Diff for "FAQ/Combinatorics" - CBU statistics Wiki
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Revision 2 as of 2015-06-30 10:17:15
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= 1 – Product (i=0, n-1) [92^6 – i] / [92 ^6] = 1 – $$ Product(i=0, n-1) [92^6 ^ – i] / [92 ^6 ^] $$
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= 1 - ( 92^6 ! / [(92^6)^n (92^6 - n)!] ) = 1 - $$( 92^6 ^ ! / [(92^6 ^)^n ^ (92^6 ^ - n)!] )

Repetition probabilities

Suppose we have a sequence of length 6 (k) taken from a possible 92 (K) stimuli and wish to consider the probability of a repetition of any single stimulus in a randomly drawn sequence then we consider two probabilities below concerning repetition of a single stimulus and an entire sequence of stimuli.

The probability of any of the six of the 92 stimuli repeating in a randomly drawn sequence of length six

= 1 – [(92 x 91 x 90 x89 x88) / 92^6] = 0.99

= 1 – (number of sequences of length six which have no repetition of any of the 92 stimuli e.g. ABCDEF) / (total number of possible sequences of length 6 chosen from 92 stimuli)

so we are almost sure to get a single stimulus repeated in a randomly chosen sequence of length 6.

Another repetition which we may be interested in is the probability of an entire sequence of length 6 taken from 92 stimuli repeating in n independent draws:

This equals 1 – P(no repetition of any sequence in the n draws)

= 1 – $$ Product(i=0, n-1) [926 – i] / [92 6 ] $$

since there are 92^6 possibly distinct sequences of 92 stimuli of length 6 and once we have used one we don’t want to use it again.

which from this website

= 1 - $$( 926 ! / [(926 )n (926 - n)!] )

None: FAQ/Combinatorics (last edited 2015-09-14 13:54:53 by PeterWatson)