure the correct way to think about it is that your EV is your advantage times the total amount of money you have on the table. If you spread that $$ over several hands, your variance will decrease (but the EV stays the same).

Now, if you are comfortable with the variance and RoR associated with a 1x$200 bet at a certain count, then you can keep the same RoR but get more money out on the table (2x$150, or $300 total). The EV in terms of % is the same, but the total expected $$ is greater.

Does that clarify it? And for other posters, is my advice 100% correct in all situations?

This does clarify the subject. So according to him betting 100 on 2 hands is better? I am worried that this takes away from my expected value on the shoe. I am playing more cards so the count will not stay as high as long. For instance, in my research I have read if you are the only person sitting at the table you should not play two hands, because you are using twice the cards. So it in effect cancels out.

So betting two hands of $150 would give me an advantage more than what I would lose from using more cards, but also have the same variance as using one $200 bet?

 

[Bendtackle]

If your max bet is the size of your entire bankroll, you're doing it wrong.

Don't be ridiculous. I never stated that anywhere.

 

[cc218]

ok, so if your max bet is 50, then play two hands of ~37.5 is what everybody else is saying... i'm not seeing the point of confusion... also, schelesinger touches on this on several occassions in Blackjack Attack

 

[Bendtackle]

ok, so if your max bet is 50, then play two hands of ~37.5 is what everybody else is saying... i'm not seeing the point of confusion... also, schelesinger touches on this on several occassions in Blackjack Attack

Honestly I do not get why no one is answering my question straight up. Lets say you stake your friend at random points in time when the count is favorable to you. Lets say you always give him $200. Would you rather have him put them on two hands or one? In other words, is the decrease in variability worth it for the decrease in expected value? Is this worth it if you do it 100 times? 1 time? 1,000,000 times?

 

[Deathclutch]

Honestly I do not get why no one is answering my question straight up.

Therefore your answer is invalid, because

This could be why.

 

[Bendtackle]

This could be why.

No harm meant Deathclutch, I was just clarifying that was not the answer I was quite looking for. I appreciate everyones insight on this topic, including yours Deathclutch.

 

[assume_R]

Honestly I do not get why no one is answering my question straight up. Lets say you stake your friend at random points in time when the count is favorable to you. Lets say you always give him $200. Would you rather have him put them on two hands or one? In other words, is the decrease in variability worth it for the decrease in expected value? Is this worth it if you do it 100 times? 1 time? 1,000,000 times?

We were all trying to explain to you WHY so you could understand exactly and come to the same conclusions.

If you gave your friend $200, his EV shouldn't decrease if your friend has $200 on the table (not considering some effects of using up cards if you're heads up or not - let's say there are 2 other people at the table). Therefore, playing 2 hands would be better since the variance is decreased.

What we were saying is that you could potentially give your friend $300, and his RoR wouldn't increase if he played 2x$150.

Is this clear yet?

 

[Bendtackle]

We were all trying to explain to you WHY so you could understand exactly and come to the same conclusions.

If you gave your friend $200, his EV shouldn't decrease if your friend has $200 on the table (not considering some effects of using up cards if you're heads up or not - let's say there are 2 other people at the table). Therefore, playing 2 hands would be better since the variance is decreased.

What we were saying is that you could potentially give your friend $300, and his RoR wouldn't increase if he played 2x$150.

Is this clear yet?

Maybe EV is not a good word to use. Lets say you are playing a six deck shoe and there are 3 decks gone. Lets say the running count is 30, so therefore the true count would be 10. The count will only be this high for so long. There is a higher chance the count will decrease than increase. Also the shoe will last longer the lesser the amount of people in the hand (If you are lucky and the count stays high). So when the count is high it is always better to have less hands on the table. This being said if you are playing two hands you are adding another hand to the table, therefore sucking up the good cards. Is it worth it to suck up the good cards just to decrease variability?

 

[SleightOfHand]

Maybe EV is not a good word to use. Lets say you are playing a six deck shoe and there are 3 decks gone. Lets say the running count is 30, so therefore the true count would be 10. The count will only be this high for so long. There is a higher chance the count will decrease than increase. Also the shoe will last longer the lesser the amount of people in the hand (If you are lucky and the count stays high). So when the count is high it is always better to have less hands on the table. This being said if you are playing two hands you are adding another hand to the table, therefore sucking up the good cards. Is it worth it to suck up the good cards just to decrease variability?

I'm sorry, but you are the one that has made the error here. Your base assumption that playing multiple hands will make the TC faster is incorrect. The True Count Theorem states that the expected true count for the next round will be the same as the true count of the round preceding it. The true count does not dictate whether the count will increase or decrease, rather it tells you the composition of the deck. If there are other players at the table, it is better to have (at least) 2 hands rather than 1 during positive counts. The more players there are, the more hands it is optimal to play (This is only the mathematical answer. Practically, playing more than 2 will most likely raise suspicion). You made an incorrect assumption about an aspect of the game, which led you to create an incorrect conclusion. This is why the answers given did not satisfy the criteria for your question.

Heads up, it is slightly better mathematically to play 1 hand than 2, although in a practical setting, I would argue that 2 will almost always be better (much higher wr at the cost of a little bit of SCORE).

So to answer your question, in general, 2x100 > 1x200, although as others have stated, if you want to keep your RoR the same, 2x150 > 2x100 > 1x200