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The Flooble Code (Posted on 2003-03-18) Difficulty: 4 of 5
A book titled The Bible Code introduced the topic of equidistant letter sequences (ELS), described below, for finding words “hidden” in text. That book referenced the Hebrew Bible, but prompts a question about finding any given word in any, say, English-language text.

For simplicity, and to better match the Hebrew, spaces and punctuation are removed. A particular text that I have in mind, thus crunched, has 284,939 characters remaining (letters and digits). How many times would you expect to find the word FLOOBLE as an equidistant letter sequence in the text? Ignore case. The word can start at any of the 284,939 characters and proceed by skipping any constant number of letters forward or backward. So, for example, if the 11,000th character were an F and the 10,000th an L, and the 9,000th an O, etc. that would be one occurrence. Of course we don’t expect always to find such decimally round spacings. The question again, How many do we expect to find?

The absolute and relative frequencies of the relevant letters in the text are:

B  4771 0.016744
E 36232 0.127157
F  7167 0.025153
L  9563 0.033562
O 22486 0.078915
that is, for each letter is shown the number of occurrences in the text and that number divided by the total of characters in the text.

See The Solution Submitted by Charlie    
Rating: 3.5000 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: to Cheradenine | Comment 9 of 12 |
(In reply to to Cheradenine by Cory Taylor)

Since you asked, it's 'probability' with an A.

With a mathematical expectation of 5.08 I rounded it to 5, but did of course use 5.08 as the value for the mean of the Poisson distribution. The expected value, being a technical term for average after many expected experiments, is of course a mathematical abstraction.
  Posted by Charlie on 2003-03-25 15:33:51

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