Sorry, The last three lines should read:
all <- apply(results, 1, function(x) length(intersect(x,
seq(sides)))==sides)
sum(all)/reps
results <- as.data.frame(results)
To generalize them for values of sides other than 6.
On Fri, Apr 22, 2022 at 11:05 PM Paul Bernal wrote:
> Thank you so much
Thank you so much David!
El El vie, 22 de abr. de 2022 a la(s) 11:04 p. m., David Carlson <
dcarl...@tamu.edu> escribió:
> Since the rolls are independent, it is not necessary to separate the rolls
> into two stages:
>
> sides <- 6
> months <- 12
> reps <- 100
>
> set.seed(2022)
> results <-
Dear Bert,
Thank you for your kind reply.
That is fine, I appreciate your feedback anyway.
Have a great day/night.
Best,
Paul
El mié, 20 abr 2022 a las 23:31, Bert Gunter ()
escribió:
> I believe I gave you sufficient information (the vector of dice roll
> results would replace 1:36 in my
I believe I gave you sufficient information (the vector of dice roll
results would replace 1:36 in my example). Furthermore, this sounds
like homework, which we try not to do here. But even if it is not, I
expect you to fill in the details based on what I have provided. If I
have misunderstood
If I understand you correctly, it's simple.
Matrices in R are vectors with a dimension attribute. By default, they
are populated column by column. Use 'byrow = TRUE to populate by row
instead. For example:
> matrix (1:36, ncol = 12, byrow = TRUE)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
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