=  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) / $$92^6 ^$$ ] = 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).

[[FAQ/combinatorics/mreps | 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 – $$ Product(i=0, n-1) [92^6 ^ – i] / [92 ^6 ^] $$

since there are $$92^6 ^$$ 
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 [[http://www.johndcook.com/blog/2012/04/16/random-number-sequence-overlap/ | this website]] the above probability

= 1 -  $$( [92^6 ^ !] / [(92^6 ^)^n ^ (92^6 ^ - n)!] )