First base or third base?

sagefr0g

sagefr0g

Well-Known Member
#21
#21
zengrifter said:
Simplified, it means that the TC has no tendency towards zero (except
right at the end of shoe, unlike the RC which always tends towards zero. zg
Click to expand...
it's interesting the relation ship between the true count and the running count. how the true count 'sluggishly' follows the more 'volatile & intense' running count. the running count soars to great heights and dismal lows while the true count just kind of sloggs along as if 'reluctantly' but 'faithfully' following the RC. the RC gets to go by however the high or low cards come out of the pack but the TC has to pay attention to the high or low cards comming out and the size of the pack remaining to be dealt. lol .
when the high & low cards are arranged in a pack symmetrically the RC & TC tend to follow one another closely and are virtually indistinquishable but if the high and low cards are arranged asymetrically the RC can go shooting up dramatically with the TC trailing along behind with less dramatic flair. :rolleyes:
 
K

Knox

Well-Known Member
#22
#22
How could it not be more advantageous to sit at 1st base? When the count is high, it would seem the next card out would be more likely to be a 10 or A. The further down the line you are, at a crowded 6D table, it seems you would be more likely to be watching the players to your right get the good cards then you would be to get them.
 
Sonny

Sonny

Well-Known Member
#23
#23
Knox said:
When the count is high, it would seem the next card out would be more likely to be a 10 or A.
Click to expand...
Why? If the cards are in random order then there is no reason to expect the tens to come out sooner rather than later. For example, let's say that I take a single deck of cards and remove all the fives. Now we have a nice positive TC. Would you rather have me deal you a hand from the top, middle or bottom of the deck? What is the probability that the top card is an ace? What is the probability that the second card is an ace? What is the probability that the bottom card is an ace? They're all the same. :)

-Sonny-
 
K

Knox

Well-Known Member
#24
#24
I was figuring if you had 10 cards left and 9 out of 10 were aces, the next card is 90% likely to be an Ace. So if the ace comes out, the next guy only has an 8/9 probability of getting an Ace. Of course, if someone gets that one crap card, then everyone else gets an Ace. So hell if I know, it just sounded intuitive to me!

:grin:
 
Ferretnparrot

Ferretnparrot

Well-Known Member
#25
#25
I dont think thats true as the farther the running count gets from zero the less likely it is to go any further, and thus the more likely it is to go the opposite way. I think at any given time as one pool of cards becomes more dominant in the remaining cards, even if by only one card, it is still MORE likely that the next card out will be one of the larger pool than one of the smaller pool. This holds true for a balanced level one system but possibly not for level 2 or 3 since the rc is based on the weight of cards not the quantity

The only time i used this concept in play was when i was playing multi action blackjack, and the count was high enough by only one card to justify taking insurance. I had to draw cards before the dealer and i had to make my insurance decision before i draw cards so Rather than taking insurance for all three dealer pulls i only took it for the first one. I dont know if it was mathematically sound, but this concept is what swayed my decision, the cool part was the dealer got blackjack ONLY on the first pull and not the other two so everybody thought i was some kinda genious.
 
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godeem23

Well-Known Member
#26
#26
Knox said:
I was figuring if you had 10 cards left and 9 out of 10 were aces, the next card is 90% likely to be an Ace. So if the ace comes out, the next guy only has an 8/9 probability of getting an Ace. Of course, if someone gets that one crap card, then everyone else gets an Ace. So hell if I know, it just sounded intuitive to me!

:grin:
Click to expand...
Sometimes intuition misguides us. Even in your example, it makes no difference which card you take, even if you wait for cards to be "removed" first.
 
Ferretnparrot

Ferretnparrot

Well-Known Member
#27
#27
Sonny said:
Why? If the cards are in random order then there is no reason to expect the tens to come out sooner rather than later. For example, let's say that I take a single deck of cards and remove all the fives. Now we have a nice positive TC. Would you rather have me deal you a hand from the top, middle or bottom of the deck? What is the probability that the top card is an ace? What is the probability that the second card is an ace? What is the probability that the bottom card is an ace? They're all the same. :)

-Sonny-
Click to expand...
cmon this is the fundamental principal behind card counting, as the beneficial cards become more dominant in the remainder of the deck they are MORE liekly to be drawn Take out any number of cards that arnt aces and the probibility of drawing an ace WILL go up, no more cocain ya hear?

Your betting on probibility not the actual value of the next card out of the deck, thats what ploppies due, play their hands on what they think the next card is, let there be no ploppies on this forum.
 
EasyRhino

EasyRhino

Well-Known Member
#28
#28
Whoah now.

A shoe with a high count is more likely to have high cards come out, sure. But you don't know if they'll be early, middle, or late in the shoe (too late, i.e. behind the cut card). You can pretty much average itassume that the cards will come out "evenly" distributed throughout the shoe.

So in such an evenly-distributed shoe, more high cards are coming out, lowering the running count. But the number of decks remaining is also dropping, meaning that the true count will only decline very slowly, until the last card in the deck is finally reached.
 
zengrifter

zengrifter

Banned
#29
#29
sagefr0g said:
it's interesting the relation ship between the true count and the running count. how the true count 'sluggishly' follows the more 'volatile & intense' running count.
Click to expand...
According to the TCT the TC does not even slugishly follow the RC. zg
 
sagefr0g

sagefr0g

Well-Known Member
#30
#30
zengrifter said:
According to the TCT the TC does not even slugishly follow the RC. zg
Click to expand...
i guess that's true where it's stated that the TC and the RC follow differant rules.
in my mind (version) it does though, lol. i mean heck you gotta figure the RC first then the TC but i guess it's a which came first the chicken or the egg sorta thing....... but then the RC is what it is and the TC is what it is regardless, right? but if you look at a graph you can see a correlation going on there. all those little ups and downs look corelated to me. correlated by the math involved, the TC that we can know is dependant on the RC that we can know. :rolleyes:
 

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Dopple

Well-Known Member
#31
#31
The thing I hate about first is that you are still going the calcs for your index plays after a round of cards and "boom" the dealer is on you to decide if you want a hit or not.
 
blackjack avenger

blackjack avenger

Well-Known Member
#32
#32
Very Well Stated Sonny

Sonny said:
Why? If the cards are in random order then there is no reason to expect the tens to come out sooner rather than later. For example, let's say that I take a single deck of cards and remove all the fives. Now we have a nice positive TC. Would you rather have me deal you a hand from the top, middle or bottom of the deck? What is the probability that the top card is an ace? What is the probability that the second card is an ace? What is the probability that the bottom card is an ace? They're all the same. :)

-Sonny-
Click to expand...
Sonny is right!
 
blackjack avenger

blackjack avenger

Well-Known Member
#33
#33
The Running Count Changes While the True Count Remains the Same

Ferretnparrot said:
I dont think thats true as the farther the running count gets from zero the less likely it is to go any further, and thus the more likely it is to go the opposite way. I think at any given time as one pool of cards becomes more dominant in the remaining cards, even if by only one card, it is still MORE likely that the next card out will be one of the larger pool than one of the smaller pool. This holds true for a balanced level one system but possibly not for level 2 or 3 since the rc is based on the weight of cards not the quantity

The only time i used this concept in play was when i was playing multi action blackjack, and the count was high enough by only one card to justify taking insurance. I had to draw cards before the dealer and i had to make my insurance decision before i draw cards so Rather than taking insurance for all three dealer pulls i only took it for the first one. I dont know if it was mathematically sound, but this concept is what swayed my decision, the cool part was the dealer got blackjack ONLY on the first pull and not the other two so everybody thought i was some kinda genious.
Click to expand...
I am afraid you fell into the trap of multi action blackjack. The trap is to somehow be influenced and play your hands differently. The correct play was to take insurance on all three hands if it was at the correct indice.

Read Sonny's post regarding the removal of 5s from a single deck to perhaps clear things up. Think about remvoing all the 5s and 6s from a deck and then deal a hand and face the insurance decision. It would not matter where from the deck the dealers hole card came from. Insurance would be the correct play.

You need to stop trying to reinvent the wheel on the fly. Doing things incorrectly can be very costly.

The running count changes while the TC tends to stay the same. You have 3 remaining decks. Your RC is 6. Your TC is 2. A deck goes by. On average you now have a RC of 4. Your TC is still 2. If you continue this till the end of the shoe and the very last card you will get your balanced count. 0 RC and 0 TC.
 
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