= 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).