FAQ/combinatorics/repeval52015-07-01 13:55:53PeterWatson42015-07-01 13:55:07PeterWatson32015-07-01 13:54:14PeterWatson22015-07-01 13:52:45PeterWatson12015-07-01 13:52:29PeterWatsonA few worked examples evaluating some repetition probabilitiesFor 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) / ($$926 $$) = 0.847. 1 instance of 1 replicate There are 15 possible pairs of positions of any of the 92 repeated stimuli. These are 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) / $$926 $$= 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 / $$926 $$ = 1.5172629 x 10(-10).