FAQ/combinatorics/Rreps - CBU statistics Wiki
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# R code computing the replication probabilities for the example of 92 stimuli randomly drawn with replacement into sequences of length 6

```#
# From Daniel Molinari July 2015
#

# S.x.y = y+1 instances of x replicates
# 0 replicates   abcdef

S.0 <- choose(92,6)*factorial(6)

# 1-0 replicates aabcde

S.1.0 <- choose(6,2)*92*91*90*89*88 +
choose(6,2)*choose(4,2)*92*91*90*89 +
choose(6,2)*choose(4,2)*choose(2,2)*92*91*90

# 1.1 PW addition using Laurence S formula aabbcd

S.1.1 <- choose(92,2)*choose(90, 2)*15*6*2

# 1-2 replicates aabbbc

S.1.2 <- choose(6,2)*choose(4,3)*92*91*90

# 1-3 replicates aabbbb

S.1.3 <- choose(6,2)*choose(4,4)*92*91

# 2-0 replicates aaabcd

S.2.0 <- choose(6,3)*92*91*90

# PW ADDING IN LAURENCE S FORMULA 2-1 replicates aaabbb

S.2.1 <- factorial(2)*choose(92,2)*choose(6,3)/92^6

# 2-2 replicates aaabbb

S.2.2 <- choose(6,3)*choose(3,3)*92*91

# 3-0 replicates aaaabc

S.3.0 <- choose(6,4)*92*91*90

# PW addition
# Yes, the order counts.  The correct computation is
# Sequences( 3 instances 1 replicate) =  3 ! (92 Choose 3)*(6 choose 2)*(4 choose 2)

S.3.1 <-  factorial(3)*choose(92,3)*choose(6,2)*choose(4,2)

# 4-0 replicates aaaaab

S.4.0 <- choose(6,5)*92*91

# 5-0 replicates aaaaaa

S.5.0 <- choose(6,6)*92

# Compute cumulative probability

S <- S.0 +
S.1.0 + S.1.2 + S.1.3 +
S.2.0 + S.2.2 +
S.3.0 +
S.4.0 +
S.5.0

P  <- S / (92^6)

S <- S.0 +
S.1.0 + S.1.1 + S.1.2 + S.1.3 +
S.2.0 + S.2.1 + S.2.2 +
S.3.0 + S.3.1
S.4.0 +
S.5.0

P1 <- S / (92^6)```

yields

```  P    # 1.002883
P1   # 1.007972  ```

None: FAQ/combinatorics/Rreps (last edited 2015-07-02 10:57:14 by PeterWatson)