Re: Measuring Turquoise Underwear
- From: Stig Holmquist <stigfjorden@xxxxxxxxxxx>
- Date: Sun, 06 May 2007 16:49:12 -0400
On Sun, 06 May 2007 11:44:12 +0100, Evil Nigel <useweb@xxxxxxxxxx>
wrote:
Stig Holmquist wrote:
On Sat, 05 May 2007 18:42:08 +0100, Evil Nigel <useweb@xxxxxxxxxx>
wrote:
Stig Holmquist wrote:
Please explain the formula used for std.dev. and what book you usedDo you have Excel?
The std.dev. for sums in the 6/49 game is 32.8, and the std.dev. for
49 integers is 14.14. Where does 12 come from?
Stig Holmquist>
A B C D E F G
1 2 3 4 5 6 =STDEVPA(A1:F1)
1 2 3 47 48 49 =STDEVPA(A2:F2)
=AVERAGE(G1:G2)
The value in G3 = 12.36, a little higher than I calculated the average
population standard deviation of the 14M combinations.
Make that 'a little lower' - my calculated average is approx. 12.72.
Your calculation of std.dev.for F1 and F2 is based on treating the
data as a population, but they are just samples from a 14 million
large population. Thus if you treat each set as a sample you'll get
1.871 for F1 and 25.211 for F2 with a mean of 13.541.
Since I didn't really know what I was doing and I needed a measure of
diverseness, I assumed I could use either population standard deviation
or sample standard deviation provided I was consistent throughout. I
leaned towards population rather than sample because the combo
1-2-3-4-5-6 is a complete population - there are no more members, the
values of which are unknown.
But there are 43 sets of samples with std.dev. of 1.871 and only one
with 25.211. Also, there is a distribution curve for the std.dev. of
all possible combinations. The shape of that curve is not known.
That's what I said. However the stats book didn't specify that the
distribution had to be normal. Apart from the end bits, which are rather
small in comparison to 14 million, it probably is very bell-curve-ish.
Thus it seems to me that any calculation based on 12 is poinless.
I'm open to suggestions for better methods of analysis.
BTW, I owe you substantial thanks. You're the first person to have a
good read of what I'm trying to calculate and make intelligent feedback.
Evil Nigel
My memory is not what it used to be so I just recalled that I
corresponded some time ago with Harry Schneider, who wrote the
book "Lottery Numbers". It is about the UK 6/49 game and has
stats for about 52 draws. He was the first to calculate the std.dev.
for the numbers and came up with 14.1.
I then modified a formula by J.E Freund from his book "Mathematical
statistics" (1962) on p. 184 where he claims the variance for the
mean is (N+1)(N-n)/12n. I suspect this is a misprint and should not
include the n in the denominator. The revised formula would yield
50x43/12=179.17 ,which equals 13.4 as std.dev. Keep in mind that the
formula for the variance of 49 integers taken one at a time is
(N^2-1)/12 and thus close to the above formula.
His book is now available with a co-author.
Stig Holmquist
.
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