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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(3): incorrect solution? | Comment 11 of 12 |
(In reply to re(2): incorrect solution? by Cheradenine)

There are two places where the posted solution can be said to be wrong in terms of the lack of independence of the trials.

1. The probability of a given set of 7 letters being FLOOBLE was given as, in effect:
(7167/284939)(9563/284939)(22486/284939)^2(4771/284939)(9563/284939)(36232/284939)
As each letter is drawn from the same stock, the figures should be:
(7167/284939)(9563/284938)(22486/284937)(22485/284936)(4771/284935)(9562/284934)(36232/284933), with the decreasing denominators resulting from one less letter available in the pool, and decreased numerator for the two instances of a repeated letter, as one had been presumed used up previously.

The result of this is 3.7563 x 10^-10, rather than the 3.7567 x 10^-10 given in the solution.

The other will be posted on the next comment...


  Posted by Charlie on 2003-03-26 13:59:28

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