View Full Version : Calculating standard deviation of X hands
LordMushroom2
11-19-2007, 03:56 PM
If I know the standard deviation of 1 tournament/hand, how do I calculate the standard deviation of X tournaments/hands?
tame_deuces
11-19-2007, 05:36 PM
SD is the square root of the average squared difference between data and the mean.
So if you got {8,16,4,20}, the mean is 12. The deviations are {-4,4,-8,8} which squared are {16,16,64,64} and the square root of that is {4,4,8,8} and the average of that is 6. So your SD is 6.
I just usually think of it as the average of all the (absolute) distances to the mean - in your case for all X hands. Since I haven't done proper numbers in years and I probably even botched this minimal thing up.
LordMushroom2
11-19-2007, 08:13 PM
I play HU STTs and using my winrate I have calculated that the standard deviation for an STT is almost 1 buy-in.
What I wanna know is how do I then calculate how big the standard deviation is over a set of say 10 STTs using the knowledge of what the standard deviation is for 1 STT.
I am pretty sure there is a simple formula for this, but a quick google didnīt help me.
jay_shark
11-19-2007, 08:21 PM
Sample standard deviation for n unknown is :
s.d /sqrtn
So ~ 1/sqrt(10)
Since we use the fact that the standard deviation for 1 heads up sng is ~ 1 . This should make a lot of sense since as n approaches infinity , the standard deviation approaches 0 .
LordMushroom2
11-19-2007, 08:38 PM
Thanks.
LordMushroom2
11-19-2007, 10:07 PM
[ QUOTE ]
Sample standard deviation for n unknown is :
s.d /sqrtn
So ~ 1/sqrt(10)
[/ QUOTE ]
Am I supposed to multiply the answer with the number of STTs as well? Like this:
ns.d /sqrtn
So (10*~1)/sqrt(10)
jay_shark
11-19-2007, 10:11 PM
No just the s.d for one buy-in divided by the square root of n .
For heads up it's ~ 1/sqrtn . For a 9 player sng , it's ~ 1.7/sqrtn , etc .
LordMushroom2
11-20-2007, 12:31 AM
[ QUOTE ]
No just the s.d for one buy-in divided by the square root of n .
[/ QUOTE ]
But if I do the calculations to find the standard deviation of a high number of STTs (say 900), the answer seems wrong:
1/sqrt(900) = 0,033 buy-ins
Say the buy-in is $105 and my long run ROI is 10%. I then read the answer as "it is 68,3% certain (within 1 standard deviation) I will end up with a profit of $105*900*0,1=$9450 +/- $105*0,033=$3,47. That canīt be right.
$9450 +/- $3,47*900=$3123 on the other hand looks more realistic.
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