High card after a string of small cards at low/negative count?

I'm sure all of you have seen this scenario.
You're dealt a pair of small cards at a low/negative count.
The count is negative (or low) and small cards continue to come out as expected.
After a string of about 5 or 6 (or maybe even more) small cards, the sum of your
total hand is still low.
According to the basic strategy (and also because the count is negative or low),
you should continue to hit.
However, because a string of small cards just came out, a high card is "due".
Or is it truly "due"?
There's no such thing as, "... something is due ..." in Blackjack. Right?
Is the probability of a high card coming out strictly based of the count?
Or is "clumping" of excessive number of small cards unlikely, so a high card
should be expected even if the count is negative or low?
Is this a case of "selective memory" where I just happen to remember those
incidences that are unusual and painful?

ArcticInferno said:

I'm sure all of you have seen this scenario.
You're dealt a pair of small cards at a low/negative count.
The count is negative (or low) and small cards continue to come out as expected.
After a string of about 5 or 6 (or maybe even more) small cards, the sum of your
total hand is still low.
According to the basic strategy (and also because the count is negative or low),
you should continue to hit.
However, because a string of small cards just came out, a high card is "due".
Or is it truly "due"?
There's no such thing as, "... something is due ..." in Blackjack. Right?
Is the probability of a high card coming out strictly based of the count?
Or is "clumping" of excessive number of small cards unlikely, so a high card
should be expected even if the count is negative or low?
Is this a case of "selective memory" where I just happen to remember those
incidences that are unusual and painful?

Click to expand...

Just remember your count tells you the composition of the remaining deck, not the order. So if your count says to hit than just hit. Any of the leftover composition could be coming next, the cards don't care if a bunch of lows already came out.

ArcticInferno said:

Or is "clumping" of excessive number of small cards unlikely, so a high card should be expected even if the count is negative or low?

Click to expand...

Not sure where you play, but if the shuffles are un-shuffle-trackable, then they are probably random enough for you to ignore any perceived clumps, and truly just play based on the count.

ArcticInferno said:

Is this a case of "selective memory" where I just happen to remember those incidences that are unusual and painful?

Click to expand...

Unfortunately this does happen, and when on that rare occasion you have a 6 card 14, and the dealer shows a 7, it really might pain you to hit because you think an X is "due". But if that's what the count says, and the decks are randomly shuffled, then it's the correct move.

ArcticInferno said:

I'm sure all of you have seen this scenario.
You're dealt a pair of small cards at a low/negative count.
The count is negative (or low) and small cards continue to come out as expected.
After a string of about 5 or 6 (or maybe even more) small cards, the sum of your
total hand is still low.
According to the basic strategy (and also because the count is negative or low),
you should continue to hit.
However, because a string of small cards just came out, a high card is "due".
Or is it truly "due"?

Click to expand...

Oddly enough, I've had the same question come to mind. In this scenario the principle of the TC ratio seems to buck up against the Theory of Random Distribution. I have never allowed that to influence my play; however it does pose an interesting theoretical question.

ArcticInferno said:

However, because a string of small cards just came out, a high card is "due".
Or is it truly "due"?

Click to expand...

My wife and I watched a progression bettor at the craps table playing only the "field" about a year or two ago.

1st roll) $100 - lost
2nd) 200 - lost
3rd) $400 - lost
4th) $800 - lost
5th) $1,600 - lost
6th) $3,000 - lost (table max)
7th) he and his girlfriend each put down $3,000 - lost

$12,100 in 7 rolls. I bet they thought a number in the field was due too.


BJC

bj bob said:

Oddly enough, I've had the same question come to mind. In this scenario the principle of the TC ratio seems to buck up against the Theory of Random Distribution. I have never allowed that to influence my play; however it does pose an interesting theoretical question.

Click to expand...

I know what you mean. Take for example, when you have your Max bet out and you get XX v 6 and you notice the last 5 out 6 cards dealt were face cards, but your index is just high enough to indicate a split. Would you split, with your Max bet on the line?

Or take the same scenario with 12 v 2. Would you Stand?

Also see index deviation http://www.blackjackinfo.com/bb/showthread.php?t=16797&highlight=index+deviation

http://www.blackjackinfo.com/bb/showthread.php?t=7605

jack said:

I know what you mean. Take for example, when you have your Max bet out and you get XX v 6 and you notice the last 5 out 6 cards dealt were face cards, but your index is just high enough to indicate a split. Would you split, with your Max bet on the line?

Click to expand...

I'd say split assuming the following:

• If the last 5 out of 6 cards were face cards, and the count is still high enough to split a XX v 6
• You are using RA indices (if you're worried about variance)
• The decks are thoroughly shuffled (and you don't know where the clumps of face cards are)

Can you give any possible mathematical reason why you wouldn't split? (fyi, that's not meant to be an insulting tone to my question)

ThInk of it this way. If there are 25 cards left in the deck, 10 small, 10 big, and 5 neutral, you're just as likely to get all 10 smalls in a row (without extra information like sequencing or tracking) as you are getting them all mixed together. If the count (composition) of the deck is all you have you must follow your indices. Worrying that you'll get a small on a double down after 3 players in front of you get 10's is just voodoo (hey, look where I am!)

Yes, your odds dropped, but doubling (if count says so) is the most profitable play.

assume_R said:

I'd say split assuming the following:

• If the last 5 out of 6 cards were face cards, and the count is still high enough to split a XX v 6
• You are using RA indices (if you're worried about variance)
• The decks are thoroughly shuffled (and you don't know where the clumps of face cards are)

Can you give any possible mathematical reason why you wouldn't split? (fyi, that's not meant to be an insulting tone to my question)

Click to expand...

I didnt take it that way guy, but thanks for being open minded.


No, i have no mathematical proof, but i do have a proposition for an experiment that could be conducted to prove whether there is a difference(however so small) or not. Perhaps Qfit could or another programmer could set it up, idk.


Since the little cards or face-cards, that come out previously are basically irrelevant, because of the number of decks in play, using TCs would have to be implemented.

A level 2 count.

The first example, would be to take a TC that drops from +12 to +4

The second example, would be to take a TC that rises from -4 to +4

Seperate simulations(i guess) would have to be performed.

My guess is, the TC that rises from -4 to +4 will show a "ever so slightly" greater performance in EV, verses the TC that dropped.

I would be very curious to see this, maybe i should start a thread on it. Ok, lemme have it lol

Like any good scientist wanting to test a hypothesis, you present an experiment. This seems like a valid experiment to test your theory.

So to summarize your proposed experiment:

Right now, the EV at a given TC is dependent only on that TC, mathematically written as EV = f(TC).
Your hypothesis: The EV at a given TC is dependent on the TC, but also on the rate of change of the TC, mathematically written as EV = f(TC, TC') where TC' is the "speed" of TC changing.


Step 1: Start at TC=+12. Run a simulation (with randomness) to drop to TC=+4. Measure EV only at the TC=+4. Repeat billlions of times, and find the average EV at all TC=+4's. (we'll call this EV_1) EV_1 is measuring the EV if TC' < 0 (since it's decreasing)

Step 2: Start at TC=-4. Run a simulation (with randomness) to increase to TC=+4. Measure EV only at the TC=+4. Repeat billlions of times, and find the average EV at all TC=+4's. (we'll call this EV_2)
EV_2 is measuring the EV if TC' > 0 (since it's increasing)

Step 3. See if EV_1 < EV_2 to any statistical significance. This would tell whether TC' (or how the TC "got" to +4) has an effect on the EV.

My only question for this experiment would be how is this done? One single hand in which the TC drops or increase? Or allow to fluctuate over several rounds, and only measure at the first time the TC=+4. Something to think about.

assume_R said:

Like any good scientist wanting to test a hypothesis, you present an experiment. This seems like a valid experiment to test your theory.

So to summarize your proposed experiment:

Right now, the EV at a given TC is dependent only on that TC, mathematically written as EV = f(TC).
Your hypothesis: The EV at a given TC is dependent on the TC, but also on the rate of change of the TC, mathematically written as EV = f(TC, TC') where TC' is the "speed" of TC changing.


Step 1: Start at TC=+12. Run a simulation (with randomness) to drop to TC=+4. Measure EV only at the TC=+4. Repeat billlions of times, and find the average EV at all TC=+4's. (we'll call this EV_1) EV_1 is measuring the EV if TC' < 0 (since it's decreasing)

Step 2: Start at TC=-4. Run a simulation (with randomness) to increase to TC=+4. Measure EV only at the TC=+4. Repeat billlions of times, and find the average EV at all TC=+4's. (we'll call this EV_2)
EV_2 is measuring the EV if TC' > 0 (since it's increasing)

Step 3. See if EV_1 < EV_2 to any statistical significance. This would tell whether TC' (or how the TC "got" to +4) has an effect on the EV.

My only question for this experiment would be how is this done? One single hand in which the TC drops or increase? Or allow to fluctuate over several rounds, and only measure at the first time the TC=+4. Something to think about.

Click to expand...

Yep! Very well put, if i say so myself. And to answer your question of "how this would be done, is a good question as well. As I mentioned it would have to be done by a "experienced" programmer. Perhaps I could start a new thread on it, and propose the challenge to someone who may want to tackle it.

I would suggest that you also calculate the standard error, so that you know if the difference (if any) between the two calculated EVs is statistically significant.

Nynefingers said:

I would suggest that you also calculate the standard error, so that you know if the difference (if any) between the two calculated EVs is statistically significant.

Click to expand...

Yes, I actually meant that to be implied when I asked about if they are statistically significantly different. Whatever the appropriate method is would be used.

Not sure if you're aware of this, but standard error is not the only thing to be used to test statistical significance. The most common being a t-test. Depending on the distribution of EV_1 and EV_2 there may be other more applicable methods, such as Fischer's LSD test. Entire statistics books have been written on different methods for statistical significance tests.

JJ, If you are serious about finding somebody to undertake this, and do make a thread about this, make sure you link back to this thread

In an ASM game I will follow the count's mandate no matter what. In a handshuffle game, I will make adjustments in borderline situations due to strings.

Blue Efficacy said:

In an ASM game I will follow the count's mandate no matter what. In a handshuffle game, I will make adjustments in borderline situations due to strings.

Click to expand...

Is that because in a hand-shuffled game, there is a higher probability of strings of low or high cards due to non-random shuffles, and the way people normally play blackjack from the previous shoe? Just curious did you learn this from a book or from norm's software?

assume_R said:

Is that because in a hand-shuffled game, there is a higher probability of strings of low or high cards due to non-random shuffles, and the way people normally play blackjack from the previous shoe? Just curious did you learn this from a book or from norm's software?

Click to expand...

No I didn't learn it from anywhere, just figured a machine is more random than a hand shuffle, I figure a hand shuffle is more likely to have a big card breaking up a string of little cards during the riffling. I know it's pure voodoo.

Deathclutch said:

If there are 25 cards left in the deck, 10 small, 10 big, and 5 neutral, you're just as likely to get all 10 smalls in a row (without extra information like sequencing or tracking) as you are getting them all mixed together.

Click to expand...

Intuitively, that does not seem right. Leave out the 5 neutral. If you have 20 cards left, 10 small and 10 large, then there is only 1 way all small cards can come up, and there are 10 ways a mixture of small and large cards or all large cards can come up in the first 10 off the top, if we don't worry about the order of the particular cards.

If you do consider the particular order, then its 100 ways for small cards and 300 ways for mixed and all large cards.

So the odds of the first 10 cards coming up all small off the top in this scenario is 10 to 1 or 9% of the time. Somebody check my assumptions and math please.

No criticism intended. These ideas are tricky.

aslan said:

Intuitively, that does not seem right. Leave out the 5 neutral. If you have 20 cards left, 10 small and 10 large, then there is only 1 way all small cards can come up, and there are 10 ways a mixture of small and large cards or all large cards can come up in the first 10 off the top, if we don't worry about the order of the particular cards.

If you do consider the particular order, then its 100 ways for small cards and 300 ways for mixed and all large cards.

So the odds of the first 10 cards coming up all small off the top in this scenario is 10 to 1 or 9% of the time. Somebody check my assumptions and math please.

No criticism intended. These ideas are tricky.

Click to expand...

Hey, you're absolutely right. My wording was terrible in that post. Let me try again.

You have the same chance of getting the sequence of the 10 lows in a row as any other sequence. There will be more sequences that have a combination of the highs and lows but with only counting we won't have enough info to change our play.

Low, low, low, low . . .
Low, low, low, high . . .
Low, low, high, high . . .
Low, high, high, high . . .
And so on and so forth.

Good catch!

Deathclutch said:

Hey, you're absolutely right. My wording was terrible in that post. Let me try again.

You have the same chance of getting the sequence of the 10 lows in a row as any other sequence. There will be more sequences that have a combination of the highs and lows but with only counting we won't have enough info to change our play.

Low, low, low, low . . .
Low, low, low, high . . .
Low, low, high, high . . .
Low, high, high, high . . .
And so on and so forth.

Good catch!

Click to expand...

I think what you're trying to say is that there are 25! equally likely orders (or permutations) that 25 cards can be in. (25! = 25*24*23*.........*4*3*2*1)

Now if you subdivide those 25 cards into 3 subgroups - low, med, high and further say there are 10 low, 5 med, and 10 high that comprise those 25 cards, you can figure the probability of a subset of 10 cards comprised of a given subgroup composition.

Let L = number of low cards in low subgroup of subset
Let M = number of med cards in med subgroup of subset
Let H = number of high cards in high subgroup of subset
Let N = total number of cards in subset (N = L + M + H)

Let P_comp = prob of drawing L consecutive low followed by M consecutive med followed by H consecutive high from total cards (25 in this case)

Let P = probability of subset
Let W = weight of subset consisting of a specific subgroup composition
Since each permutation of low, med, high are equivalent
W = N!/L!/M!/H!

Then
P = W * P_comp

Most likely this is like a foreign language so we'll compare the probability of a subset of 10 low cards to the probability of a subset of 4 low, 2 med, and 4 high as an example:

(10 low)
W = 10!/10!/0!/0! = 1 (note 0! = 1)
P_comp = 10/25*9/24*8/23*7/22*6/21*5/20*4/19*3/18*2/17*1/16
P = W * P_comp = 3.0592640634368996194275505084497e-7

(4 low, 2 med, 4 high)
W = 10!/4!/2!/4! = 3150
P_comp = 10/25*9/24*8/23*7/22*5/21*4/20*10/19*9/18*7/17*6/16
P = W * P_comp = 0.13491354519756727321675497742263

Assuming I didn't make any errors, the probability of a subset of 10 low cards is very close to 0 and the probability of a subset of 4 low, 2 med, and 4 high is about .135 or 13.5%.

If you're still with me this far, you can use the above to compute the probability of any possible subset drawn from 25 cards consisting of 10 low, 5 med, and 10 high. As a check the probabilities of all of the possible subsets should sum to 1. (You could even go further, changing starting number of cards and/or subgroup composition.)


Basically this is what the program I am working on does except that it adapts to any number of cards and any counting system and also allows for cards known to be specifically removed be incorporated and in addition to subset probability it computes probability of each rank based upon the count/specific removal information, making it possible to accurately determine insurance strategy mathematically.


On a side note having nothing to do with this post, not that my website is such a great site but I have been unable to connect to it for 3 or 4 days although my email still works. So far I've received no response from my web host and I am wondering if it may have gone belly up.

:devil::whip: