FAQ/Combinatorics - CBU statistics Wiki

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# 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)!] )