Talk:Index of coincidence

message: this article needs formulas
: It now has one; I improved it and added text about generalization of the concept. — DAGwyn 15:45, 20 September 2006 (UTC)
replies: text:  A while back I added a second formula, for kappa I.C. — DAGwyn 23:08, 25 November 2006 (UTC)

text: Since IC computation depends on context, it is better to give the general rule than just a specific computation. — DAGwyn 15:45, 20 September 2006 (UTC)

message: Are the first three formulae in this article correct to say '1/(1/c)' (or some form thereof)? Doesn't this just cancel to 'c' in the numerators, or is this perhaps just a style issue rather than syntactic?
94.193.216.167 (talk) 04:13, 24 January 2010 (UTC)
replies: text: I restored the original formulas.  The person who changed them to the form you saw was evidently trying to reorganize them in terms of probabilities rather than number of coincidences. — DAGwyn (talk) 20:57, 23 July 2010 (UTC)
thread_title:  Formulae Corrections 

message: The current version of this article says:
   "If all c letters of an alphabet were equally distributed, the expected index would be 1.0. The actual monographic IC for telegraphic English text is around 1.73, reflecting the unevenness of natural-language letter distributions."
Unfortunately, this statement is not true for the formulas given.  For example, the following text uses a uniform distribution of letters:
   "abcdefghijklmnopqrstuvwxyz
    abcdefghijklmnopqrstuvwxyz"
The existing formula yields an index of coincidence of 0.5098 for the above text.  But since the letters are uniformly distributed (each letter is used exactly twice), we should compute an index of coincidence of 1.0.
The formula approaches 1.0 as the length of the text increases: 2x alphabet -> 0.5098, 4x -> 0.7573, 16x -> 0.9398.
If the formula did not use the '-1' for both numerator and denominator, then we would properly compute an index of coincidence of 1.0 for equally distributed text.  So instead of this idea:
   "The chance of drawing that same letter again (without replacement) is (appearances - 1 / text length - 1)"
We'd compute the chance of drawing the same letter again (after replacing the original letter).
This article is fairly lengthy, and I'm not an expert in this subject.  So I hesitate to make such an edit.  I suspect the formula was intended as an introduction to the idea, and not intended to be the final formula (as the Generalization section suggests).  But I know the current article to be inconsistent.  For short texts, the expected index is not 1.0 for a uniform distribution of letters, given the current formula.  — Preceding unsigned comment added by Mbb25 (talk • contribs) 22:19, 6 July 2017 (UTC)
replies: text: There is nothing wrong with the formula. The expected value is the limit of the IC as the message length increases. The value for any specific message will in general not be the expected value. The IC can be wildly off the mark for short messages, as this is a statistical concept.--Bill Cherowitzo  (talk) 02:42, 7 July 2017 (UTC)
replies: text: I just fixed the actual error; it should have said equiprobable, not equidistributed.  Then the formula is correct.  Note that the paragraph after the derivation of the formula restates the meaning of the formula using "uniform random distribution" in the sense of "probability distribution".  The -1 is needed for the delta I.C. because we don't count a character as matching itself; it is not needed for kappa I.C.  — DAGwyn (talk) 12:47, 7 July 2017 (UTC)
thread_title:  Current formula doesn't give 1.0 for equally distributed texts.  Consider omiting the '-1'.