FAQ/Combinatorics - CBU statistics Wiki

Upload page content

You can upload content for the page named below. If you change the page name, you can also upload content for another page. If the page name is empty, we derive the page name from the file name.

File to load page content from
Page name
Comment
WHat wOrd is made by the captiaL lettErs?

location: FAQ / Combinatorics

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 stimulus repetition in a randomly drawn sequence of length n. We consider two probabilities, below, concerning repetition of (i) a single stimulus and (ii) 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 x 89 x 88 x 87) / $$926 $$ ] = 0.15

= 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).

We can similarly work out probabilities for 1,2,3,4 and 5 repetitions (from Laurence Shaw and Daniel Molinari) taking into account that there can be multiple instances of a repetition in a sequence e.g. the sequence AABBCD represents two instances of a single repetition (of A and B).

Another repetition which we may be interested in is the probability of at least one 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 $$926 $$ distinct sequences of 92 stimuli of length 6 and once we have used one sequence we don’t want to use it again.

Using the result from this website the above probability

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