standard deviation

sagefr0g

sagefr0g

Well-Known Member
say:
we have two games
one game has SD = X and is played f% of the time
the other game has SD = Y and is played z% of the time

mathematically what is the proper way to calculate the 'combined' standard deviation?
 
sagefr0g said:
say:
we have two games
one game has SD = X and is played f% of the time
the other game has SD = Y and is played z% of the time

mathematically what is the proper way to calculate the 'combined' standard deviation?
Click to expand...

Assuming f and z are fractions rather than percents for simplicity.

Variance=(f*X^2+z*Y^2)

Sandard_Deviation=SquareRoot(Variance)
 
lol, so a rehash

Canceler said:
I would start with this thread, which seems like it was active only a couple months ago, but is actually two years old, WTF?
Click to expand...

lol, thanks as always, Canceler :)
but yeah, two years old, good reason i guess for not being able to remember how to do it, but hazily suspecting it had something to do with taking a square root before adding.... edit: (geesh, yeah two years, took that much time to realize how much i care about standard deviation) :confused::whip:
some stuff you can just add, like i think EV is additive, other stuff, like standard deviation, errhh a bit trickier, lol

so this looks key here to me:
Yes. First of all, there's a question of how you're calculating SD to begin with. If you're playing on a computer which is tracking your results by hand, that's probably okay. SD(total) = sqrt(SD(1)^2 + SD(2)^2 + SD(3)^2).

But more often, SD is calculated from session wins, which means you calculate SD(total) = sqrt((ActualWin(1)-ExpectedWin(1))^2 + (ActualWin(2)-ExpectedWin(2))^2 + ... (ActualWin(n)-ExpectedWin(n))^2).
Click to expand...
 
iCountNTrack said:
Assuming f and z are fractions rather than percents for simplicity.

Variance=(f*X^2+z*Y^2)

Sandard_Deviation=SquareRoot(Variance)
Click to expand...

ok, thank you ICNT.
so digressing a bit......
say you have in the case of blackjack
SD = Z for a hand of blackjack
so to get the standard deviation for N hands would it be:
Z*SQRT(N) ?
 
sagefr0g said:
ok, thank you ICNT.
so digressing a bit......
say you have in the case of blackjack
SD = Z for a hand of blackjack
so to get the standard deviation for N hands would it be:
Z*SQRT(N) ?
Click to expand...

yep that is why accumulated expectations overcome accumulated standard deviations as the number of hands increases, because the former (expectation) is proportional to the number of hands, while SD is proportional to the square root of the number of hands
 
iCountNTrack said:
yep that is why accumulated expectations overcome accumulated standard deviations as the number of hands increases, because the former (expectation) is proportional to the number of hands, while SD is proportional to the square root of the number of hands
Click to expand...

heh, heh, funny you should mention that, cause that's what i just noticed fooling around in excel with this stuff. kewl!
 
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