How to calculate how often a specific combination occurs in random design?

#1
Hi everybody,

I hope someone can help me out! We designed an experiment in which 24 short clips are shown to particpants. These clips belong to four different categories (so we have 6 clips per categorie). Let's call the categories A, B, C and D. Each participants gets to see all 24 clips in random order without replacement.

Since I am particularly interested in the effects if a C clip is shown after a B clip I would like to know (1) for how many participants we can expect that at least two times a C clip is shown directly after a B clip; and (2) what the expected average times of a C clip being shown after a B clip per person is.

We hope for 3000 participants in total.

I would like to do these kinds of calculations myself so I hope someone can explain to me what the steps are. Thanks!
 

hlsmith

Omega Contributor
#2
So you have not conducted the experiment yet? You randomized the order of clips for each person or just once? If you randomized the clips with a randomization algorithm you should be fine? So I am not understanding your question.
 
#3
Thanks for your reply. We are conducting the experiment right now. We randomized the order of the clips for each person separately and indeed randomized the clips with a randomization algoritm. When the experiment was designed we did not know that the effects of clip C shown after clip B would be most interesting to us and thereofre we did not put restrictions to the randomization method to guarantee sufficient cases of C being shown after B. Right know I would like to make an estimation whether if we go on like this we will end up with a sufficient number or not. That's the background of my questions (1) for how many participants we can expect that at least two times a C clip is shown directly after a B clip; and (2) what the expected average times of a C clip being shown after a B clip per person is.
 

Dason

Ambassador to the humans
#6
On average you can expect about 1.5 observations of "C directly follows B" per participant. There is approximately a 48% chance that a participant will have 2 or more cases where C directly follows B. I was too lazy to do the math so I just simulated it.
 
#7
On average you can expect about 1.5 observations of "C directly follows B" per participant. There is approximately a 48% chance that a participant will have 2 or more cases where C directly follows B. I was too lazy to do the math so I just simulated it.
Thanks! Did you by any chance simulate it in R? In that case, could you send me the code?

I am still interested in knowing how to calculate this by the way, so if anyone is willing to do so I would greatly appreciate it!
 

Dason

Ambassador to the humans
#8
I did use R. I used a throwaway script and it wasn't too efficient but it got the job done. Basically I did something like...

Code:
vals <- rep(LETTERS[1:4], 6)

cfollowsb <- function(x){
  # this should return how many times C follows B in the input vector
  # This is off the top of my head but it was something like...
  sum(outer(which(x == "B"), which(x == "C"), "-") == -1)
}

sim <- replicate(1000, cfollowsb(sample(vals)))
mean(sim)
mean(sim >= 2)