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← Revision 21 as of 2015-09-14 13:54:53 ⇥
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= Repetition probabilities = |
= Repetition probabilities = |
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| 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. | 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. |
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| = 1 – [(92 x 91 x 90 x89 x88) / 92^6] = 0.99 | = 1 – [(92 x 91 x 90 x 89 x 88 x 87) / $$92^6 ^$$ ] = 0.15 |
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| = 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) | = 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). |
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| so we are almost sure to get a single stimulus repeated in a randomly chosen sequence of length 6. | [[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). |
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| 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: | 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: |
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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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| 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. | 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. |
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| which from this website | Using the result from [[http://www.johndcook.com/blog/2012/04/16/random-number-sequence-overlap/ | this website]] the above probability |
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| = 1 - ( 92^6 ! / [(92^6)^n (92^6 - n)!] ) |
= 1 - $$( [92^6 ^ !] / [(92^6 ^)^n ^ (92^6 ^ - n)!] ) |
Repetition probabilities
Suppose we have a sequence of length 6 (k) taken from a possible 92 (K) stimuli and wish to consider the probability of stimulus repetition in a randomly drawn sequence of length n. We consider two probabilities, below, concerning repetition of (i) a single stimulus and (ii) an entire sequence of stimuli.
The probability of any of the six of the 92 stimuli repeating in a randomly drawn sequence of length six
= 1 – [(92 x 91 x 90 x 89 x 88 x 87) / $$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)!] )