Without Counting, best point in a Shoe?

[KenSmith]

Good catch. It should read:
"The running count must be zero at the end of the deck. Therefore, drawing cards in high counts tends to cause the running count to fall, and drawing cards in low counts tends to cause the running count to rise. But the expected true count is unchanged."

 

[dacium]

I don't think this is a question of the true count not effecting the house edge.

Its is obviously that the distribution of cards at the end of the shoe is random but balanced in the low run. There can be no doupt that the average count is over many shoes would be zero.

However this does not prove that the house edge has not declined. Unless you can prove a direct relationship between count and house edge you cannot say that a average count though out the deck means that the house edge should stay constant thoughout the deck. This is clearly not the case because of the variable number of cards each hand can end up drawing.

 

[sagefr0g]

I don't think this is a question of the true count not effecting the house edge.

Its is obviously that the distribution of cards at the end of the shoe is random but balanced in the low run. There can be no doupt that the average count is over many shoes would be zero.

However this does not prove that the house edge has not declined. Unless you can prove a direct relationship between count and house edge you cannot say that a average count though out the deck means that the house edge should stay constant thoughout the deck. This is clearly not the case because of the variable number of cards each hand can end up drawing.

apparently it's a statistical thing. for each individual shoe the edge does change either in favor of the house or player. but over many shoes those advantages, disadvantages cancel one another out leaving a net affect of no change in advantage through out the average shoe.
thats my take on it but i really haven't tryed to understand the proof just glossed it over. it looked pretty well layed out thought.
best regards,
mr fr0g

 

[Sonny]

Here’s a little experiment that might clear things up:

Take a shuffled deck of cards, remove the top card and place it face down on the table in front of you. Without looking at the card think about what the count for the remaining 51 cards is. It is the same! Since you don’t know what card was removed you cannot alter your count or update the house edge. You cannot estimate the value of the card either because it is equally likely to be a high or low card (assuming you use a balanced count).

Now do the same experiment but flip the card face up on the table. With enough trials you will see that each card is removed the same number of times so the overall count cancels itself out.

Now take six decks of shuffled cards and “burn” the first 260. What is the count? It is the same! It is equally likely that high and low cards have been removed. In the long run the counts will cancel out to zero.

This is what BS play is like – playing blindly.

However, I don’t think that the actual house edge would quite cancel like the count does. The card tags (values) are based on the effects of removal, but the effects of removal do not sum to zero the way many count systems do. Even though the average count anywhere in the shoe should be zero, the actual house edge may not correspond exactly. By my estimation, the average change in house edge after the removal of one card from a single deck is roughly -0.003% (-0.0001% fo 6D) even though the average TC is still zero. In this case the average house edge would actually climb slowly as more decks are dealt, although the change would be negligible.

-Sonny-

 

[sagefr0g]

Here’s a little experiment that might clear things up:

Take a shuffled deck of cards, remove the top card and place it face down on the table in front of you. Without looking at the card think about what the count for the remaining 51 cards is. It is the same! Since you don’t know what card was removed you cannot alter your count or update the house edge. You cannot estimate the value of the card either because it is equally likely to be a high or low card (assuming you use a balanced count).

Now do the same experiment but flip the card face up on the table. With enough trials you will see that each card is removed the same number of times so the overall count cancels itself out.

Now take six decks of shuffled cards and “burn” the first 260. What is the count? It is the same! It is equally likely that high and low cards have been removed. In the long run the counts will cancel out to zero.

This is what BS play is like – playing blindly.

kewl experiment i'm gonna try that :bomb:

However, I don’t think that the actual house edge would quite cancel like the count does. The card tags (values) are based on the effects of removal, but the effects of removal do not sum to zero the way many count systems do. Even though the average count anywhere in the shoe should be zero, the actual house edge may not correspond exactly. By my estimation, the average change in house edge after the removal of one card from a single deck is roughly -0.003% (-0.0001% fo 6D) even though the average TC is still zero. In this case the average house edge would actually climb slowly as more decks are dealt, although the change would be negligible.

yeah they talk about the running count doesn't obey the same laws as the true count. drawing cards does effect how the running count is expected to behave in that it will tend to rise when cards are removed if the true count is negative and fall when cards are removed if the true count is positive. so it's really where we get advantageous results is from the fluctuation of the running count but it is the value of the true count that tips us off as to the degree of volatility of the running count. i'm thinking the house edge obeys similar rules as the running count. that is to say it goes up and down like the running count for the same reasons. this is for one single shoe at a time. so for a given shoe the house edge is all over the place volatile as the running count is. but the frequency distributions for true counts is what tells the story as far as what this all means in the long run. the frequency distributions for true counts (like those found in Wong's Professional Blackjack appendix C pages 286-290) show a normal distribution of negative and positive values centered around a TC=0 . these values are for many, many shoes not just one. but the point is that with such a normal distribution one would expect that the average true count for the average shoe would be zero. i suppose this is the gist of what they talk about in referance to the true count theorem when they state: "Expected value is a precise mathematical term defined as the mean average, which is computed by summing the probability of an event times the value of that event, over all possible events. So the expected value of the true count after drawing a card is the summation of the probability of drawing each card times the value of the true count after drawing that card." see link: http://www.bjmath.com/bjmath/counting/tcproof.htm (Archive copy)
in other words i guess the true count theorem has more meaning to us from the long term perspective than the short term. so for an individual shoe it's not going to mean the true count is really zero at any number of decks left in a pack hence there is no reason to think that the house advantage is going to improve for us once there is only one deck left in the pack to be dealt. the true count is just as likely to be negative, neutral or positive for any given shoe from our short term perspective. but just playing a pack in the latter near one deck stages is not going to afford us a advantage similar to that found for playing single deck blackjack because the effects of all those card combination possibilities of the original pack is going to be felt. that being the range of true count frequencies that are possible and will happen in the long run. we got just one sample of those happening in any given shoe. so the true count could be any where for a given shoe and so could the house advantage. in the long term there is no effect on the house advantage because the true count frequencies either cancel out or are equal to zero. the only time we could say that the house edge is lowered by reaching one deck left to be dealt in the pack is when we in fact know that the true count is zero at that point. but it would not be true that in general we see a decrease in advantage once a pack has been dealt out to one deck remaining.

best regards,
mr fr0g

 

[dacium]

I think all this is clear:

The average count is always zero.

A count that is not zero always tends to zero after drawing a certain amount of cards. But the true count will remain the same.

What isn't clear is that do you actually have the same chance of winning with a certain true count at the start of a shoe as you do at the end of a shoe? If the ratio of highs to low is the same does that mean on average the hand is played exactally the same? The answer seems to be no, as we all know there are subtle difference between 4 deck and 6 deck games. I think this is to do with the values assigned to the counts. We all know +1 for 2 though 6, -1 for T through Ace. But this isn't prefect and suits a certain number of cards more than others.

 

[sagefr0g]

I think all this is clear:

The average count is always zero.

A count that is not zero always tends to zero after drawing a certain amount of cards. But the true count will remain the same.

What isn't clear is that do you actually have the same chance of winning with a certain true count at the start of a shoe as you do at the end of a shoe? If the ratio of highs to low is the same does that mean on average the hand is played exactally the same? The answer seems to be no, as we all know there are subtle difference between 4 deck and 6 deck games. I think this is to do with the values assigned to the counts. We all know +1 for 2 though 6, -1 for T through Ace. But this isn't prefect and suits a certain number of cards more than others.

your point about "We all know +1 for 2 though 6, -1 for T through Ace. But this isn't prefect and suits a certain number of cards more than others."is true how ever i don't believe that justifies changing how you would bet if a given TC presents early in a shoe or late in a shoe. such an effect would carry through all stages of a shoe.
the fact is if you neglect putting out your optimal bet for a given TC when it presents early in shoe you slight your self the same EV that you would credit your self with should you bet on that same TC in the latter stages of the shoe.
regardless of the fact that there are subtle difference between 4 deck and 6 deck games the determinant factor is what type of shoe do you start out with a 4 deck or 6 deck or what ever. it is with iin the original shoe that the TC frequency distributions spring forth and the reason that they spring forth as they do has to do with the unique nature of that original shoe not from the point at which you happen to find your self in the shoe as it is being dealt out.

best regards,
mr fr0g

 

[sagefr0g]

I think all this is clear:

The average count is always zero.

A count that is not zero always tends to zero after drawing a certain amount of cards. But the true count will remain the same.

What isn't clear is that do you actually have the same chance of winning with a certain true count at the start of a shoe as you do at the end of a shoe? If the ratio of highs to low is the same does that mean on average the hand is played exactally the same? The answer seems to be no, as we all know there are subtle difference between 4 deck and 6 deck games. I think this is to do with the values assigned to the counts. We all know +1 for 2 though 6, -1 for T through Ace. But this isn't prefect and suits a certain number of cards more than others.

hmm after i initially responded to your post i realized something that i had been overlooking when considering this issue. that being the fact that the true count frequency distributions as tabulated by Wong in Professional Blackjack indicates that the zero true count is the highest value frequency wise of all the tabulated true counts and the only value that does not have an opposite counterpart (example tc=-1 & tc=+1).
well there are -0 & +0 but there is no opposite counterpart for a pure tc = 0.
the point is that a pure true count of zero is essentially the only true count value that doesn't have a counterpart value to cancel it's self out in the long run. it seems logical that the frequency distributions for a given pack would be reflected in the frequency distributions of a partial pack. to me that would mean that the true count value of zero is going to have the predominate effect in the long run over all the other true count values as they would cancel each other out. that would make a long run effect of any partial pack left to be dealt to be predominated by the true count value of zero. that would mean that in the long run there would be a slight advantage for the basic strategy non-counting player to enter a pack when it is down to one deck to be dealt out. of course the problem with that is there are virtually no games that deal out the last deck in a pack of a multi-deck game.

best regards,
mr fr0g