- #1

- 352

- 1

## Main Question or Discussion Point

I have been researching the bible code theory, and I have a few problems with it.

(btw, and ELS value is how many letter's you skip to spell out a word, like a crossword puzzle.)

When the statistics are presented, they are usually worded like "The changes of this particular encoded word occuring with this specific ELS value in this specific chapter of the bible is 1 / 65000). But the only problem is I think the writer twisted these statistics by not taking into account that you can search at multuiple ELS values (I.E. every -10000 letter to every 10000 letter), and they probally didnt take into account that there is ALOT of text in the bible.

I know that there's alot of advanved mathematicians here and I would like their input on the probability of finding a particular code (perhaps an equation). Out of the limited knowledge in math that I have, this is what I derived :uhh: I am in no way saying its correct lol. Everything below here is probably extremely inaccurate.

The changes of any one particular letter occuring in a certian place (in the english language) would be 1 / 26. This is assuming that all 26 letters occur the same number of times.

So the change of finding another letter a little later would be 1 / (26 * 2) because you are searching for 2 letters. So let's make this into an equation.

[tex]P = \frac {1} {26 \cdot L}[/tex] where P is the probability, 26 is how many combination if letter's there are, and L is the length of the word you are searching for. So if we take into consideration that we are searching at multiple ELS values, we have

[tex]P = E \cdot \frac {1} {26 \cdot L}[/tex] where E is the ELS range. I.E if you are searching from -1000 to +1000, E would equal about 2000 (even though you wont be searching ELS values of 0 and 1).

Now this is just a guess, but if we were to take into account how many letters we are searching (the bible probally has millions), we would have

[tex]P = EC \cdot \frac {1} {26 \cdot L}[/tex] where C is the total count of how many letters you are searching, but maybe you would have to divide E by L first. Im probally way off :rofl:

Any ideas?

(btw, and ELS value is how many letter's you skip to spell out a word, like a crossword puzzle.)

When the statistics are presented, they are usually worded like "The changes of this particular encoded word occuring with this specific ELS value in this specific chapter of the bible is 1 / 65000). But the only problem is I think the writer twisted these statistics by not taking into account that you can search at multuiple ELS values (I.E. every -10000 letter to every 10000 letter), and they probally didnt take into account that there is ALOT of text in the bible.

I know that there's alot of advanved mathematicians here and I would like their input on the probability of finding a particular code (perhaps an equation). Out of the limited knowledge in math that I have, this is what I derived :uhh: I am in no way saying its correct lol. Everything below here is probably extremely inaccurate.

The changes of any one particular letter occuring in a certian place (in the english language) would be 1 / 26. This is assuming that all 26 letters occur the same number of times.

So the change of finding another letter a little later would be 1 / (26 * 2) because you are searching for 2 letters. So let's make this into an equation.

[tex]P = \frac {1} {26 \cdot L}[/tex] where P is the probability, 26 is how many combination if letter's there are, and L is the length of the word you are searching for. So if we take into consideration that we are searching at multiple ELS values, we have

[tex]P = E \cdot \frac {1} {26 \cdot L}[/tex] where E is the ELS range. I.E if you are searching from -1000 to +1000, E would equal about 2000 (even though you wont be searching ELS values of 0 and 1).

Now this is just a guess, but if we were to take into account how many letters we are searching (the bible probally has millions), we would have

[tex]P = EC \cdot \frac {1} {26 \cdot L}[/tex] where C is the total count of how many letters you are searching, but maybe you would have to divide E by L first. Im probally way off :rofl:

Any ideas?