Consistency in Lists of Numbers
By Tom Brown
- 34 reads
A measure of probability distribution for consistency in large lists of numbers is explained. If there are discrepancies it would be a sign of tampering.
I will try to explain in much detail how my method works and why, but without notation.
Given a number N, how many numbers up to N begin with the digit One?
-
Imagine a clock with one hand (a dial), marked from 1 to 9 (without a ten, or Zero). Starting from one imagine the clock hand is ticking from 1 to 9, as 1-2-3-4- … 8-9- followed again by 1-2-3 … 9 - … in other words the clock does not count a Zero.
N counts the total amount of ticks, and Q counts the number of cycles of the hand (revolutions).
A cycle is 9 ticks. So after 9 ticks I will have a One again, that is for each cycle One is counted once, or, 1 is added to Q each time the clock reads a One. So you count like that from 1 and up to the given number N while also counting the cycles.
After N ticks altogether you've had Q cycles of 9 ticks, or N = 9Q ( + R) , that is 9Q plus the R remaining ticks.
The desired formula, our final answer is therefore Q = (N - R) / 9 with R < 9
QED.
-
The result can be easily extended to the other digits by symmetry, it is also easy to directly calculate the probabilities themselves. The probability for a given number n to begin with a One is then 1/9, since for each digit it has to the same, and the nine probabilities added must give 1.
This then must apply to large lists of random numbers.
The idea of a test for fraud is to physically count L, those beginning with a One that are in your given list, find the ratio L/N, and compare with how closely it corresponds to the actual probability of 1/9.
-
This could indicate if the lists have been tampered with (been manipulated). I assume there are other such tests too for auditing. This might also apply to such as suitable data in mathematical or scientific tables.
You may say the reason the method actually works is that a number n with a digit Zero in is not counted. Actually I think this could give a simpler proof.
A friend told me of this long ago, it might have been published.
- Log in to post comments