Benford's law - Generalization to digits beyond the first -- Is there a VBA solution?

Juggler_IN

Active Member
Joined
Nov 19, 2014
Messages
358
Office Version
  1. 2003 or older
Platform
  1. Windows
I am looking for a VBA led output for 2nd, 3rd, 4th, etc, (Generalization to digits beyond the first).

The refrence link is : Benford's law - Wikipedia

Essentially the formula to code is below the text "And the probability that d (d = 0, 1, ..., 9) is encountered as the n-th (n > 1) digit is"

Thanks
 

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Press Ctrl+End to move to what Excel thinks is the last used cell.
To replicate the table,

[Table="width:, class:grid"][tr][td="bgcolor:#C0C0C0"][/td][td="bgcolor:#C0C0C0"]
A​
[/td][td="bgcolor:#C0C0C0"]
B​
[/td][td="bgcolor:#C0C0C0"]
C​
[/td][td="bgcolor:#C0C0C0"]
D​
[/td][td="bgcolor:#C0C0C0"]
E​
[/td][td="bgcolor:#C0C0C0"]
F​
[/td][td="bgcolor:#C0C0C0"]
G​
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H​
[/td][td="bgcolor:#C0C0C0"]
I​
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J​
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K​
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L​
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5​
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Pos\Digit
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0​
[/td][td]
1​
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2​
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3​
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4​
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5​
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6​
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7​
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8​
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9​
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6​
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1​
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n/a​
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30.10%​
[/td][td="bgcolor:#CCFFCC"]
17.61%​
[/td][td="bgcolor:#CCFFCC"]
12.49%​
[/td][td="bgcolor:#CCFFCC"]
9.69%​
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7.92%​
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6.69%​
[/td][td="bgcolor:#CCFFCC"]
5.80%​
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5.12%​
[/td][td="bgcolor:#CCFFCC"]
4.58%​
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7​
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2​
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11.39%​
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10.88%​
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10.43%​
[/td][td="bgcolor:#FFFFCC"]
10.03%​
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9.67%​
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9.34%​
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9.04%​
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8.76%​
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8.50%​
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8​
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3​
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10.18%​
[/td][td="bgcolor:#FFFFCC"]
10.14%​
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10.10%​
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10.06%​
[/td][td="bgcolor:#FFFFCC"]
10.02%​
[/td][td="bgcolor:#FFFFCC"]
9.98%​
[/td][td="bgcolor:#FFFFCC"]
9.94%​
[/td][td="bgcolor:#FFFFCC"]
9.90%​
[/td][td="bgcolor:#FFFFCC"]
9.86%​
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9.83%​
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9​
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4​
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10.02%​
[/td][td="bgcolor:#FFFFCC"]
10.01%​
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10.01%​
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10.01%​
[/td][td="bgcolor:#FFFFCC"]
10.00%​
[/td][td="bgcolor:#FFFFCC"]
10.00%​
[/td][td="bgcolor:#FFFFCC"]
9.99%​
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9.99%​
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9.99%​
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9.98%​
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5​
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10.00%​
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10.00%​
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10.00%​
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10.00%​
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10.00%​
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10.00%​
[/td][td][/td][/tr]
[/table]
 
Last edited:
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