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