Standard Deviation - Multiple Hands

I am doing more work on probability modelling and EV in MS Excel. I have more questions on standard deviations. Sorry!!
a) How do I calculate the standard deviation for one round of cards in blackjack? Is it true that the VARIANCE for one hand is equal to 1.33 – therefore will the standard deviation of one hand be equal to the square root of 1.33 i.e. 1.153?
b) So let’s say that I will play 4 hands against the dealer and bet $10 on each hand. How do I calculate the standard deviation for the one round of 4 hands? Is co-variance relevant in this situation?
c) Where can I find the relevant co-variance figure for playing for 2, 3, 4, 5, 6 or 7 hands at a time?
d) If i play 4 hands of $10 and then 4 hands of $1,000, will the theoretical $ standard deviation in the second round be 100 times that of the first round?
I’d appreciate any helpful comments to any of the above questions! Thanks in advance,
Matt

Yes, the average variance for a hand of BJ is around 1.33 so the standard deviation would be 1.153.

The covariance is very important since you are playing each hand against the same dealer hand. The covariance is about 0.5 per hand, so for four hands the variance would be about 1.33 + (3*0.5) = 2.83. There is a good chapter on this in Blackjack Attack, and here’s a link about bet sizing for multiple hands that uses this concept:

http://www.blackjackinfo.com/bb/showthread.php?p=16433

The covariance is around 0.5 for each additional hand you add. You can calculate the total variance per round using the formula we used above:

Var = 1.33 + (NumHands-1)*0.5

No, the VARIANCE will be 100 times higher, the STANDARD DEVIATION will only be 10 times higher because it is a square root function. With four hands of $10 the variance is $10 * 2.83 = $28.30 (SD = $5.32). With four hands of $1,000 the variance is $1000 * 2.83 = $2830 (SD = $53.20).

-Sonny-

Sonny thanks for your comments.

i understand your comment regarding co-variance.

in calculating the variance for 1 round of 4 hands - i thought that we would multiply 2.83 by $40 (the total amount wagered) rather than by $10 (the amount wagered per hand). SD of $5.32 seems low for 1 round of 4 hands of $10 each. Can you please confirm?

and also got you as to multiple of variance vs multiple of SD.

matt

Yeah, that was my mistake. I should have been using the squared betting unit for the variance calculation. The correct numbers should be:

Squared Bet * Variance * Number of Hands = Total Variance

One hand of $40
$40^2 * 1.33 * 1 = $2,128 (SD = $46.13)

Two hands of $20
$20^2 * 1.83 * 2 = $1,464 (SD = $38.26)

Four hands of $10
$10^2 * 2.83 * 4 = $1,132 (SD = $33.65)

Alternatively, I could have saved myself some trouble by finding the SD first, then multiplying it by the bet amount like this:

One hand of $40
sqrt(1.33 * 1) * $40 = $46.13

Two hands of $20
sqrt(1.83 * 2) * $20 = $38.26

Four hands of $10
sqrt(2.83 * 4) * $10 = $33.65

Sorry for the confusion. This is a good example of what can happen when you do square root functions without paying attention!

-Sonny-

ok, perfect.

so if i played 5 hands of $100, then my SD for that one single round would be:

sqrt ($100^2 x 3.33 x 5)
= sqrt($166,500)
= $408

or sqrt(3.33 * 5) * $100
= 4.08 x $100
=$408

where variance =
1.33 + (number of hands-1)*0.5
=1.33 + (5-1)*0.5
= 1.33 + 2 = 3.33

so let's assume further that
in round 1 we bet 4 hands of $10 and EV is -$7
in round 2 we bet 5 hands of $100 and EV is $21

then for the two rounds consolidated (note here the two rounds are in succession)
EV = -$7 + $21 = $14
SD
SD for 5x$100 is $408
SD for 4x$10 is $33.65
thus consolidated SD would be sqrt ($33.65^2 + $408^2)
= $409.39

further, if we only know that there is a certain likelihood of us betting a certain amount, how would we then calculate SD?
i.e. we will only play 1 round and there is a 50/50 chance of how we will bet, depending on the occurence of a previous event.

if there is 50% chance of us betting 4x$10 and a 50% chance of us betting 5x$100 then how would we calculate the SD for that round?
would it be equal to 0.5* EV(A) + 0.5* EV(B)
and SD would be equal to sqrt (0.5*$33.65^2 + 0.5* $408^2)
ie. EV would be 0.5*-$7 + 0.5*21 = -$3.50+$10.50 = $7
and SD would be $289.48.

is that correct?

A+

Right on!

I get $409.43 but the difference is just from rounding. Using the actual variance instead of squaring the SD is a little more precise, but obviously you understand what you’re doing so I’ll let it slide this time.

Yup, that’s it. It would be easier if you were always playing the same number of hands, but for a strategy where you are switching back and fourth the method above is the best way.

-Sonny-

No big deal. I do agree, fwiw, with all that you and Sonny are saying.

Hewre are my areas of confusion.

The avg co-variance will vary from, from what I've seen, from 0.43 to 0.48 (as opposed to Sonny's very good general estimate of 0.5 depending on rules.) I always use 0.48 as co-variance in cvcx sims and usually seem to get close to sim results but who cares.

Like DAS will have higher co-variance than D10 but no DAS.

Also, why I say "avg co-variance" rather than just "co-variance" is because I believe, but have little actual proof of my belief, that the actual co-variance will vary a little bit with each TC and would definitely be different at TC -10 than TC +10, both of which would be different than covariance at TC=0.

Also, it has always been very frustrating to me that I have absolutely no way that I know of of actually computing an avg co-variance for a given set of rules let alone a co-variance at each TC.

I'm not even sure what software could break this out but, it seems to me, that QFIT's software, while, as far as I know and believe can cure the common cold, will not do this and that it is invisibly calculated in sim results in his software. Both co-variance at each TC and avg covariance over all TC's.

All of this just a way of saying to you Matt21 of other slight variables to think about including how you determine what your covariance actually is other than using an estimate lol.

How do you calculate the SD for a round with 2 different bets, let's say $10 and $20? The examples given were all with equal bets.

this is a good question - and i am not sure i know the answer. could it be as follows:


If SD for One hand of $40
sqrt(1.33 * 1) * $40 = $46.13

And if SD for Two hands of $20
sqrt(1.83 * 2) * $20 = $38.26

then one hand of $40 and one hand of $10 in the same round:
sqrt(1.83 * 1) * $40 + sqrt(1.83 * 1) * $10

Sonny could you comment on this for us?

I guess you must first calculate the variance taking care of the covariance and after that obtain the SD.

Mike

that's correct - i used 1.83 - ie to incorporate the co-variance.
and by doing the sqrt you are transforming the variance into the SD.

Matt

That formula will give you the wrong answer ($67.63). The SDs are not additive so you have to do all the work using the variance before you take the square root. The formula above should be changed to:

SD = sqrt(1.83 * $40^2 + 1.83 * $10^2)
SD = sqrt(1.83 * ($1600 + $100))
SD = $55.78

That will give us a ballpark estimate. As others have mentioned, the covariance will change as the bets drift apart. If we always leave the first hand at $10 and we raise the second bet to $20, the covariance drops from 0.5 to around 0.4. If we raise the second hand up to $40 then the covariance is closer to 0.22. Using these new numbers we can change our original formula to:

SD = sqrt((1.33 + 0.22) * ($1600 + $100))
SD = sqrt(1.55 * $1700)
SD = $51.33

This number seems reasonable if we look at the situation. The SD of a single hand of $50 is $57.66 and two hands of $25 each is $47.83, so our number falls between those two. It is more risky than an even split but less risky than a single hand.

As the difference between the bets gets larger, the covariance decreases since the total variance is mostly influenced by the hand with the larger bet. Since the value of the hands are not equally weighted, the variance will begin to converge on the variance of the larger hand. Imagine betting $5 on one hand and $500 on the other. That little $5 hand isn't going to make much of a difference, right? The total variance will be very similar to a single hand of $500.

-Sonny-

Sonny,

How do you calculate the values of the covariance for the different bets that are mentioned?

I'd like to know the method you use to estimate these different covariances.

Thanks,

Mike

If you are using different bets the general equation will not apply because the variance per hand is different. The more money you bet the higher the variance. So for instance if you are playing two hands, one flat-betting (h1) and the other hand with a bet spread (h2) you cannot use,

Var(h1+h2)=2*1.26+2*Cov(h1,h2), it will be rather: Var(h1+h2)= 1.26+Var(h2)+2*Cov(h1,h2) because Var(h2) is not equal to 1.26

As far as calculating Covariance when the bets are different, covariance is linear in the sense that Cov(h1,a*h2)=a*Cov(h1,h2) where a is a constant.

But you can see that using different bets can add unnecessary mathematical complications so why bother.

i have come across another playing situation that is similar to the above and wanted to get feedback whether i am doing the calculation correctly?

Assume Event 1 has EV of $46 and SD of $219.
Assume Event 2 has a 25% chance of EV of -$7 and SD of $166 and a 75% chance of not occurring at all i.e. EV and SD of $0.

In that case would the EV and SD of Events 1 & 2 be as follows:

EV = $46 + 0.25*-$7 = $44.25

SD - sqrt(sum of variances)
Event 1 Variance - 219^2 = 47,961
Event 2 Variance - 166^2 multiplied by 0.25 = 6,889
= 47,961 + 6,889 = 54,850
Thus SD = sqrt (54,850) = $234.20

Could anyone comment on whether I have done this correctly?

Many thanks in advance!