Sanity check with high school math

The calculation

In the example under discussion, we consider the blue ELS to be in a fixed position (it is our axis), and we count approximately how many configurations are available for the red ELS (in the same column) that are at least as interesting as the configuration observed. The red ELS can appear above or below the blue one, and its direction can be either up or down the page. So there are 4 configurations immediately adjacent to the blue ELS, 4 more positions with a gap of one letter, 4 more with gap 2, and 4 more with gap 3 (the gap seen in the original picture). This is a total of 16 configurations. [ This estimate simplifies the criteria for competitiveness, ignoring factors such as letter-sharing and "weaved" pattern ELSs ].

We now multiply the 16 configurations by P, the probability of the red ELS occurring at one of these particular locations. This probability is the product of the letter frequencies of the red ELS, that is, the product of how often each letter of the ELS appears in the overall text. The red ELS has P = 3.6E-07. Finally, we multiply by 160, corresponding to the fact that the blue ELS is the 160th minimal in Torah for the key word "waves of light". Our result is 16(160)(3.6E-07) = .00092, or 1 in 1090, about the same as our one-column protocol result.

As a rule of thumb, for even quicker assessments, the product of the letter frequencies, on average, is about .0625 to the power of n, where n is the number of letters in the ELS.


Actually the whole sanity check is a simplification, and must be used with care - only as a tool to get a general check on the more accurate result obtained by comparing to monkey texts.

Also, we must always adjust the results to account for alternative key word choices which may or may not also be significant.

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