Multiple Hands Variance in $$

A

assume_R

Well-Known Member
Can somebody please check my math and let me know where I am mistaken? And I'm going to assume that the frequencies of each count don't matter, "eating up cards" don't matter, etc. etc.

Situation: EV = +1%, Var = 1.3, Cov = 0.5, Bet = $20

Situation #1: 1 hand of $20
EV = 1% * $20 = $0.20 / round.
Var = 1.3 * $20^2 = $^2 520 / round. Std = $23 / round

Situation #2: 2 hands of $10 each
EV = 1% * $10 + 1% * $10 = $0.20 / round
TotalVar = Var * n + Cov * n * (n - 1)
TotalVar = 1.3 * 2 + 0.5 * 2 * 1 = 3.6
TotalVar = 3.6 * $20^2 = $^2 1440 / round. Std. = $38 / round

Shouldn't the std be less when 2 hands are played with the same $$ on the table?? Perhaps my last line should have read 3.6 * $10^2??? which would make the std. $19 / round??
 
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Sonny

Sonny

Well-Known Member
assume_R said:

Situation #2: 2 hands of $10 each
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TotalVar = 1.3 * 2 + 0.5 * 2 * 1 = 3.6
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Click to expand...
Don't multiply the variance by 2. The total variance for each hand should be 1.3 + 0.5 = 1.8.

-Sonny-
 
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assume_R

Well-Known Member
Thanks Sonny.

Yet that means that the EV is the same both situation #1 and #2 ($0.20), yet the variance for situation #2 (multiple hands) is 1.8 * $20^2 while the variance of situation #1 is 1.3 * $20^2? Or do I multiply that 1.8 by $10^2?

I only did what I did because of:

"According to Professional Blackjack by Stanford Wong (page 203), the variance for similar rules is 1.32 and the covariance is 0.48. The total variance of n hands would be 1.32*n + 0.48*n*(n-1). Take the final square root to get the standard deviation."

from http://wizardofodds.com/blackjack/appendix4.html

So what about if there are 3 hands??? The standard deviation according to wizardofodds would be sqrt(1.3 * 3 + 0.5 * 3 * 2) = 2.63

Sorry for the confusing post I'm just trying to make sure I understand this 100%.
 
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assume_R

Well-Known Member
Okay I think I finally answered my own question from http://www.bjmath.com/bjmath/ror/tripror.htm (Archive copy)

1. Variance = 1.3 * n + 0.5 * n * (n-1)
2. Variance/hand = Variance / n = 1.3 + 0.5 * (n - 1)
3. Variance/hand in $$^2 = [1.3 + 0.5 * (n - 1)] * (Bet/hand)^2
4. Variance/round = Variance/hand * n = Var/hand * n = [1.3 * n + 0.5 * n * (n - 1)] * (Bet/hand)^2
5. Std/round = (Bet/hand) * sqrt(1.3 * n + 0.5 * n * (n - 1)) for n hands

So in my case, std/ round = $10 * sqrt(3.6) = $19 / round while my EV is $0.20 / round

Regarding Sonny's post, I see now that 1.8 is per hand, with a $10 bet. I multiplied by 2 to get the variance per round. But that $10, though, should stay at $10 (not $20) based on the equations i derived above.
 
blackjack avenger

blackjack avenger

Well-Known Member
The Important Points

If you bet approx 73% over each of 2 hands that you would bet on 1 hand then your long term ror is the same.

Example:
Instead of one hand of $100 you can bet two hands of $73 or $75 and have the same approx long term ror.

However:joker::whip:

If playing 2 hands over a session you will need more cash in order to keep the trip ror the same, approx 20% to 33% more $ then you would need if you were to just play one hand.:joker::whip:
 
Sonny

Sonny

Well-Known Member
assume_R said:
Regarding Sonny's post, I see now that 1.8 is per hand, with a $10 bet.
Click to expand...
Right. So for 3 hands the variance would be (1.3 + 0.5 * 2) = 2.3 per hand. For all 3 hands the SD would be sqrt(2.3 * 3) = 2.63, which confirms your numbers above. My version just rearranges the formula a bit:

(1.3 * 3) + (0.5 * 3 * 2) = (1.3 + 0.5 * 2) * 3

-Sonny-
 
A

assume_R

Well-Known Member
Sonny said:
Right. So for 3 hands the variance would be (1.3 + 0.5 * 2) = 2.3 per hand. For all 3 hands the SD would be sqrt(2.3 * 3) = 2.63, which confirms your numbers above. My version just rearranges the formula a bit:

(1.3 * 3) + (0.5 * 3 * 2) = (1.3 + 0.5 * 2) * 3

-Sonny-
Click to expand...
Okay, thanks for confirming for me. I finally got it down pat. I was confused in part also because on your previous posts (from 2009 I think) I never noticed you including the (n-1) term. I think we're on the same page now, and the equations are just rearrangements of each other.

And P.S. thanks for gettin rid of that spammer haha
 
A

assume_R

Well-Known Member
Perfectly understandable. However, the equation in the post you linked to (BR * EV / Var = Bet) doesn't take into account the fact that you might also play some -EV hands.

To take that into account, you must multiple that number ("Bet" above) by a factor which Brett Harris calls "kb". And to calculate kb, you need to know your total variance at max bets. So if your max bet is at TC = +5, and you always play 2 hands at TC = +5, you'd need to know that total variance for the TC = +5 round, which is why I wanted it in that form instead of per hand.
 
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