posted January 25, 2007 04:23 PM            

See http://mathworld.wolfram.com/WeakLawofLargeNumbers.html

posted January 26, 2007 12:50 PM            

The Wolfram web site on the Weak Law of Large Numbers also lists
a number of other interesting sites, including:

The Law of Truly Large Numbers: "With a large enough sample, any
outrageous thin is likely to happen..."

The Frivolous Theorem of Arithmetic: "Almost all natural numbers are
very, very, very large numbers."

None of the above, coincidentally, is a joke.

But if we consider "50-50" to be a natural number 0.5..... it is
among those numbers that so qualify as very large and indeed it may
be said to have an infinite number of decimal places. And in
that case the Law of Large Numbers gets into trouble, does it not?
I.e., can even an infinitely large random sample of data arrive
at an infinitely precise estimate of 0.5....? Two decimal places
was good enough for Grandpa and it is usually good enough for me if
I'm talking about everyday things, but that is irrelevant in the
"hard sciences" or math.

Here's a quote from one theoretical physicist that I think relevant.
It should, however, be rated "PG" if not "R" for David since it
does drop the G** word once.
http://edge.org/q2005/q05_8.html#susskind

posted January 26, 2007 12:54 PM            

where p= probability of getting exactly 50:50 in n flips

given
p=(2/(pi*n))^.5
and
p=n!/( (n!/2)^2 * (2^n) )

n*pi/2 = ( ( (n/2)^2*2^n )/n! )^2

pi = ( ( (n/2)!^2*2^n ) / n! )^2*2/n

Since the PC gags on large factorials very rapidly, this has to be turned into a
computer-friendly algorithm. With n>=100,000 how would you do this in PB?

Hint: use logs wherever you can!

posted January 26, 2007 01:21 PM            

quote:

---

It should, however, be rated "PG" if not "R" for David since it does drop the G** word once.

---

posted January 26, 2007 01:41 PM            

Equally, given a probability then the equation will give a good approximation to n.

posted January 26, 2007 01:49 PM            

By the way, there are 2 G words in the article, are you still feeling alright?

posted January 27, 2007 11:49 AM            

'Pi_est.BAS 
'Language: PBCC4.02
'Note that N must be even or answer will be zero
'Max N is not much more than 10,000
'but Stirling's estimate of log factorial might improve on that.
'Try different values of N & draw a graph of the program's various 
'estimates of N. And of course remember the Law of Large Numbers.
#COMPILE EXE
#DIM ALL
DECLARE FUNCTION LogFactorial(num AS EXT) AS EXT
DECLARE FUNCTION PiEst(N AS EXT) AS EXT
FUNCTION PBMAIN () AS LONG
    DIM A AS STRING, N AS EXT, Pi AS EXT, Pi2 AS EXT
    LOCATE 10,1
    LINE INPUT "N? ";A$
    N=VAL(A$)
    Pi=ATN(1)*4
    a$ = USING$("&=#.##############", "PB's best Pi", Pi)
    ' returns "Pi=3.14159265358979"
    PRINT A$
    Pi2=PiEst(N)
    PRINT
    PRINT
    PRINT "Pi by Charles Pegge's finagle factor"
    a$ = USING$("&=#.##############", "Estimated Pi", Pi2)
    PRINT A$
    WAITKEY$
END FUNCTION
FUNCTION PiEst (N AS EXT) AS EXT
    DIM LogPi AS EXT
    DIM X1 AS EXT,X2 AS EXT,X3 AS EXT, X4 AS EXT, X5 AS EXT
    X1=LogFactorial(N/2)
    X2=LOG(2^N)
    X3=LogFactorial(N)
    X4=((X1*2+X2)- X3)
    X5=LOG(2/N)
    LogPi=X4*2+X5
    'pi = ( ( (n/2)!^2*2^n ) / n! )^2*2/n  'cf Charles Pegge PB forums
    FUNCTION=EXP(LogPi)
END FUNCTION
FUNCTION LogFactorial(Num AS EXT) AS EXT
    DIM I AS EXT, X AS EXT, XX AS EXT
    XX=0
    FOR I=1 TO Num
        XX=XX+LOG(I)
    NEXT
    FUNCTION=XX
END FUNCTION

posted January 27, 2007 12:26 PM            

2 2 4 4 6 6 8 8
- - - - - - - - ... = pi/2
1 3 3 5 5 7 7 9

See http://mathworld.wolfram.com/WallisFormula.html

posted January 27, 2007 12:49 PM            

James:
Does your earlier formula for p also have a name?
Isn't the Wallis formula an exact one rather than an estimate
based on random numbers? (Of course one's answer will be only an
approximation, insofar as N falls short of infinity.)

Formula 12 in the Wallis reference comes closest to BASIC language
and I'm glad that it was included.

------------------

posted January 27, 2007 05:00 PM            

This is a method I came up with last night. It follows Wallis's Product quite closely.
Just by combining the logfactorials and 2^ in one loop. I am road-testing my scripting
language at the same time. I hope it makes sense. This is what it looks like:

new n=1e6, m=n*2, nn=n, b=0
{
b+log(n--/m); m-4; if m GT 0 then repeat
}
$ pi by factorials result: `2/(exp(b*2)*nn)` compare with `atan(1)*4`

[result]
pi by factorials result: 3.14159419221193 compare with 3.14159265358979

I could call it Lazy-Fingers.

posted January 27, 2007 05:28 PM            

arctan(1/2)/Pi in base 2 is 0.0010010111001000....

the binary expansion of arctan(1/2)/Pi by Simon Plouffe
http://www.cs.uwaterloo.ca/journals/JIS/construction503.gif

also see, http://www.cs.uwaterloo.ca/journals/JIS/compass.html

posted January 27, 2007 05:34 PM            

The formula I gave for the probability of a 50:50 split in terms
of pi does not have a name per se. However, most of the posts
are dancing around the same idea - we are really all talking
about the central binomial coefficient (2n) Choose (n). The
above formula comes from using Stirling's approximation on the
factorials in that coefficient. Wallis's formula is really just
another form for the same coefficient squared and rearranged
slightly. Walli's formula is exact in the limit.