The Great Debate, With Meaning

blackjack avenger · Mar 17, 2011

You walk by a 6 deck shoe half dealt and on the table is RC 12

Option 1
RC 12 TC 2 with 6 decks unseen

Option 2
You are assuming those 12 high cards are split between the cards before the hand you have seen and half are in the cards yet to be played.
RC 6 TC 2 with 3 decks unplayed

Option 2 would have very high variance; mostly due to the low divisor?, but it could be done.

So let's try to lower the variance with another example.

6 deck shoe
Your partner misses the first deck and informs you.
Your partner also informs you the rc is 10 with 1 deck to go.
1 deck to go rc 10

Option 1
rc 10
2 decks unseen
tc 5

Option 2
Since you know the 10 good cards are split between the first deck missed and the deck yet to be played. You can make an assumption about the remaining deck.
rc 5
1 deck to go
tc 5

The variance of the second example would be less then the first example because you have knowledge of more cards and are making assumptions on fewer cards.

In all situations above your expectation is positive. Also, in the real world where many use level I counts, perhaps unbalanced counts, grouped indices, negative expectation bets, camo betting, non perfect TC conversions, combined BS and the effect of the float being off a few cards in one's assumptions in both option 2's is NOT going to break you! Whenever we play we can miss the good cards.

:joker::whip:
This should be good

Shoofly · Mar 17, 2011

blackjack avenger said:
You walk by a 6 deck shoe half dealt and on the table is RC 12

Option 1
RC 12 TC 2 with 6 decks unseen

Option 2
You are assuming those 12 high cards are split between the cards before the hand you have seen and half are in the cards yet to be played.
RC 6 TC 2 with 3 decks unplayed

Option 2 would have very high variance; mostly due to the low divisor?, but it could be done.

So let's try to lower the variance with another example.

6 deck shoe
Your partner misses the first deck and informs you.
Your partner also informs you the rc is 10 with 1 deck to go.
1 deck to go rc 10

Option 1
rc 10
2 decks unseen
tc 5

Option 2
Since you know the 10 good cards are split between the first deck missed and the deck yet to be played. You can make an assumption about the remaining deck.
rc 5
1 deck to go
tc 5

The variance of the second example would be less then the first example because you have knowledge of more cards and are making assumptions on fewer cards.

In all situations above your expectation is positive. Also, in the real world where many use level I counts, perhaps unbalanced counts, grouped indices, camo betting, non perfect TC conversions, combined BS and the effect of the float being off a few cards in one's assumptions in both option 2's is NOT going to break you! Whenever we play we can miss the good cards.

:joker::whip:
This should be good

OMG not so soon. I'm still bloody from the last go around.

MangoJ · Mar 17, 2011

Option 1
RC 12 TC 2 with 6 decks unseen

I'll go for that, because it is exactly what you see. 6 decks unseen (say 4 in shoe, 2 in discard). For unseen cards it doesn't matter where they are, behind cut card, discard, in front of the shoe, or my dog ate them.

Option 1 rc 10 2 decks unseen tc 5

I'll also go for that.

Gamblor · Mar 18, 2011

Option 1 and Option 2 are both right.

Option 1 probably has less variance, and its the "safe" assumption.

Option 2 would probably have greater variance. However, the question is does Option 2 provide better EV (or worse, or same)?

Just guessing, I would think it might provide same EV or slightly better (since you'll tend to bet more because of more variance).

A sim would probably answer this

Sucker · Mar 18, 2011

Well; yes - This would work ok for the next hand or two. But as the deck progresses, it starts to break down quite rapidly. Take the first example: Suppose after 2 more decks are played out of that shoe, the RC is still the same. According to option 1, the RC is +12 with 4 decks unseen, or TC = 3. But according to option 2, the RC is still +12, but with only one deck unplayed, the TC would have to be measured at +12!

This is not a matter of variance, but a case of inaccurate TC.

But as you say - it COULD be done. You would have to start over, and using this method, recalculate after EVERY hand. I think I'll just stick to the tried & proven option 1.

Methinks you're trying to reinvent the wheel.

MangoJ · Mar 18, 2011

Sucker said:
Well; yes - This would work ok for the next hand or two. But as the deck progresses, it starts to break down quite rapidly. Take the first example: Suppose after 2 more decks are played out of that shoe, the RC is still the same. According to option 1, the RC is +12 with 4 decks unseen, or TC = 3. But according to option 2, the RC is still +12, but with only one deck unplayed, the TC would have to be measured at +12!

This is not a matter of variance, but a case of inaccurate TC.

I agree completely on that. Think of the funny consequences when in the first example you would play the full shoe but the very last card (which remains unseen).
If option 2 were true, you would be able to PIN DOWN this very last card based on the TC (at least being a high,neutral, or low card). But this is surely impossible with 2 full decks unseen. Hence option 2 must be wrong.

iCountNTrack · Mar 19, 2011

I can't believe this nonsense is still going, I had my scotch and I am in a good mood, so I will try to point out the absurdity of option 2.

Suppose the casino add an extra "weird" final shuffling step where the dealer removes decks 3 and 4 from the middle of the shoe AFTER the player cuts the shoe.
Now suppose you count the first two decks of a shoe for an RC of +12, saw your long lost love and missed two decks, can anyone please tell me how this is different than the scenario i just described where you basically counted the first two decks had a count of +12 do i calculate the TC any differently for the weird shuffling procedure OF COURSE NOT, basically the two situations are identical if the dealer grabbed the two decks from the middle of the shoe and placed them in the back of the shoe or i missed those two decks. The remaining shoe has an excess of 12 high cards in both cases and a true count of +3.

Finally True count is always given by the following:

Gamblor · Mar 19, 2011

Another argument against Option 2 that makes it seem stranger, is that it would similar to doing something like this: in a 6 deck shoe, dealer cuts off 2 decks (just about your typical shoe game). And then we would assume RC 0, TC 0, and count assuming 4 decks remaining. Of course most CC would say this is crazy.

Or put another way, imagine we pull 52 cards randomly from an infinite deck. Should we start counting this is a single deck game (with RC 0, TC 0 initially)?

Having said that, in terms of probability it seems valid to do, as things will average out in the long run. As others mentioned earlier, the variance would be higher, and I'm thinking EV would be same or could be better in Option 2 - but in all likelihood the gain in EV would not be worth the increased variance.

Of course, theoretically, if you didn't care about bankroll, you would bet table max with a mere TC +2 every chance you got

MangoJ · Mar 19, 2011

Gamblor said:
Or put another way, imagine we pull 52 cards randomly from an infinite deck. Should we start counting this is a single deck game (with RC 0, TC 0 initially)?

Having said that, in terms of probability it seems valid to do, as things will average out in the long run. As others mentioned earlier, the variance would be higher, and I'm thinking EV would be same or could be better in Option 2 [...]

Drawing 52 cards from an infinite deck shoe is clearly distinct from a single deck game. In the infinite case, you know exactly nothing about the remaining cards. You could try to count that game, and you get a nice RC - but TC conversion is always zero (because the number of unseen cards is infinity).
This is quite the opposite for the single deck game.

Your statement, that both cases are identically in the long run is simply wrong. The infinite deck game does have an even worser EV than a not-counted single deck game. Not to say of a counted single deck game.

blackjack avenger · Mar 19, 2011

Gamblor said:
Another argument against Option 2 that makes it seem stranger, is that it would similar to doing something like this: in a 6 deck shoe, dealer cuts off 2 decks (just about your typical shoe game). And then we would assume RC 0, TC 0, and count assuming 4 decks remaining. Of course most CC would say this is crazy.

I don't think it is the same for this reason:
In my examples there are counted cards, you do have information. In your example you have not counted any cards, you do not have information.

good cards:joker::whip:

blackjack avenger · Mar 19, 2011

Why Does Everyone Not Assume the Average

Sucker said:
Well; yes - This would work ok for the next hand or two. But as the deck progresses, it starts to break down quite rapidly. Take the first example: Suppose after 2 more decks are played out of that shoe, the RC is still the same. According to option 1, the RC is +12 with 4 decks unseen, or TC = 3. But according to option 2, the RC is still +12, but with only one deck unplayed, the TC would have to be measured at +12!

This is not a matter of variance, but a case of inaccurate TC.

But as you say - it COULD be done. You would have to start over, and using this method, recalculate after EVERY hand. I think I'll just stick to the tried & proven option 1.

Methinks you're trying to reinvent the wheel.

With both option 2s, on average the TC and RC will be accurate. A shoe always ends at 0.

Given both option 2s if you count down:
the decks seen
those yet to be played and the
cards you assume

you will get rc 0 and tc 0. Whenever we bet the cards that come out can be different then what we expect.

It's obvious any example that states:
All the good cards could be in the assumed section which would be bad
can be countered with
All the bad cards could be in the assumed section which would be good

:joker::whip:
good cards

psyduck · Mar 19, 2011

iCountNTrack said:
I can't believe this nonsense is still going, I had my scotch and I am in a good mood, so I will try to point out the absurdity of option 2.

Suppose the casino add an extra "weird" final shuffling step where the dealer removes decks 3 and 4 from the middle of the shoe AFTER the player cuts the shoe.
Now suppose you count the first two decks of a shoe for an RC of +12, saw your long lost love and missed two decks, can anyone please tell me how this is different than the scenario i just described where you basically counted the first two decks had a count of +12 do i calculate the TC any differently for the weird shuffling procedure OF COURSE NOT, basically the two situations are identical if the dealer grabbed the two decks from the middle of the shoe and placed them in the back of the shoe or i missed those two decks. The remaining shoe has an excess of 12 high cards in both cases and a true count of +3.

Finally True count is always given by the following:

LOL! Hard to miss your point! That is dZ2!

Gamblor · Mar 19, 2011

blackjack avenger said:
I don't think it is the same for this reason:
In my examples there are counted cards, you do have information. In your example you have not counted any cards, you do not have information.

good cards:joker::whip:

I think in the example I gave, you do have information in a sense, similar to your example. On average I know that of the 2 decks cut off, the high and low cards removed are the same (in other words TC 0 for the 2 decks cut off).

And MangoJ, for the example of infinite decks, similarly I'm assuming that on average, cards drawn from an infinite deck will have an equal number of high and low cards over the long haul.

I give these examples to point out similar examples that make Option 2 seem strange. I frankly admit, maybe there's something wrong with my thought processes in thinking about average. But off hand I cannot really think of a reason why its wrong.

There's probably some disconnect from a theoretical average to the real world. However, if I know the average percentage of people who die, I can set up a an life insurance company, and sell people insurance, priced just right so the price isn't too exorbitant, and where I can make a little profit

MangoJ · Mar 19, 2011

Gamblor said:
I frankly admit, maybe there's something wrong with my thought processes in thinking about average. But off hand I cannot really think of a reason why its wrong.

I'm sorry to disappoint you, but your understanding of "average" should be overthought. Your example of an infinite deck shoe is brilliant though, because it servers us an academic example to test our thought-process.

On an infinite deck shoe, you gain no information with counting. Why is that ? I gave the reason before: Regardless of any RC of any fancy counting system, the true count is always zero and hence the same at the beginning of the infinite shoe.
Here is a second reason: An infinite deck shoe is equivalent to a 13-sided dice, where you roll the dice instead of drawing a card. Although you can assign count values to each side, and (for a balanced count) they add up to zero. And on average (the real average) you expect to keep a RC=0 before your first card/roll.
But you must agree, rolling a dice 100 times (or even a billion times) doesn't give you the tiniest hint for the next roll.

The problem is a mis-interpretation of "average". Although the RC of an infinite deck shoe is 0 on average, this doesn't mean your RC will somehow drift towards 0. Being 0 on average just means, that there is an equal chance of rising or falling for the RC.
In fact the infinite deck shoe is a random walk, and you are expected to find any RC in such a shoe. If you would record the peaks of positive and negative RCs during your draws on the infinite shoe, on average you are expected that those peaks will rise during time (both positive and negative peaks!). That is quite the contrary to your expectation that you will level at RC=0.

Gamblor · Mar 20, 2011

MangoJ said:
I'm sorry to disappoint you, but your understanding of "average" should be overthought. Your example of an infinite deck shoe is brilliant though, because it servers us an academic example to test our thought-process.

On an infinite deck shoe, you gain no information with counting. Why is that ? I gave the reason before: Regardless of any RC of any fancy counting system, the true count is always zero and hence the same at the beginning of the infinite shoe.
Here is a second reason: An infinite deck shoe is equivalent to a 13-sided dice, where you roll the dice instead of drawing a card. Although you can assign count values to each side, and (for a balanced count) they add up to zero. And on average (the real average) you expect to keep a RC=0 before your first card/roll.
But you must agree, rolling a dice 100 times (or even a billion times) doesn't give you the tiniest hint for the next roll.

The problem is a mis-interpretation of "average". Although the RC of an infinite deck shoe is 0 on average, this doesn't mean your RC will somehow drift towards 0. Being 0 on average just means, that there is an equal chance of rising or falling for the RC.
In fact the infinite deck shoe is a random walk, and you are expected to find any RC in such a shoe. If you would record the peaks of positive and negative RCs during your draws on the infinite shoe, on average you are expected that those peaks will rise during time (both positive and negative peaks!). That is quite the contrary to your expectation that you will level at RC=0.

Thanks MangoJ for your thoughts. Your are correct of course in your points. An infinite deck is a whole different animal than a finite deck, and makes the infinite deck example different then the others mentioned. Of course you lose any memory with an infinite deck.

Having said that, of course on a million flips of the coin, it should be close to 50% heads and 50% tails. Probably not much you can do with that info in a casino, as all propositions are bets on what the next event will be

BMDD · Mar 21, 2011

If we are entering the game with a positive expectation wouldn't the True Count Theorem suggest we take the game with more remaining cards, because we will be able to play more hands with the same advantage? Waaiiiiit a second, I might be overlooking some details.. I think the question in scenario 1 may be flawed, how can you have a half-dealt 6-deck shoe with 6 decks remaining?!

tthree · Mar 31, 2011

When you assume you are an ASSh and U lose MonEy

The thing about any assumption like option 2 is your assumption is almost certainly wrong. The advantage to AP is only as good as his information is accurate. Options 1 are the only way to play out the shoes and truly call what you are doing AP. If you choose option 2 in either case your information starts out reasonable but with each card you see after that your incorrect assumption will cause your assessment of your playing conditions to deviate from the actual situation you are playing in(unless by some miracle you won the lottery with a correct assumption). Why give away your advantage when you dont have to. When a counter assumes he/she is an ASSh and U loses MonEy.

blackjack avenger · Mar 31, 2011

Yep

Agree
It appears the variance of option 2 is to great, even though we would know the ratio of high to low cards on average. If the distribution were perfect we would be fine, but that would be very very rarely the case.

Koz1984 · Apr 1, 2011

I had a similar experience at my local store whilst I was there with a friend. She's been showing interest in the game, including counting, so I took her down to show her the basic format of Blackjack. Whilst casually observing a table which was half through an 8 deck shoe, I saw 3 rounds dealt to 5 boxes with absolutely ZERO face cards. The RC was around the +50 mark as people were drawing 3-4 cards/hand. I threw some money out on the felt, my friend did also, and we won every single hand until it was shuffle time. I questioned myself too as to how to approach approximating the count, I even posted it on the message boards here. I was told you would simply divide the RC by 8, considering 8 decks were unseen. Seems logical to me.

blackjack avenger · Apr 1, 2011

Exactly

All unseen cards are unseen. As others including myself have stated many times. It's good you realized that you could play with an advantage.

The Great Debate, With Meaning

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Banned

Well-Known Member

Well-Known Member

Well-Known Member