Ploppy would do whatever I want him to do

[assume_R]

For me, if ploppy asked, 80% of the times I told them the correct BS play. 20% I said I don't know. Only 1% of the times I told the truth. But in this case that 1% became my mistake of the day!

Eh, I don't buy that logic. The only mistake would be if you did something to affect your odds. Which you didn't. The fact that advising a player how to play his hand hurts you isn't logical, because it could have just as easily helped you. When the count is high, and you get a 17 but the dealer gets a 20, do you think it was a mistake betting too much? No. Just because something happens to hurt you once means nothing. All that matters is if you made the right decision.

Thanks for that example - it is a perfect one. I'm happy you came up with it, so there will be no question whether the example is good enough.


So there are two tens, and a deuce left.
What is the probability that I catch the ten, if the player before stands ? Obviously it is 1/3 exactly.

Now what happens if the player double his hand also?
With chance of 2/3, he will catch a ten. Leaving you with T,2 for drawing, from which you draw a ten with 50%.
This is your argumentation. Your mistake is, although this a highly probable scenario, it is not the only scenario you must account for.
Because: With chance of 1/3, the player will catch the deuce!. Then the remaining cards are T,T. For which drawing a ten happens with chance of 100%.

Since we don't know which card the player will draw, we need to average over all possible scenarios, with the probability of them happening. Then you immediatly get:
2/3 * 50% + 1/3 * 100% = 2/3. That is exactly the chance of drawing a ten, if the other player had stood on his hand.


There is another way to verify this, by deck composition. Say, one of the Ten is in hearts, while the other ten is in diamonds. Of course suits doesn't matter, but it helps if we can distinguish all three cards.
Then those three cards arrange to exactly 3!=6 possible deck arrangements:
2ThTd, 2TdTh, Th2Td, ThTd2, Td2Th, TdTh2. (first card is top card on the pile)

What happens if the other player stands, or doubles ?
2ThTd: You get a 2 (if he stands), or 10 (if he doubles)
2TdTh: 2 (stand), 10 (double)
Th2Td: 10 (stand), 2 (double)
ThTd2: 10 (stand), 10 (double)
Td2Th: 10 (stand), 2 (double)
TdTh2: 10 (stand), 10 (double)

The important thing to remember is: A pile of cards is randomly shuffled, when each arrangement of cards has equal probability. Since there are 6 different arrangements, each arrangement has the probability of 1/6.

We now see from the table above, we get a 10 in 4 out of 6 arrangements if the player stands, that is 4/6 = 2/3.
We also see that if the player doubles, we still a 10 in 4 out of 6 arrangements, that is still 4/6 = 2/3.

Whatever the other player does (the only exception is, if he depletes the pile, then the PB will reshuffle), it does not affect the probability of you getting a ten.

Those are 3 proofs. Do you need another one ?

If the argument above where true, you could never influence the probabilities of what other player would draw. Since the dealer is nothing different, you cannot influence the probabilities of the dealer drawing any card.
In fact, you cannot even influence what his downcard is. This is obvious (as it is drawn before your action). However, an undrawn card and the downcard is completely identical in terms of probability (unless the dealer has peeked, and since he did not turn up his hand you know it is not a BJ hand).
As you cannot influence the probability of the downcard, you cannot influence the probability of the card he will draw.

That makes it proof #4.

Seriously, thank you so much for posting this. This summarizes the TC Theorem. The RC has a higher chance of dropping, but your odds (which are based on the TC) on average won't change. You just have more information. That's it. Close the book. Another player's decisions won't change your odds. If the TC is high, don't try to manipulate the other players, unless it will significantly affect the amount of information you are getting (i.e. at the very end of a pitch game you want better information), or give you the chance at another round.

 

[MangoJ]

Yes, TC theorem and no-bust-card theorem (let's call it this way) are strongly connected.
Once you accept the no-bust-card theorem, one has to accept the TC theorem if one understands what TC actually is:
TC is not just RC divided by remaining cards. TC is not about running count conversion, TC is all about estimating deck composition. This is the intrinsic reason why TC helps you to beat the game.
RC is about cards from the past (information you have), TC is about cards to come (information you don't have).

Back to the strong connection of those theorems:

If the expected composition does not change when drawing a random card (no-bust-card theorem), and TC is a representation of deck composition, expected TC won't change.

Most people have problem with TC theorem, because all they understand is RC. They know RC will tend to zero (which it does), and conclude that TC must also tend to zero (because it is RC divided by "some number"). However this "some number" is the remaining deck (or more precisely number of cards left / 52) which will also tend to zero.
From school people should know, if two quantities tend to zero, you don't know anything about their fraction tending towards anything.
In fact RC tends towards zero linearly with the number of cards left (if the count is balanced), and so does the number of decks (also linearly).
If both quantities tend towards zero linearly, the fraction keeps constant.
Surprise - the fraction is the TC.

 

[aslan]

Yes, TC theorem and no-bust-card theorem (let's call it this way) are strongly connected.
Once you accept the no-bust-card theorem, one has to accept the TC theorem if one understands what TC actually is:
TC is not just RC divided by remaining cards. TC is not about running count conversion, TC is all about estimating deck composition. This is the intrinsic reason why TC helps you to beat the game.
RC is about cards from the past (information you have), TC is about cards to come (information you don't have).

Back to the strong connection of those theorems:

If the expected composition does not change when drawing a random card (no-bust-card theorem), and TC is a representation of deck composition, expected TC won't change.

Most people have problem with TC theorem, because all they understand is RC. They know RC will tend to zero (which it does), and conclude that TC must also tend to zero (because it is RC divided by "some number"). However this "some number" is the remaining deck (or more precisely number of cards left / 52) which will also tend to zero.
From school people should know, if two quantities tend to zero, you don't know anything about their fraction tending towards anything.
In fact RC tends towards zero linearly with the number of cards left (if the count is balanced), and so does the number of decks (also linearly).
If both quantities tend towards zero linearly, the fraction keeps constant.
Surprise - the fraction is the TC.

My take is different. To me, both are about cards already dealt, and both are about cards to come. What TC does is reduce it to a single deck basis, because the convention is to measure advantage on a single deck basis.

 

[assume_R]

My take is different. To me, both are about cards already dealt, and both are about cards to come. What TC does is reduce it to a single deck basis, because the convention is to measure advantage on a single deck basis.

Yes, RC / Decks = TC
is the same as
RC / Decks = TC / 1 (Deck)

It's a fine way of looking at it, but doesn't change the truth of his points at all.

 

[21gunsalute]

Only 1% of the times I told the truth.

Yes, I know I'm taking this comment out of context, but it's just screaming "Geniusism, Geniusism, Geniusism"! :laugh:

 

[jnrwilliam]

The expected running count will drop, that is very true. But the expected true count will stay, simply because the true count is a representation of the remaining deck composition (the running count isn't). Since the expected deck composition does not change when drawing a random card, the expected true count won't change.
agree, pls explain more.
i heard a cc assured repeatedly not to hit at high TC if u take insurance,
as u may 'waste' a 10.

 

[tthree]

Bad idea. Play every play to full advantage.

 

[MangoJ]

i heard a cc assured repeatedly not to hit at high TC if u take insurance,
as u may 'waste' a 10.

I'm not sure what you mean. Unless it is a no hole card game, you immediatly get from the dealers peek whether she has a 10 or not.
If it is a ENHC game (no hole card, or no peek), yes you must account for the possibility of a ten in the hole. But that does only affect doubling and splits (for the very reason that you place additional money on the table) - hits are unaffected since you lose the hand whether or not you hit/stand if dealer has a BJ.

 

[21gunsalute]

The expected running count will drop, that is very true. But the expected true count will stay, simply because the true count is a representation of the remaining deck composition (the running count isn't). Since the expected deck composition does not change when drawing a random card, the expected true count won't change.

This scenario is exactly equivalent to "taking the bust card".

I think on the board is a lengthy discussion about this, even with simulated data proving that the expected true count will not change after removing random cards.

IMO the way the True Count Theorum has been used in many of these discussions is nothing but smoke and mirrors and basically irrelevant to any actual scenario. Note the use of the word "expected" in the above statement. Terms like "expected" and "on average" may be useful for some type of statistical analysis, but they certainly do not reflect the wide range of possibilities that can and do occur in real life shoe games.

Rarely does the True Count remain constant or static throughout a shoe. Yesterday I had a shoe with a running count of -15 about a deck in for a TC of -3. Later the TC ws positive and by the end of the shoe the RC was again -15 for a TC of -15! In another shoe I had an RC of 18 with 2 decks to go for a TC of 9. When the end of the shoe came 3 hands later the RC was -6 for a TC of almost -6. In one other shoe the RC and TC kept increasing throughout the shoe. The RC ended at 18 for a TC of almost 18. You certainly are not going to get a TC of 18 early on in the shoe and yet according to the TCT...

The removal of a card may not affect the TC according to the TCT, but once again that is looking at the situation in the wrong light, IMO. The fact is that there are a finite # of cards in the shoe and a finite # of 10 value cards in the shoe. Removal of a 10 value card means there is one less 10 value card remaining to make your double down or whatever the situation may be.

The TCT may say that removing a card makes no difference, but in actual play removal of a card, especially a 10 value card, can and does often make a difference.

 

[tthree]

The true count theorem is most applicable with small numbers of cards. The larger number of cards you talk about the more likely the TC will move with the play of those cards. When you talk about 1 card the TC theorem is totally applicable to the situation.

 

[21gunsalute]

The true count theorem is most applicable with small numbers of cards. The larger number of cards you talk about the more likely the TC will move with the play of those cards. When you talk about 1 card the TC theorem is totally applicable to the situation.

Theorem:
The expected true count after a card is revealed and removed from any deck composition is the same as before the card was removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after any number of cards are revealed and removed from any deck composition is the same as before the cards were removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after a round is the same as before the round, for any balanced count, provided you do not run out of cards.

Once again, the problem is we're dealing with expected values or averages which rarely reflect actual, specific situations. Removal of cards can and often does affect the True Count despite what the expectation is according to the above corollary. And yes, the removal of even 1 card can very well make a difference despite what "expected outcome" is.

IMO the True Count Theorum is rendered meaningless for most of the discussions where it has been used here. Basically it's saying "the expectation is what the expectation was, even when what actually happens does not reflect these expectations, which is most of the time in specific, individual situations". Basically it's just a bunch of double talk.

 

[paddywhack]

Theorem:
The expected true count after a card is revealed and removed from any deck composition is the same as before the card was removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after any number of cards are revealed and removed from any deck composition is the same as before the cards were removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after a round is the same as before the round, for any balanced count, provided you do not run out of cards.

Once again, the problem is we're dealing with expected values or averages which rarely reflect actual, specific situations. Removal of cards can and often does affect the True Count despite what the expectation is according to the above corollary. And yes, the removal of even 1 card can very well make a difference despite what "expected outcome" is.

IMO the True Count Theorum is rendered meaningless for most of the discussions where it has been used here. Basically it's saying "the expectation is what the expectation was, even when what actually happens does not reflect these expectations, which is most of the time in specific, individual situations". Basically it's just a bunch of double talk.

Keep beating that horse.......

 

[assume_R]

Terms like "expected" and "on average" may be useful for some type of statistical analysis, but they certainly do not reflect the wide range of possibilities that can and do occur in real life shoe games.

The removal of a card may not affect the TC according to the TCT, but once again that is looking at the situation in the wrong light, IMO. The fact is that there are a finite # of cards in the shoe and a finite # of 10 value cards in the shoe. Removal of a 10 value card means there is one less 10 value card remaining to make your double down or whatever the situation may be.

The TCT may say that removing a card makes no difference, but in actual play removal of a card, especially a 10 value card, can and does often make a difference.

It isn't a matter of opinion. The theorem is proven, and you are not fully understanding it. The point is, yes, getting dealt a 10 changes the composition and the count. But you are forgetting the chance of drawing that 10 beforehand.

The whole point of "expected" and "averages" is that it does reflect the wide range of possibilities. In fact, it, by definition, reflects the entire range of possibilities.

You are the one looking at it in the wrong light. The fact that a 10 came out indeed means one less 10 for you. But if the TC was +10 beforehand, there was a higher chance of getting dealt that 10.

The TC's in your examples changed, yes. Nobody is arguing that. But would you be able to predict how it would have changed? No, because on average it has as much a chance of decreasing as it does of increasing. That's all the TCT says. On average it stays the same. Not that you should predict it to remain the same. But that it won't tend towards one way or another.

If I flip a coin 50 times, the expected value is 25 times heads and 25 times tails. But that doesn't mean I can predict it will happen to be exactly 25 and 25. Yet that doesn't change the usefulness of knowing the expected value.

 

[MangoJ]

"Average" and "expected" are used in a statistical context. The statement "on average X will happen" (and similar) doesn't mean that you are most likely to observe X, in fact X may never happen.

The average birth rate per women is - say- 1.7. That means, "each woman has given birth to an expected number of 1.7 children". This may be a true statement in purely statistical terms, although nobody has ever observed a women with 1.7 children.
You could then argue, that the value of 1.7 has absolutely no meaning for practical life. It may be for your life (you don't plan your life for 1.7 children).
However this statistical number is important for future planing of comunities, pensions, health care ...


Now there is the statement of the TCT "drawing any number card will not change the expected TC". This is a true statement. You may or may not be able to observe it. But nevertheless it is a true statistical statement (which can be - and have been - proven). You might think "I don't care, this is the shoe I'm playing, and I don't want someone else to draw a card since TC could change". This is your perfect right of your opinion. I also don't care if my wife is expected to have 1.7 children. I rather like to have 2 children with her regardless of average birth rate.

I'm not sure you fully catch the term "expected", it is: in a (imaginary) arbitrarly large number of identical situations (those doesn't exist of course), if you look what happens in each of those situations, the average value of X in those situations is the "expected value of X".
Nothing more, nothing less.

If the other player draws a 10, it drops the TC. If the other player draws a non-10, it increases the TC. You can quite simple construct those large number of identical situations: draw a card, calculate new TC, return the drawn card, and reshuffle the deck. The TC theorem just states: The expected TC is the TC before the draw exactly.
The actual draw might help you, it might hurt you. If you prefer to keep the favourable TC (in terms of risk aversion), then you are better of not drawing the card. However on average it doesn't make a difference.

 

[aslan]

Yes, RC / Decks = TC
is the same as
RC / Decks = TC / 1 (Deck)

It's a fine way of looking at it, but doesn't change the truth of his points at all.

Maybe only to me.

 

[aslan]

"Average" and "expected" are used in a statistical context. The statement "on average X will happen" (and similar) doesn't mean that you are most likely to observe X, in fact X may never happen.

The average birth rate per women is - say- 1.7. That means, "each woman has given birth to an expected number of 1.7 children". This may be a true statement in purely statistical terms, although nobody has ever observed a women with 1.7 children.
You could then argue, that the value of 1.7 has absolutely no meaning for practical life. It may be for your life (you don't plan your life for 1.7 children).
However this statistical number is important for future planing of comunities, pensions, health care ...


Now there is the statement of the TCT "drawing any number card will not change the expected TC". This is a true statement. You may or may not be able to observe it. But nevertheless it is a true statistical statement (which can be - and have been - proven). you might think "I don't care, this is the shoe I'm playing, and I don't want someone else to draw a card since TC could change". This is your perfect right of your opinion. I also don't care if my wife is expected to have 1.7 children. I rather like to have 2 children with her regardless of average birth rate.

I'm not sure you fully catch the term "expected", it is: in a (imaginary) arbitrarly large number of identical situations (those doesn't exist of course), if you look what happens in each of those situations, the average value of X in those situations is the "expected value of X".
Nothing more, nothing less.

If the other player draws a 10, it drops the TC. If the other player draws a non-10, it increases the TC. You can quite simple construct those large number of identical situations: draw a card, calculate new TC, return the drawn card, and reshuffle the deck. The TC theorem just states: The expected TC is the TC before the draw exactly.
The actual draw might help you, it might hurt you. If you prefer to keep the favourable TC (in terms of risk aversion), then you are better of not drawing the card. However on average it doesn't make a difference.

. So there may be an actual advantage but not a statistical advantage?

 

[MangoJ]

. So there may be an actual advantage but not a statistical advantage?

Your expected value of the double won't change with an additional card. That means you will double your hand regardless of the card drawn before. There will be no advantage or disadvantage by that card if you double in any case. The reason is, that the expected TC won't change, and since you double in all possible scenarios, the expected value of your double won't change.

If you actually allow to hit instead, if a helping card has been drawn before, it does give you an advantage in your "decision", because now you have an option to hit or to double. If you know the drawn card, you have a better estimate when a double or when a hit would be more efficient, you hit if the deck turns unfavourable, and you double if it turns favourable.

By knowing the drawn card, you do have an advantage. However if will only be a non-zero advantage if the drawn card could actually change your decision. If you double regardless of the drawn card, there will be no advantage increase.

What follows from that: A additionally known card can never turn into a disadvantage (for that hand), since you can always chose to ignore this cards drawn, which in turn gives you a "strategy" that doesn't give a advantage (but also no disadvantage). However this ignoring cannot be "selective" in any way, because if you chose to ignore that card after you have seen the value of the card. you're not ignoring that card, but just making a bad strategy decision.

 

[assume_R]

. So there may be an actual advantage but not a statistical advantage?

Your expected value of the double won't change with an additional card. That means you will double your hand regardless of the card drawn before. There will be no advantage or disadvantage by that card if you double in any case. The reason is, that the expected TC won't change, and since you double in all possible scenarios, the expected value of your double won't change.

If you actually allow to hit instead, if a helping card has been drawn before, it does give you an advantage in your "decision", because now you have an option to hit or to double. If you know the drawn card, you have a better estimate when a double or when a hit would be more efficient, you hit if the deck turns unfavourable, and you double if it turns favourable.

By knowing the drawn card, you do have an advantage. However if will only be a non-zero advantage if the drawn card could actually change your decision. If you double regardless of the drawn card, there will be no advantage increase.

What follows from that: A additionally known card can never turn into a disadvantage (for that hand), since you can always chose to ignore this cards drawn, which in turn gives you a "strategy" that doesn't give a advantage (but also no disadvantage). However this ignoring cannot be "selective" in any way, because if you chose to ignore that card after you have seen the value of the card. you're not ignoring that card, but just making a bad strategy decision.

Or, put another way, it's like sitting at third base instead of first. Slightly more accurate decisions if you choose to use indices. But barely noticeable in a shoe game, but it exists. That's the advantage of having the player before you hit. But that could go away if it forces the shoe to end a round early

 

[tthree]

. So there may be an actual advantage but not a statistical advantage?

They explained it better than I did. Actually these two, Assume_R and MangoJ, are the people who I respect the most when it comes to math from the posts I have read. Look at it this way, all parts of the shoe that are unseen comprise the TC. Any block of cards is expected to have this TC value as much as any other block. The count being high doesn't make the next block of cards any more likely than any other to be high or low cards. Until they are seen all unseen cards are equally likely to be a high or low card as predicted by the count.

I really explained it about as bad as can be before. What I was trying to explain was the way people were misapplying the TC theorem, that effect is seen most in small numbers of cards not the theorem itself.

 

[assume_R]

Look at it this way, all parts of the shoe that are unseen comprise the TC. Any block of cards is expected to have this TC value as much as any other block.

I like this way of thinking about it!

Yes, all the blocks will probably be different, but they are all have an "expected TC" of the current TC.