FAQ/combinatorics/repeval - CBU statistics Wiki

Upload page content

You can upload content for the page named below. If you change the page name, you can also upload content for another page. If the page name is empty, we derive the page name from the file name.

File to load page content from
Page name
Comment
Finzd thee wrang lelters ino eacuh wosrd

location: FAQ / combinatorics / repeval

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) / ($$926 $$) = 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) / $$926 $$= 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).