Expected probabilities:

   (1+i.9),. 10^.1+1%1+i.9
1   0.30103
2  0.176091
3  0.124939
4   0.09691
5 0.0791812
6 0.0669468
7 0.0579919
8 0.0511525
9 0.0457575
   
Linda

-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Linda Alvord
Sent: Wednesday, May 15, 2013 1:32 AM
To: [email protected]
Subject: Re: [Jprogramming] Benford's law

Here's the actual distribution (rather than a percentage):

   newfib2=: (,{: + [: {: }:)^:((2 -~ ])`([: * 2 # ]))
   FIBS=:newfib2 1000
   F=:("."1) 1{."1 ":"1 |."1 ,.FIBS 
   $F
1000
   20{.F
1 1 2 3 5 8 1 2 3 5 8 1 2 3 6 9 1 2 4 6
   fd=: [: /:~ ({. , #)/.~
   fd F
1 301
2 177
3 125
4  96
5  80
6  67
7  56
8  53
9  45
 
At the moment I've forgotten how to get the expected frequency.  Maybe in
the morning...

Linda

   

-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of David Ward
Lambert
Sent: Tuesday, May 14, 2013 5:58 PM
To: programming
Subject: [Jprogramming] Benford's law

http://rosettacode.org/wiki/Benford%27s_law#J

When trying to recover a padlock combination my father told me to start with
1.  "Most combinations start with 1."


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