
Just the FAQs
Readers have asked us some very interesting questions about code research. We thought you might enjoy seeing a few of their questions along with our answers, presented below.

Q: Can you use a simple example to explain your process of calculating the odds that ELSs appear by chance in the Bible?
A: We use the mathematical system of probability to determine these odds. It’s basically the same analysis used by code breakers attempting to understand an enemy’s secret messages. Perhaps the following will help.
Steve and Marilyn Jones firmly believe in using the fickle finger of fate when it comes to making decisions. They depend on a unique and unusual method of determining what to do. Looking for codes that will give them the direction they are seeking, they go for a drive and scan the license plates of other cars on the road.
The Joneses live in a state where each license plate first shows four letters, then three numbers. Forget the numbers. The Joneses just check the letters. Whenever they need guidance, they look for their answer in the codes on license plates. Only “YES” and “NO” as codes will do. The first one of these they see decides it for them.
“NO” doesn’t have to show up as two consecutive letters, but the N must be first. So if they find a license plate with *N*O, where the *’s can be any letter, that will do as a sure sign. So, what are the chances that any particular license plate they scan will give them a “NO” message?
Solving this problem is very much like what we have to do to figure out the odds that a selected word will show up as an equidistant letter sequence (ELS) in a selected text.
Counting Up
To begin, we need to count up how many two letter ELSs can be found in any given license plate. We do this by designating each ELS by the numbers of its letter positions (from left to right). So 12 represents an ELS that appears as the first and second letters on the license plate. The possible letter positions are: 12,13,14,23,24,34. That’s six possible ELSs.
What are the chances that any given ELS is “NO”? For the sake of this simple example, we assume all letters appear equally often. So, the probability that any given ELS is “NO” is the product of the probability the first letter is N times the probability the second letter is O. That’s 1/26 times 1/26, or 1/(26x26) = 1/676. That means the odds are only 1 in 676. And the odds that any selected ELS won’t be “NO” are 675 out 676.
(In practice, we actually calculate the specific letter probabilities from how often each letter appears in the text.)
What are the odds that at least one of the ELSs will be “NO”? We can figure that out by rearranging a formula that is fairly simple.
What are the possible events that can occur? We can slot them into one of two situations. First is the situation where at least one of the ELSs is “NO.” Second is the situation where none of the ELSs are “NO.” Now these two situations include all possible things that could happen, so we know that the total of the probability that the first will occur and the probability that the second will occur must be 100%. This can be represented by the following equation:
Prob[At Least One ELS is “NO”] + Prob[No ELS is “NO”] = 100%.
Rearranging this equation, we get
Prob[At Least One ELS is “NO”] = 100%  Prob[No ELS is “NO”].
As you will recall, we already figured out what Prob[No ELS is “NO”] is for any single ELS. It is 675/676. In order for none of the six ELSs to be “NO”, it must be true that every one of them will not be “NO.” So the probability of that is
(675/676) x (675x676) x (675x676) x (675x676) x (675/676) x (675/676), or 99.12%.
So the probability that at least one of the six ELSs will be “NO” is 0.88%, or about one out of 114 (= 1 / 0.88%). So it’s quite likely Steve and Marilyn would have to look at a goodly number of license plates before they found one where “NO” was “encoded.”
Thinking it Over
What have we done to solve this problem? We first counted up how many ELSs we could find in the four letters of any license plate. Then we figured out the odds that any given ELS would be “NO” and we blended this information together to figure out the odds that any selected license plate would have “NO” encoded on it.
What if we added the additional limitation that one of the letters had to be the second letter on the license plate? We approach this the same way. We count up the number of ELS positions where this would be true. They would be: 12,23,24, out of the original set of 12,13,14,23,24,34. So there are only three ways this can happen. The odds that any selected ELS will be “NO” are still the same, but the probability that “NO” will not appear as an ELS that has at least one letter in the second position involves multiplying (675/676) times itself fewer times:
(675/676) x (675x676) x (675x676), or 99.56%.
So, this is the probability that at least one of the ELSs will be “NO” and that one of its letters will be second letter will be 0.44%. This kind of calculation is very similar to what we go through when we calculate the probability that a given ELS with a skip less than N will cross a selected section of text. In this simple example, the selected section of text is the second letter of the license plate.
In Chapter Nine of Breakthrough, we display sample calculations for a cluster of ELSs about Hitler that appear in Genesis 8. Part of our calculation is to figure out how many times we would expect that Hitler would show up as an ELS with a skip of 31 or fewer letters (31 is picked because that’s the actual size of the skips between the letters in the Hitler ELS as it appears in Genesis 8).
We do this by first using a formula that tells us how many possible ELSs can be found in the Torah that have skips of 31 or fewer letters (18.9 million). This formula is based only on the length of the Torah in letters, the number of letters in the selected ELS, and the size of the maximum skip we will accept. We then figure out the odds that any given ELS would be the Hitler ELS. It is simply the product of the probabilities that the first letter will be Heh, the second, Yod, the third, Tet, the fourth, Lamed, and the fifth, Resh. That gives us odds of only one in 4.6 million that any selected five letter ELS will spell Hitler in Hebrew. By dividing the 18.9 million possible ELSs by the 4.6 million, we get that Hitler is expected to show up as an ELS 4.1 times in the Torah with a skip of 31 or fewer letters.
The Joneses’ method is a little more complicated than flipping a coin, which would be a lot less trouble and expense, but it is the same idea.
Q: Should we expect that real Bible codes will only express truth?
A: We don’t think so. Our view is that real Bible codes express a coherent viewpoint, which could come from a truth teller or a liar. We suspect that Bible codes are no different than the literal text of the Bible after removing all attribution. Take Genesis 3:45. The serpent speaks a lie, so clearly there are lies in the Bible if you remove attribution.
It is also possible that portions of a real Bible code are coincidental, so we don’t think people should go around hanging their hat on something discovered as an intelligible code.
Q: A distraught woman called me and asked me to do a search on herself and her husband. You see, the husband had left home claiming that they were never really married. So, I asked her if there was anything unusual about the marriage. She told me that they were married under three palm trees. So I put in 3Palm and tree and the date of their marriage, etc., a total of 19 pieces of information. I had everything but the husband's last name. Well, I figured, the 18 marital facts wouldn't mean anything if the husband's last name didn't show up. But it did! The six Hebrew letters that spelled his name showed up in the mix. Was it just chance that his name landed on Proverbs 11:29 “He that troubleth his household inherits the wind.”
A: The skeptics’ example of the Hanukah cluster found in a Hebrew translation of Tolstoy’s War and Peace, which we all know is sheer coincidence, would likely be rated as more improbable than your cluster. That may be surprising to you but understanding why is interesting.
The main problem with your cluster is that you need to find many more ELSs with much higher R Values. To do this, you would need to find a number of ELSs that are seven or eight or nine or more letters in length. The longer an ELS is, the more likely it is that you won’t find it. For every letter you add, your reducing the chances of finding that ELS by a factor of about 20.
Here’s what is going on. Search for virtually any 3 or 4letter long ELS and you are almost sure to find it in your matrix. So if you find an ELS of that length, you haven’t really found anything that really means anything. This is also largely true of almost any 5 letter long ELS you may search for. Search for a 6 letter long ELS and your chances of finding it drop to something in the range of 10% to 50%still quite likely to happen by chance.
In contrast, suppose you search for a 10 letter long ELS in your matrix. Odds are about 1 in 1 million to 1 in 2 million you won’t find it. So if you do find it, it is likely you really have found something that was intentionally placed there. So what you need to do if you really believe there is something to the matrix you have presented is to search before and after each of the ELSs you have found to see if they can be extended into longer ELSs that make sense in terms of your search topic. If you come up with a number of them, then you may have the start of something. Otherwise, all your matrix shows is that you’ve put a lot of time into searching for things, most of which could be there by chance, and you found a number of them.
Now, on to a more general philosophical problem. Trying to use codes to predict the future or to make decisions is downright hazardous at best. Even if you find something you think is relevant to your topic, do you really know that that word was intended to be there? Not really. There’s always a chance it could just be there because there are an awful lot of ways for it to appear there by chance. So you’re really taking an unwarranted risk to depend on any specific thing you might think the codes are telling you.
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