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I am trying to find a way to calculate all possible combinations of a sequence that have a certain length of long run.

When answering questions regarding sequences of heads and tails, sometimes participants will consider a sample space of longest run.
For a sequence of five coin flips, this is easy enough to calculate using a brute-force method - writing out all 32 possible sequences and then categorising them based on their longest run.
So, for example, a sequence with the longest run of 5 has a probability of 1/32 as there is only one way to have a longest run of 5 with a sequence of length 5.

However, I now have a sequence of length 10. I want to find out exactly how many sequences out of the 1024 possible sequences that have a longest run of 2 or 3.
I am assuming that it doesn't matter if the longest run is of heads or tails.

Is there code that could be used to calculate this?

So far I have:

x <- c(0, 1) # heads = 0, tails = 1

p = permutations(n = 2, r = 10, v = x, repeats.allowed = T)

p.df = as.data.frame(p)

To create a data frame of all possible combinations. Now I need to find the longest run of consecutive zeros or ones in each row and count how many rows have the same longest run.

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Aha! Figured it out!

Here is the code I have:

install.packages('gtools')
library(gtools)

x <- c(0, 1) # heads = 0, tails = 1

p = permutations(n = 2, r = 10, v = x, repeats.allowed = T)

p.df = as.data.frame(t(p))

q = apply(p.df, 2, function(x){
max(rle((x == 0))$lengths)
}
)

length(which(q == 10)) #where q == the length of the longest run
```
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