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# 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).

None: FAQ/combinatorics/repeval (last edited 2015-07-01 13:55:53 by PeterWatson)