Question About Standard Deviation/Gambler's Fallacy

[LovinItAll]

This thought occurred to me when I was down 30 units in 30 minutes recently. I'm not a mathemetician, so my fundamental understanding of the following statements/comments may be incorrect. Some of the statements are just curiousities, so if they seem absurd to someone with a solid understanding of the subject, my apologies.


Let's say that a player is wagering on a game that has a 10 unit/hour SD. After 30 minutes, the player is down 30 units, or 3 standard deviations. Are the following questions/comments flawed?

- The player is really down 6 standard deviations, as 1 SD per hour = .5 SD's per 1/2 hour.

- The player hasn't played an hour, so the results re: SD are unknown as of yet.

- The player says. "It doesn't matter what just happened. I could easily lose another 11 units over the next 30 minutes (4+ SD's over 1 hour). I'm having a bad night. Bye."

- The player decides to increase his bet, as the chances of being 4+ standard deviations after an hour are slim.

- The player decides to keep his bet the same, as the chances of being 4+ standard deviations after an hour are slim and the player should have a decent shot at recovering some of his losses without increasing his bet.

- Based on the last statement, how does the Gambler's Fallacy apply to time and standard deviation? For example, we know that knowing past results doesn't impact future odds. With that, after the half hour of play and the resulting -3 SD's in results, do we simply 'reset the clock' after every wager as it relates to standard deviation? If the player loses 11 units during the next 30 minutes (common) the player is now down 4.1 SD's/hour, a rare occurence (.006%, I think).

To me, this is where the Gambler's Fallacy might bite someone.

The player is stuck 30 units in 30 minutes. He knows that the chance of being down 40+ units in an hour is supposed to be a fairly rare event so he manipulates his wager accordingly to recoup losses. He then loses and must withdraw his children from private school (he is a hopeless degen.....boo hoo).

I've never made the claim of being smart, so if my ignorance is profound, save the bullets.

Best ~ L.I.A.

 

[tthree]

That is certainly not normal but we have all been there. I would walk just in case its not a random event. I was learning a new count and was really doing well counting down a deck of cards. I had been making no mistakes even though I added a side count. I was getting pretty fast when one day I end with a count of 4. I realize you often make the same mistake over again but decide to count again without shuffling. I get 4 again and again. I count how many cards in the deck 52. I shuffle and get 4 again and again. Finally I confirm I was forgetting to count the twos which were neutral in my old system.

My point is you may have a reason why you are down so much. The dealer may be cheating you. You may be making a systematic mistake. You may be making an error in math that once you did it once you kept repeating the same error.

If you are sure that none of these are the cause you might want to stay. I say why risk it maybe it is just an unlucky day for you. Maybe just an unlucky 30 minutes. The math says no reason to leave but the math doesnt believe you can make a mistake or be cheated. Silly math.

 

[LovinItAll]

That is certainly not normal but we have all been there. I would walk just in case its not a random event. I was learning a new count and was really doing well counting down a deck of cards. I had been making no mistakes even though I added a side count. I was getting pretty fast when one day I end with a count of 4. I realize you often make the same mistake over again but decide to count again without shuffling. I get 4 again and again. I count how many cards in the deck 52. I shuffle and get 4 again and again. Finally I confirm I was forgetting to count the twos which were neutral in my old system.

My point is you may have a reason why you are down so much. The dealer may be cheating you. You may be making a systematic mistake. You may be making an error in math that once you did it once you kept repeating the same error.

If you are sure that none of these are the cause you might want to stay. I say why risk it maybe it is just an unlucky day for you. Maybe just an unlucky 30 minutes. The math says no reason to leave but the math doesnt believe you can make a mistake or be cheated. Silly math.

I tried to add a spoiler to my post, but this forum or version of vBulletin doesn't support it.

In fact, I thought that maybe I was making some fundamental error, though it didn't really seem likely. Rather than leave, I flatted my min. bet for the next 30 minutes to sort of regroup and think about my play (B.S. is like auto-pilot, so....). That's when I started thinking about SD. I lost another 14 units, bringing the session loss to 44 units in 65 minutes. Good times.

I can't imagine the casino was cheating - it's a big, LV joint. That .006% seems SO unlikely on paper, but when one is on a cooler, you start expecting the dealer to draw out every hand. Kinda like someone catching a 2-outer on the river three times in an hour (but not as tilt-ish in BJ, as the dealer is rarely going to say something that makes one want to hit them in the head with a ball peen hammer).

Everyone has crappy sessions, and this wasn't a record loss in dollars, but I can't remember losing so many units so fast with my max bet out so few times. I've won that many units in an hour, though, but it was a case of hitting the right split(s) with the right double(s) at the right times.

Take care ~ L.I.A.

 

[tthree]

I had a run of dealer blackjacks so bad one time I tried to insure against a king. I saw the face card and thought great another dealer blackjack. To bad they didnt offer 12:1 for insurance of a ten value card. I would have won.

 

[MangoJ]

Let's say that a player is wagering on a game that has a 10 unit/hour SD. After 30 minutes, the player is down 30 units, or 3 standard deviations. Are the following questions/comments flawed?

- The player is really down 6 standard deviations, as 1 SD per hour = .5 SD's per 1/2 hour.

The problem here is, that "SD per hour" is not a good choice of unit. Why ? Because standard deviation don't simply add up. For example, you must play the same game 4 times longer in order to double your total SD.
Hence you cannot divide SD into smaller chunks and expect that those chunks behave as ordinary numbers. This works with pieces of cakes, EV, but not with SD.

3 units per half-hour SD is not the same as 6 units per hour SD.

The proper unit of fluctuations is "variance", that is the square of standard deviation. Variance does add up like numbers, and you can divide variance like pieces of cakes.
If you play a game with 100 units^2 per hour (this is the equivalent of your 10 units per hour SD), then this is identical to a variance of 50 units^2 per half-hour.
Since SD is the square root of variance, this is identical to a SD ~7 units "per half-hour".
If your player is down 30 units after half an hour, that is 4.3 SD away (if a zero-EV game). It is not your 6, as the SD per half hour is 7 units (not your 5 units).


Ever wonder why the total session SD is the hand's SD times the square root of number of hands played ?
It is because the total session variance is the hand's variance times the number of hands. The latter is the manifestation, that variance is the "natural" unit of fluctuations. However standard deviations just "feels" better, because it has the same unit as EV, and thus can be compared to EV.

 

[LovinItAll]

3 units per half-hour SD is not the same as 6 units per hour SD.
If you play a game with 100 units^2 per hour (this is the equivalent of your 10 units per hour SD), then this is identical to a variance of 50 units^2 per half-hour.
Since SD is the square root of variance, this is identical to a SD ~7 units "per half-hour".

If your player is down 30 units after half an hour, that is 4.3 SD away (if a zero-EV game). It is not your 6, as the SD per half hour is 7 units (not your 5 units).

Re: Extrapoloating Standard Deviation: I apologize for being so lazy and not looking this up before posting. I wasn't saying that I thought that the SD was 6 in my example, I was saying I didn't know. 6 SD's is such an unlikely occurence that intuitively I felt that couldn't be right. Even the 4.3 as cited SUCKED HARD!

Thanks for clearing that up - I appreciate it.

Best ~ L.I.A.

 

[k_c]

The problem here is, that "SD per hour" is not a good choice of unit. Why ? Because standard deviation don't simply add up. For example, you must play the same game 4 times longer in order to double your total SD.
Hence you cannot divide SD into smaller chunks and expect that those chunks behave as ordinary numbers. This works with pieces of cakes, EV, but not with SD.

3 units per half-hour SD is not the same as 6 units per hour SD.

The proper unit of fluctuations is "variance", that is the square of standard deviation. Variance does add up like numbers, and you can divide variance like pieces of cakes.
If you play a game with 100 units^2 per hour (this is the equivalent of your 10 units per hour SD), then this is identical to a variance of 50 units^2 per half-hour.
Since SD is the square root of variance, this is identical to a SD ~7 units "per half-hour".
If your player is down 30 units after half an hour, that is 4.3 SD away (if a zero-EV game). It is not your 6, as the SD per half hour is 7 units (not your 5 units).


Ever wonder why the total session SD is the hand's SD times the square root of number of hands played ?
It is because the total session variance is the hand's variance times the number of hands. The latter is the manifestation, that variance is the "natural" unit of fluctuations. However standard deviations just "feels" better, because it has the same unit as EV, and thus can be compared to EV.

This is a chance to frame a question about the variance of splits. To keep things as simple as possible assume a hand of A-A versus 9 from a full single deck that is split once and hitting split aces isn't allowed so one card is dealt to each split ace.

EV(1 split) = +.2898

Drawing one card to the first hand results in the following (may be small round off error):
EV(hand 1) = +.1449
Prob(win 1 unit) = 0.534045
Prob(push) = 0.07681
Prob(lose 1 unit) = 0.389145
Variance(hand 1) = .9022 units squared

The second hand considered by itself has the exact same data as the first hand. In other words if whatever happened in hand 1 is not considered, then hand 2 EV and variance would be identical to hand 1 data in the long run.

@MangoJ
You say that variance is additive so it would seem split variance(1 round) = 1.8044. However if each possible 2 card draw is considered, the resulting probabilities for the round are:
prob(+2) = 0.366139
prob(+1) = 0.0475381
prob(0) = 0.298499
prob(-1) = 0.0856004
prob(-2) = 0.202223
variance(1 round) = 2.32262

I think the discrepancy is because of covariance. If each of the 2 split hands were played on separate tables versus identical compositions then there would be no covariance. However since they are played at the same time there is covariance. Also if they were played in the context of an infinite deck then there would be virtually no covatiance.

The question: Is there a way to get split variance of 1 round for a finite shoe composition from the variance of the individual split hands? I don't expect an answer because it may not be possible but I don't know much about statistical analysis and you seem to have some familiarity with it.

Peter Griffin gives a formula on P. 142 of Theory of Blackjack:

V(n) = 1.26*n + .50*n*(n-1)

for computing the variance of playing n simultaneous hands. 1.26 units squared is the estimated overall variance of a blackjack hand and .50 is the approximate covariance of 2 blackjack hands determined by sim.

I'm not sure but after reading about how to get covariance I got covariance(split A-A v 9) = .41998 (but this could be wrong)

Applying above formula to specific case of A-A v 9

V(2) = .9022*2 + .41998*2*1 = 2.64436, which still doesn't agree with the correct answer of 2.32262.

In order to get the correct answer from the formula then covariance must somehow compute to about .25 and it would seem that the computation of covariance would somehow need to be a function of shoe composition.

As I said ther may be no answer to this problem but at least I thought this was a good opportunity to try and state it.

 

[tthree]

covariance

I may be making a fool out of myself here but how could there not be covariance even with infinite deck if they are played against the same dealer hand. If he has an ace under or hits to 21 you are in bad shape on both hands. If the dealer busts you win both no matter what else happens. Am I missing something here.

 

[tthree]

I found the information in TOB p.142. The formula is for playing 2 spots in general. The covariance of a split hand is a more specific example of 2 spots with a higher codependence for any deck composition. Naming a specific pair v a specific dealer card is very specific with a higher codependence for the deck composition. Most splits are offensive and would favor the player when used.

 

[tthree]

your probabilities are off under @mangoj using your win, push, lose probabilities

 

[MangoJ]

Ok, first some clarifications. Part of my PhD thesis deals with fluctuations, so I know some stuff about that business.
The summing up of variance is only valid for uncorrelated events (statistically independent events are always uncorrelated). When summing up variance per hour (or half-hours) it is understood that one plays different shoes in each (half-)hour. When it comes to variance within one shoe, one might argue that correlation between different rounds may be neglectable.

But this is definitely not the case within a single round. Hence, if we want to discuss variance of a split hand, the simple summing rule is violated. As tthree correctly points out, the reason is strong correlation, resulting in covariance.
For hands within a single round of blackjack, there are two sources of correlation. The strongest source is: the dealer. Both hands play against the same dealer. If the dealer hand is strong, both player hands are likely to lose simultaneous (easiest case, a dealers Blackjack). If the dealer hand is weak, both player hands are more likely to win simultaneous. The second source is the shoe. Easiest example is a simultaneous 2 player blackjack. It is less likely in single or double deck.
Hence, if we want to discuss variance of splits, the sum rule of variance does not apply (even with infinite deck - the main source of correlation is the dealers hand).

If one would like to calculate the variance of a split hand (i.e. for risk-averse play), one would need to evaluate each pair of results of possible split hands against each dealers result. This approach would incorporate the covariance in an exact way, but is very costly to compute.
There is however a much simpler route to take. It will not give exact figures, but it will give an upper bound of split variance.

Say X and Y are two random numbers (i.e. the return of a bet), not necessarily independent. VAR is the variance, and SD is the standard deviation (square root of variance).
Then the following is true for every case:
(SD(X) - SD(Y))^2 <= VAR(X+Y) <= (SD(X) + SD(Y))^2

Application to the split AA case. X is the return of the left split hand, Y is the return of the right split hand. Both are symmetrical, and hence share the same variance (and standard deviation). If your figures are accurate, then
SD(X) = sqrt(0.9022) = 0.9498

So the combined variance of both hands VAR(X+Y) is between
(SD(X) - SD(Y))^2 = 0
and
(SD(X) + SD(Y))^2 = 3.608

In fact the variance (from the probabilities) is 2.322, which lies perfectly in this interval.

We now can calculate the correlation between the hands, knowing the total variance (2.322) of the hand, and the variance of an individual hand (0.9022).

The formula is:
VAR(X+Y) = VAR(X) + VAR(Y) + 2*COR(X,Y)*SD(X)*SD(Y)

If the correlation COR(X,Y) is zero, then variance just adds up, as originally stated when summing up playing time.
If the correlation is exactly +1, then this is the first binominal theorem, and
VAR(X+Y) = (SD(X) + SD(Y))^2 or correspondingly SD(X+Y) = SD(X) + SD(Y). Examples of +1 correlation is back betting.
If the correlation is exactly -1, then SD(X+Y) = SD(X) - SD(Y) (second binominal theorem)
Examples of -1 correlation is black&red bets on a zero-less roulette wheel, a zero-comission baccarat game (banker/player). A blackjack example of strong negative correlation is insuring a "strong hand" like 20 or blackjack.

So for the AA split hands "left" and "right":
COR(left, right) = (VAR(left + right) - VAR(left) - VAR(right)) / (2 * SD(left) * SD(right)) = (2.322 - 0.9022 - 0.9022) / (2 * 0.9498 * 0.9498) = 28.7%

Both AA split hands have a correlation of 28.7%. This is quite well comparable to the correlation of two different hands at the same round (ca. 30%). The reason is of course the dealer. Both split hands are played against the same dealer.

Hope that clarifies a lot about variance. It's not simple, but no voodoo.

 

[k_c]

I may be making a fool out of myself here but how could there not be covariance even with infinite deck if they are played against the same dealer hand. If he has an ace under or hits to 21 you are in bad shape on both hands. If the dealer busts you win both no matter what else happens. Am I missing something here.

I think you are right. I ran some data for the max number of decks my program can handle (41,297,762) and got similar data to a single deck. Initially it seemed to me that for a large number of decks covariance would disappear but evidently it doesn't.

 

[k_c]

your probabilities are off under @mangoj using your win, push, lose probabilities

As far as I know those probabilities are right. They are the probabilities ending a round of splitting A-A versus 9 and bankroll changing by -2,-1,0,+1,+2 units.

 

[k_c]

I found the information in TOB p.142. The formula is for playing 2 spots in general. The covariance of a split hand is a more specific example of 2 spots with a higher codependence for any deck composition. Naming a specific pair v a specific dealer card is very specific with a higher codependence for the deck composition. Most splits are offensive and would favor the player when used.

I'm not very familiar with statistics. I was just wondering about variance of split hands. I wondered if it's possible to figure variance for a split hand round just by being able to compute variance for the individual split hands, which I'm able to do. It's probably not possible but I thought maybe someone with more knowledge than I have might say otherwise.

 

[MangoJ]

Initially it seemed to me that for a large number of decks covariance would disappear but evidently it doesn't.

Covariance is mainly due to the dealer, not the finite deck.
Make a program where each hand plays against a different dealer (but from the same finite deck), and you see covariance will almost disappear.

Hey wait, multiple hands against a different dealer each, dealt from the same deck - this is equivalent to a shoe game when you play the hands heads up one after another. We already know that past rounds wins/losses won't affect future wins/losses (it does, however only to a very small amount).

 

[MangoJ]

I wondered if it's possible to figure variance for a split hand round just by being able to compute variance for the individual split hands, which I'm able to do.

Depends on the degree of accuracy. If you assume a 30% correlation (which is very reasonable for a non-ENHC game), then you can figure the variance of split hands from the variance of individual hands.
Then
Variance(total split hand) = 2*(1+0.3)*Variance(individual split hand)

 

[iCountNTrack]

your probabilities are off under @mangoj using your win, push, lose probabilities

Euhh what are you basing your statement on

 

[tthree]

OOPS Made another mistake

Euhh what are you basing your statement on

I just noticed the "if each possible two card draw is considered". My statement was based on a one card draw. I now realize it was made in error. Thank you for correcting me or at least questioning me.

 

[k_c]

Depends on the degree of accuracy. If you assume a 30% correlation (which is very reasonable for a non-ENHC game), then you can figure the variance of split hands from the variance of individual hands.
Then
Variance(total split hand) = 2*(1+0.3)*Variance(individual split hand)

Thank you, Mango, for your information.

Questions (which may have been answered by your explanation, but still escaped me):
I am assuming the 30% correlation figure is not specific to A-A v 9. If not, where does it come from? Is it a general figure that applies to most blackjack hands? Will it vary with shoe composition to some extent?

Using 30% correlation for A-A v 9 results in a very good approximation of its variance using your formula. Would other splits be expected to have as good of an approximation using a 30% correlation?

Thanks again.

 

[MangoJ]

The covariance would in principle depend on the split value vs. upcard, together with rules (I guess strongest on S17/H17 and on DAS/NDAS). If you are counting it will also depend on the count (although I would not expect that effect to be high).
The true answer is of course exact calculation (or simulation). I guess, if you split against dealer stiff hands, covariance should be higher (as you won't bust your hands).

If you must know covariance between split hands, you should simulate for each of them. There are only 100 different splits, this should be doable on a weekend if you must.