# A few worked examples evaluating some repetition probabilities

For drawing from 92 stimuli with replacement into a sequence of length 6.

No repetitions

P(0 repetitions) = (Number of permutations of six distinct stimuli chosen from 92 with replacement) / (total number of permutations of 92 stimuli with replacement) = (92x91x90x89x88x87) / ($$92^{6 }$$) = 0.847.

1 instance of 1 replicate

There are 15 possible pairs of positions of any of the 92 repeated stimuli. These are

1 2 1 3 1 4 1 5 1 6 2 3 2 4 2 5 2 6 3 4 3 5 3 6 4 5 4 6 5 6

The remaining four positions are assumed to be distinct and different to the stimulus that has been repeated. This gives a probability of (92 x (5+4+3+2+1)x91x90x89x88) / $$92^{6 }$$= 0.146.

5 repetitions

For 5 repetitions (the same stimulus used in all six positions of the sequence) there are 92 possibilities giving a probability of 92 / $$92^{6 }$$ = 1.5172629 x 10(-10).