Black Jack Uneven betting Standard deviation

Dear Wizard,

To understand better my question (Underlined CAPS text at the bottom of this topic) there is a bit of theory from Casino Operation Management book:


The following formula can be used for calculating the approximate
variance per round (Peter Griffin, 1996, p. 142):
V(n) = 1.26n + 0.50n(n − 1)
where n equals the number of spots played and the standard deviation
per round equals the square root of the variance:
V
(n) =
SQRT(1.26n
+ 0.50
n(n −
1))

This formula yields the following when multiple spots are played:


Hands l Variance l STDEV

1_____l__1.26__l_1.1225
2_____l__3.52__l_1.8762
3_____l__6.78__l_2.6038
4_____l__11.04_l_3.3226
5_____l__16.30_l_4.0373
6_____l__22.56_l_4.7497
7_____l__29.82_l_5.4608

Assume this scenario: A casino executive has the choice of two different
players. Player “A” bets $350 per round in blackjack and plays one spot,
whereas player “B” bets the same $350 per round, but instead of playing
only one spot he plays seven spots of $50 each. Which player should the
casino executive choose?
Player A’s standard deviation per hand is as follows:
Amount Wagered^2 x Times Wagered -

3502 × 1 = 122,500

The square root of this product is then multiplied by the standard deviation
for one spot in blackjack:

SQRT(
122,50
0) × 1.1225 = 392.875

This $392.875 represents the standard deviation of the house win amount
on a $350 bet on one spot during a single round of play. If the player were
to play 100 rounds, the standard deviation would be:


SD per round x SQRT(Total rounds played):

392.875 × SQRT(1
00) = 3,928,75

The standard deviation of the house win amount for the seven $50 bets
placed by player B would be different, because the outcomes of the bets at
the seven spots are not independent.

Amount Wagered^2 x Times Wagered -

502 × 1 = 2,500

For player B, the square root of this product is multiplied by the standard
deviation for seven spots in blackjack. This differs from the formula used
to determine the standard deviation per round for player A, since only
the standard deviation for one spot was used.

The product represents the
standard deviation of the house win on seven $50 spots for a single round
of play for player B.


SQRT(2,500) × 5.4608 = 273.04

If player B were to play 100 rounds, his standard deviation would be:

SD per round x SQRT(Total rounds played):

273.04 × SQRT(1
00) = 2,730.40

So the question is HOW TO CALCULATE THE STANDARD DEVIATION FOR THE PLAYER "C",
who bets the same $350 per round, but instead of playing
only one spot he plays seven spots with uneven bets of, let's say: $200+20$+20$+20$+20$+25+45$ ???

Is the calculation should be the same as at 50$ x 7 example? (to get average per round: 350$/7 )
What if there is a wide bet spread lets say: 2$ 2$ 2$ 2$ 2$ and 340$ So the average bet is not balanced anymore!?
Or what if sum all bets are uneven from a round to round?

Anyone? :D

Sad...

Rly...?

Quote: Winter

Sad...

It happens to most of us. You take the time to form and write a thoughtful post and get no responses. There can be many reasons why and I wouldn't want to speculate in this case. What I can say is that some may find your last three posts off putting. Typing the word sad is not the way to get a response.

Perhaps you could add something or ask the question in a different way. Good luck.

Sure, it was only a simple way to put topic in the top of list. Otherwise I can't see any benefit to create a new topic with a bit different construction of a question. Question I've asked is interesting itself and topic also provides some useful information which might be interested for some group of people. Actually I'm surprised why it hasn't replied yet.

Quote: Winter

So the question is HOW TO CALCULATE THE STANDARD DEVIATION FOR THE PLAYER "C",
who bets the same $350 per round, but instead of playing
only one spot he plays seven spots with uneven bets of, let's say: $200+20$+20$+20$+20$+25+45$ ???

Is the calculation should be the same as at 50$ x 7 example? (to get average per round: 350$/7 )
What if there is a wide bet spread lets say: 2$ 2$ 2$ 2$ 2$ and 340$ So the average bet is not balanced anymore!?
Or what if sum all bets are uneven from a round to round?

I gave this question some thought when it was first posted. I would not know how to do this except by simulation and the OP appears to want a closed formula applicable to a variety of wagers. I personally do not find the question interesting.

Anyway, thank you for reply teliot.

Quote:

The following formula can be used for calculating the approximate
variance per round (Peter Griffin, 1996, p. 142):
V(n) = 1.26n + 0.50n(n − 1)

If the variance would be purely linear in n, this would indicate that all betting spots are uncorrelated. Since there is a n² contribution, you have correlation between spots.

Just the background:

If you play 2 spots A and B, in principle you have
Variance(A+B) = Variance(A) + Variance(B) + Correlation(A,B) + Correlation(B,A).

Likewise, if you play n spots A,B,C,...Z, you have
Variance(A+B+C+...+Z) = Variance(A) + Variance(B) + Variance(C) + ... + Variance(Z)
+ Correlation(A,B) + Correlation(A,C) + ... + Correlation(A,Z)
+ Correlation(B,A) + Correlation(B,C) + ... + Correlation(B,Z)
+ Correlation(C,A) + Correlation(C,B) + ... + Correlation(B,Z)
+ ...
+ Correlation(Z,A) + Correlation(Z,B) + Correlation(Z,C) + ...

I hope you get the pattern. There are n Variance-terms and n(n-1) Correlation-terms.

Now a reasonable assumption is, that all betting slots are equal in gameplay, rules, and odds.
Then Variance(X) = X² * var and Correlation(X,Y) = X*Y * cor for two new variables "var" and "cor" representing the variance and correlation per unit bet, and X and Y are the bet sizes of different spots.

Now we need to figure out what "var" and "cor" are in value. We just plug in "n unit bets", so there are n different spots "A,B,C,...,Z", and all A=1, B=1, C=1, ... Z=1. Hence for n unit bets the total variance must be (just count the number of Variance() and Correlation() terms)
Variance(n unit bets) = n * var + n(n-1) * cor.
This coincides in structure exactly with the quoted Griffins formula, so you can now easily identify var=1.26 and cor=0.5.

Quote:

So the question is HOW TO CALCULATE THE STANDARD DEVIATION FOR THE PLAYER "C",
who bets the same $350 per round, but instead of playing
only one spot he plays seven spots with uneven bets of, let's say: $200+20$+20$+20$+20$+25+45$ ???

That said, you can simply plug in the numbers. From your seven bets, take each squared and multiply by 1.26 for the variance part.
From the correlation part, take each (different) pair and multiply by 0.5.

Variance(Player C) = 1.26 * ($200² + 4 * $20² + $25² + $45²) + 0.5 * ( $200 * (4 * $20 + $25 + $45) + 4 * $20 * (3 * $20 + $25 + $45) + $25 * ($200 + 4 * $20 + $45) + $45 * ($200 + 4 * $20 + $25) )
= $$ 86880.

If you happen to check this formula, don't be confused by the factors "4" (and "3" likewise). In your Player C bet pattern, you have 4 times the bet of $20. As you see the formula is rather long and would have 7*7 = 49 terms, so I use some abbreviation here.

Anyway, the variance (in square dollars) is 86880, so to answer your question the standard deviation of player C is exactly $294.75.

Wow! Great answer! I appreciate that!

Ghm...
MangoJ,

I have different outcome var= 94880 and SD=308.026
I've tried to calculate using all 49 numbers...

...if you look at this part of calculations you posted: ...4 * $20 * (3 * $20 + $25 + $45) + $25 * ($200 + 4 * $20 + $45) + $45 * ($200 + 4 * $20 + $25) )

Could you explain why you skipped bet of 200$ in this part of calculation? It was with reason or just type mistake?

And another question to get complete correct session SD (let's say for 100 hands with totally uneven bets)
The most accurate is to calculate variance for each particular hand (100 values) then sum all up and take SQRT?

You are right, I slipped a $200 in there. It should read "4 * $20 * ($200 + 3 * $20 + $25 + $45)". Then the result is indeed $308.

I thought about this problem more a bit, and came up with a geometrical approach to it.

Imagine a 7 (or more) dimensional euklidian space. Put in 7 vectors with your length of your betsizes, and orient them in a way that they *pairwise* enclose an angle of alpha, where cos(alpha) = 0.5 / 1.26 and could be named "correlation angle".

Once all vectors are oriented, add up all 7 vectors. The SD should then be 1.26 times the length of that resuling vector.

Oh,

Nice geometrical interpretation...I'll try it.

If we're talking with regards to multi-hands calculations...

Let's say 2 uneven hands...

4 boxes and then 6 boxes 50$ in each...

so Var(1st hand) = 27600
and Var(2nd hand) = 56400

Then SD1= 166.13
and SD2= 237.48

So the question is: SD(series) = SD1+SD2= 403.61$ or SD(series)= SQRT(Var1+Var2) = 289.82$ ?

Not sure if I understand your question.

If bets are uncorrelated, always add up variances.

I'll try to explain my question in a different way.

How to calculate SD for playing session of 2 hands

1 st hand - Player bets 6 boxes, 50$ each
2 st hand - Player bets 4 boxes, 50$ each

First I've tried to calculate it separate using your formula. Let's get Variance for the first hand and Variance for the second

Var 1 hand = 1.26*(6*(50^2))+0.5*6*50*5*50= 56400 Sq $

Var 2 hand = 1.26*(4*50^2))+0.5*4*50*3*50= 27600 Sq $

So what is correct session SD?

SQRT(VAR1+VAR2) = 289$ or SD1+SD2 = 403$ ?

I've tried to calculate in a different way (through SD coefficient):

SD per 1 playing spot so according to Peter Griffin formula:

V(n) = 1.26n + 0.50n(n − 1)

where n equals the number of spots played and the standard deviation
per round equals the square root of the variance:

SD(n) =SQRT(
( 1.26n
+ 0.50
n(n −
1) )

Then

Approx variance for 1 unit for 4 boxes is 11.04 and SD is 3.3226
Approx variance for 1 unit for 6 boxes is 22.56 and SD is 4.7497


Then

Hand 1 SD = SQRT(50^2)*3.3226= 166.13
Hand 2 SD = SQRT(50^2)*4.7497= 237.48

According to P.Griffin approach:
Total session SD = SD1+SD2 = 403.615

I'm not sure is it correct... :(

Quote: Winter

I'll try to explain my question in a different way.

How to calculate SD for playing session of 2 hands

1 st hand - Player bets 6 boxes, 50$ each
2 st hand - Player bets 4 boxes, 50$ each

First I've tried to calculate it separate using your formula. Let's get Variance for the first hand and Variance for the second

Var 1 hand = 1.26*(6*(50^2))+0.5*6*50*5*50= 56400 Sq $

Var 2 hand = 1.26*(4*50^2))+0.5*4*50*3*50= 27600 Sq $

So what is correct session SD?

SQRT(VAR1+VAR2) = 289$

Yes, that is the correct SD, provided that 1st and 2nd hand is uncorrelated. They are if there is a shuffle inbetween. If there is no shuffle inbetween, in principle they are correlated as cards come from the same deck. But if you ignore that, then $289 should be correct.

Quote:

According to P.Griffin approach:
Total session SD = SD1+SD2

That's most surely not what Griffin says.

Don't add SDs of uncorrelated bets. Never, they simply don't add. Add up variances, then from the total variance get your total SD.

Quote: Winter

where n equals the number of spots played and the standard deviation
per round equals the square root of the variance:

SD(n) =SQRT(
( 1.26n
+ 0.50
n(n −
1) )

So.... for craps.

Is it a simple calculation of squaring the standard deviation to determine the variance?