The TC Theorem is Still Denied by Some ...

21gunsalute said:

Actually it has everything to do with the discussion. What goes up must come down. A high count must trend back toward zero if a significant number of cards have been played since it must eventually reach zero, and the more cards that have been played, the more likely the count is to drop.

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How many times must it be said that the TC is not zero is when there are zero cards left in the shoe? Zero (RC) divided by zero (Decks Remaining) is not zero (TC); It is undefined; it is not small, nor is it large; it is just that, undefined.

Also note that, as I have previously said, when there is one card left in the shoe, the TC is either -52, 0 or +52. If four decks remain, and the TC is +12, then we know that there is an abundance of negative cards in those 4 decks (with respect to 2s,3s,4s,5s, and 6s). Therefore, from this point, it is more likely that the last card in the shoe will be a 10 or A, thus making the TC more likely to be -52 than +52. In some cases, however, the last card will be a 2,3,4,5, or 6, thus making the TC -52 when only one card remains.

(Note that TC = -52 to TC = 52 covers the range of all possible TC's for a level one system).

As you can see, when the TC is +12 when 4 decks remain, the TC when one card remains in the shoe is most likely to be +52, but it will not always be +52. As it will sometimes be -52 when one card remains, but not as often as it will be +52 when one card remains, it will average out to the following:

When the TC is X after Y decks have been dealt out, the average TC when there is only one card left in the shoe will, too, be X.

Spaw

An equivalent statement to the TC theorem is: The distribution of random cards is homogeneous.

Imagine you have a 8 deck shoe, and after 2 decks dealt the RC is +6. We know that there are 6 high cards in 6 remaining decks, and this distribution is called (TC=1).
What does it mean, that distribution cards is homogeneous ? That means that we expect the same distribution (=concentration) of high cards in the first deck (of those 6 undealt decks), in any given section of the shoe, and even in the last deck of the shoe. Since this concentration is called TC=1, we expect a TC=1 for the last deck in the remaining shoe. Not only to the last deck, but also down to the very last card of the shoe.


Most people unsure about TC theorem confuse RC with TC. RC is information about the dealt cards, while TC is information about the undealt cards. The connection between those two regimes is the homogeneous distribution of shuffled cards.
Of course drawing through the shoe, we won't see TC=1 for the last deck very often. When we deal new cards, we gain additional information about the undealt cards, and hence this will change TC. But this doesn't mean TC=1 wasn't expected.

For people in doubt, I suggest an experiment you can do on the kitchen table. Get 8 decks, shuffle, and draw 104 card (2 decks) from this shoe. Switch cards untill you get a RC of +6. Set those 2 decks aside.
Now repeat the following 100 times:
- Shuffle the remaining 6 decks
- Draw (and count) 5 decks of those
- Calculate the TC for the remaining 1 deck
- Make a note of the TC you got for that deck.

After you repeat this 100 (or better 1000 times), see the distribution of TCs you found for that last deck. Get the average value of that distribution, it should be +1.

Southpaw said:

How many times must it be said that the TC is not zero is when there are zero cards left in the shoe? Zero (RC) divided by zero (Decks Remaining) is not zero (TC); It is undefined; it is not small, nor is it large; it is just that, undefined.

Also note that, as I have previously said, when there is one card left in the shoe, the TC is either -52, 0 or +52. If four decks remain, and the TC is +12, then we know that there is an abundance of negative cards in those 4 decks (with respect to 2s,3s,4s,5s, and 6s). Therefore, from this point, it is more likely that the last card in the shoe will be a 10 or A, thus making the TC more likely to be -52 than +52. In some cases, however, the last card will be a 2,3,4,5, or 6, thus making the TC -52 when only one card remains.

(Note that TC = -52 to TC = 52 covers the range of all possible TC's for a level one system).

As you can see, when the TC is +12 when 4 decks remain, the TC when one card remains in the shoe is most likely to be +52, but it will not always be +52. As it will sometimes be -52 when one card remains, but not as often as it will be +52 when one card remains, it will average out to the following:

When the TC is X after Y decks have been dealt out, the average TC when there is only one card left in the shoe will, too, be Y.

Spaw

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Once again, you may call it whatever you like but for all practical purposes, zero divided by zero is zero. It's also irrelevant since we've run out of cards at this point.

Your second paragraph is also irrelevant and somewhat confusing. I've never seen a game that gets dealt to the last card, but calling the TC +52 or -52 is overkill at this point, although technically correct I guess. So I guess if there were four cards left and a running count of 1 the TC would be 13. Sounds impressive, but what good is this info? You could have 2 high cards, 1 low card and 1 that falls into the 7,8 or 9 category. You could have 1 high card and three that fall into the latter category. Such info would not seem to be very useful, but once again, I've never seen such penetration rendering such a discussion irrelevant. It doesn't matter what the last card in the shoe is because we'll never get to that point, but even if we did there would still be a 50% degree of uncertainty because of the unseen cut card.

"When the TC is X after Y decks have been dealt out, the average TC when there is only one card left in the shoe will, too, be Y."

I'm pretty sure you meant to say:

When the TC is X after Y decks have been dealt out, the average TC when there is only one card left in the shoe will, too, be X.

This info is also irrelevant due to my argument above, but more glaring is the fact that it totally contradicts the argument you just made above:

"Also note that, as I have previously said, when there is one card left in the shoe, the TC is either -52, 0 or +52."


I would argue that the average TC when there is one card left in the shoe is zero no matter what the TC is after Y decks have been dealt. It will either be a high card (-1), a low card (+1) or a zero card (7, 8 or 9). It matters not how the cards randomly came out earlier in the shoe, these are the only possibilities and the number of times a high card comes out on the last card should = the number of low cards that come out on the last card, thereby canceling each other out. Once again though this info useles because cards are not dealt down to the last card.

21gunsalute said:

I would argue that the average TC when there is one card left in the shoe is zero no matter what the TC is after Y decks have been dealt.

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This is true, but under different conditions than the application of the TC theorem. The average TC of the last card from a complete shoe (after shuffling) is indeed 0.

The TC theorem is about a different, more special, scenario.The scenario is that the first Y cards of the shoe have a RC of X for sure (i.e. you counted those cards yourself). Under this (special) conditions, the TC of the last card of the shoe is nonzero.

MangoJ said:

This is true, but under different conditions than the application of the TC theorem. The average TC of the last card from a complete shoe (after shuffling) is indeed 0.

The TC theorem is about a different, more special, scenario.The scenario is that the first Y cards of the shoe have a RC of X for sure (i.e. you counted those cards yourself). Under this (special) conditions, the TC of the last card of the shoe is nonzero.

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I would argue that it still doesn't matter, unless you have irrefutable evidence on say the next to last card, such as a count of 2 with one card to go (+ the burn card). In any other scenario the last card should average zero since the cards still come out randomly.

21gunsalute said:

Once again, you may call it whatever you like but for all practical purposes, zero divided by zero is zero.

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It is also every number (including imaginary ones) above and below zero. All of them. Also, none of them. It is NOT DEFINED! For a system based on math, the ignorance of such a basic (you should've learned this in high school, I know I did) concept truly boggles the mind. When people mess up math and then defend their obvious error, it gets to me really really bad.

Once again 21, I, respectfully, tell you that zero divided by zero is not zero for "all practical purposes." Contending so will get you into more trouble than betting 2.5 kelly in practical applications. Equations must be worked in such a way to remove the problem to arrive at the correct solution. Think limits or the case of extraneous solutions.

The reason I discuss nonsensical items such as shoes dealt to the last card is not because they are realistic but because you have made the incorrect argument that the TC must decline in magnitude since the TC becomes zero at the end of the shoe. I have already shown you that the TC will not be zero when zero cards remain in the shoe, as it is undefined. The next thing most will bring to the table is what happens when there is only card remaining.

As you still seem to not be getting my point about predicting the average TC of the last card from the vantage point of there being 4 decks remaining with a TC of +12, I am going to use a simplified example first.

Say you have a barrel of apples. Within the barrel, there are 8 red apples and 4 green apples. You decide to randomly pull out apples until only one remains. Is it more probable that the last apple will be red or green? Of course, red. This is no different than asking what the probability of what any randomly picked apple will be; 8/12.

Now apply this to the case of a shoe with 4 decks remaining when the TC is +12. There are more 10s and A's (red apples) than 2s, 3s, 4s, 5s, and 6s (green apples). When we get down to the last card in the shoe, it will be more probable that the last card in the shoe will be a 10 or A, thus making the TC +52. But it will not always be a 10 or A, so it will average out to be lower. In fact, because of the homogeneous distribution principle, it will average out such that the TC will average out to be +12 as it was 4 decks previously.

Indeed the argument can be extended to any point (not just when one card remains) in the shoe when the TC was known at a previous point.

Spaw

BrianCP said:

It is also every number (including imaginary ones) above and below zero. All of them. Also, none of them. It is NOT DEFINED! For a system based on math, the ignorance of such a basic (you should've learned this in high school, I know I did) concept truly boggles the mind. When people mess up math and then defend their obvious error, it gets to me really really bad.

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You're missing the point. It doesn't matter what you call it. It's zero in my book and it's zero for all practical purposes. Zero is nothing. It doesn't matter if nothing divided by nothing is nothing, infinity or undefined. But just for arguments sake, you just defined it as every number and no numbers. How can you define something that you claim is undefined? And why are we arguing about nothing?

Southpaw said:

Once again 21, I, respectfully, tell you that zero divided by zero is not zero for "all practical purposes." Contending so will get you into more trouble than betting 2.5 kelly in practical applications. Equations must be worked in such a way to remove the problem to arrive at the correct solution. Think limits or the case of extraneous solutions.

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Once again, it doesn't matter what you call it. If you have a TC of 6 with 3 decks left you won't have a TC of 6 with zero cards left. You won't have a TC of 6 with 1 card left. You won't have a TC of 6 with 2 cards left. You won't have a TC of 6 with 3 cards left. If that's what the average should be then it's of little to no use because it isn't a reflection of any real world scenario. You said with one card left the TC would either be 52, -52 or zero and none of the figures is remotely close to 6.

But I dispute almost your entire argument. Please do me a favor and run some sims. Take a certain TC at various places in the shoe and see if the last card averages close to zero. I would imagine the higher (or lower) the count and the deeper into the shoe you get the further from zero the last card will tend to be. But I'll bet it's not a slam dunk until you get well into the last deck at extremely high (or negative) counts. That's because the cards still come out randomly and the last high card (or low card) should have (nearly) the same chance of coming out on the next to last card as on the last card. That's my theory anyway.

21gunsalute said:

But I dispute almost your entire argument. Please do me a favor and run some sims. Take a certain TC at various places in the shoe and see if the last card averages close to zero.

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If you take a deck with TC x and deal it out a billion times, the TC will average x at every point for the rest of the deck. Even with one card left.

I would imagine the higher (or lower) the count and the deeper into the shoe you get the further from zero the last card will tend to be. But I'll bet it's not a slam dunk until you get well into the last deck at extremely high (or negative) counts.

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This is true. As the # of decks gets smaller and even becomes fractions, your TC numbers will be further from zero. Going back to the sims you are proposing, when you average the results out you will see that the TC = TC wherever you started measuring.

That's because the cards still come out randomly and the last high card (or low card) should have (nearly) the same chance of coming out on the next to last card as on the last card.

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It is true that the cards come out randomly, but in a positive count the last card is more likely to be a high card. In a negative count the last card is more likely to be a low card. The same can be said for the last 2 cards, etc. Saying that the TC approaches zero at the end of the deck is incorrect.

apex said:

If you take a deck with TC x and deal it out a billion times, the TC will average x at every point for the rest of the deck. Even with one card left.



This is true. As the # of decks gets smaller and even becomes fractions, your TC numbers will be further from zero. Going back to the sims you are proposing, when you average the results out you will see that the TC = TC wherever you started measuring.



It is true that the cards come out randomly, but in a positive count the last card is more likely to be a high card. In a negative count the last card is more likely to be a low card. The same can be said for the last 2 cards, etc. Saying that the TC approaches zero at the end of the deck is incorrect.

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Once again I see "facts" called out but no proofs. Where's the beef? And I really don't see what good an average does when actual results vary so widely.

21gunsalute said:

Once again, it doesn't matter what you call it.

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No! You've got this backwards - it doesn't matter what YOU call it. You're wrong; the facts are the FACTS! The True Count Theorem is based upon mathematical PROOF. You ARE entitled to your opinion, but you're NOT entitled to the FACTS. You can use your twisted logic to pretend anything you WANT, but keep it to yourself.

Since I joined this site, I've been VERY patient with people, and gone out of my way to and try keep from denigrating others; but there IS a limit to my patience. Some people, even after being shown absolute proof; will STILL insist upon denying the facts right before their eyes!

21gunsalute said:

Once again I see "facts" called out but no proofs. Where's the beef? And I really don't see what good an average does when actual results vary so widely.

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Show us YOUR proof!

Why are we trying to reason with an obvious troll?

21gunsalute said:

You're missing the point. It doesn't matter what you call it. It's zero in my book and it's zero for all practical purposes. Zero is nothing. It doesn't matter if nothing divided by nothing is nothing, infinity or undefined. But just for arguments sake, you just defined it as every number and no numbers. How can you define something that you claim is undefined? And why are we arguing about nothing?

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The only case where you have a division by zero is, when you try to estimate the remaining card distribution of ZERO remaining cards.
Physically this is nonsense, and mathematics are "allowed" (and moreover expected) to diverge when applied to non-applicable problems.

Again: The RC will drop to zero with the last card. That is true, and will ever be true for any balanced counts.
But a decreasing RC does not mean the TC will decrease. The reason is that the number of unseen cards DO decrease, and the ratio of the decreasing RC and decreasing number of cards can be any value.

21gunsalute said:

Once again, you may call it whatever you like but for all practical purposes, zero divided by zero is zero. It's also irrelevant since we've run out of cards at this point.

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For the practical purpose in this case, after the last card is dealt the TC is zero/# of decks in the new shoe. If the cards run out during play, they will shuffle and deal the cards from a fresh shoe... So in this case I would have to agree that the TC after the last card is dealt is actually zero.

If one examines the shoe at the start when TC = 0, shouldn't the TC approach zero at the end with TC theorem? You cannot pick a point where TC is positive and say the TC at the end tends to be positive. These positive TCs do exist, but in the long run they will be balanced out with negative TCs. As a result, in the long run TC does tend to end at zero more than any other counts. What am I missing?

farmdoggy said:

For the practical purpose in this case, after the last card is dealt the TC is zero/# of decks in the new shoe. If the cards run out during play, they will shuffle and deal the cards from a fresh shoe... So in this case I would have to agree that the TC after the last card is dealt is actually zero.

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Technically it's the TC of another game which is zero once you shuffle.

psyduck said:

If one examines the shoe at the start when TC = 0, shouldn't the TC approach zero at the end with TC theorem? You cannot pick a point where TC is positive and say the TC at the end tends to be positive. These positive TCs do exist, but in the long run they will be balanced out with negative TCs. As a result, in the long run TC does tend to end at zero more than any other counts. What am I missing?

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You are correct. A full deck will tend to average zero at all points. When you get to a point where a deck is say, +4, then if you take +4 as your starting points and run your sims, your average TC will be +4 at all remaining points.

psyduck said:

If one examines the shoe at the start when TC = 0, shouldn't the TC approach zero at the end with TC theorem? You cannot pick a point where TC is positive and say the TC at the end tends to be positive. These positive TCs do exist, but in the long run they will be balanced out with negative TCs. As a result, in the long run TC does tend to end at zero more than any other counts. What am I missing?

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What you are missing is that once the TC becomes large in magnitude you now have new information that is more reliable than what you knew at the beginning of the shoe. In fact, the shoe with a TC of +12 does not know that it has had more small cards taken out of it and that it needs to adjust itself to accord with what you knew at the beginning of the shoe (i.e., that the TC was 0); rather, on average, each round will decrease the RC by an amount that accords with the TC remaining constant, as your divisor becomes smaller.

Spaw