SAS and R

%let n = 5;
data h;
array val [&n] v1 - v&n;
 do r = 1 to &n;
 do c = 1 to &n;
 val[c] = 1/(r+c-1);
 end;
 output;
 end;
run;

proc print data = h noobs; 
var v1 - v5;
format _all_ fract.; 
run;

 v1 v2 v3 v4 v5
 1 1/2 1/3 1/4 1/5
1/2 1/3 1/4 1/5 1/6
1/3 1/4 1/5 1/6 1/7
1/4 1/5 1/6 1/7 1/8
1/5 1/6 1/7 1/8 1/9

n = 5
1/sapply(1:n,function(x) x:(x+n-1))

n=5
h1 = matrix(nrow=n,ncol=n)
for (r in 1:n) {
 for (c in 1:n)
 h1[r,c] = 1/(r+c-1)
} 

> library(MASS)
> fractions(h1)
 [,1] [,2] [,3] [,4] [,5]
[1,] 1 1/2 1/3 1/4 1/5 
[2,] 1/2 1/3 1/4 1/5 1/6 
[3,] 1/3 1/4 1/5 1/6 1/7 
[4,] 1/4 1/5 1/6 1/7 1/8 
[5,] 1/5 1/6 1/7 1/8 1/9 

/* the following macro will write out equal probabilities for selecting 
each integer between 1 and the argument, in the format needed for the 
rand function. E.g., if the argument is 3, 
it will write out
1/3,1/3,1/3
*/

%macro tbls(n);
%do i = 1 %to &n;
1/&n %if &i < &n %then ,
%end;
%mend tbls;

/* then we can use the %tbls macro to create the randomization
via rand("TABLE") (section 1.10.4). */ 
data kids;
do family = 1 to 10000;
 nkids = 2; chosen = rand("TABLE",%tbls(2)); output;
 nkids = 3; chosen = rand("TABLE",%tbls(3)); output;
 nkids = 4; chosen = rand("TABLE",%tbls(4)); output;
 nkids = 5; chosen = rand("TABLE",%tbls(5)); output;
end;
run;

/* check randomization */
proc freq data = kids;
table nkids * chosen / nocol nopercent;
run; 

 nkids chosen

 Frequency|
 Row Pct | 1| 2| 3| 4| 5| Total
 ---------+--------+--------+--------+--------+--------+
 2 | 50256 | 49744 | 0 | 0 | 0 | 100000
 | 50.26 | 49.74 | 0.00 | 0.00 | 0.00 |
 ---------+--------+--------+--------+--------+--------+
 3 | 33429 | 33292 | 33279 | 0 | 0 | 100000
 | 33.43 | 33.29 | 33.28 | 0.00 | 0.00 |
 ---------+--------+--------+--------+--------+--------+
 4 | 25039 | 24839 | 25245 | 24877 | 0 | 100000
 | 25.04 | 24.84 | 25.25 | 24.88 | 0.00 |
 ---------+--------+--------+--------+--------+--------+
 5 | 19930 | 20074 | 20188 | 20036 | 19772 | 100000
 | 19.93 | 20.07 | 20.19 | 20.04 | 19.77 |
 ---------+--------+--------+--------+--------+--------+
 Total 128654 127949 78712 44913 19772 400000

proc format;
value chosen
1 = "oldest"
2 = '2nd oldest'
3 = '3rd oldest'
4 = '4th oldest'
5 = '5th oldest';
run;
 
/* now, make a text variable the concatenates (section 1.4.5) the variables 
and some explanatory text */
data k2;
set kids;
if nkids eq 2 then
 t1 = "If there are " || strip(nkids) ||" children then choose the " ||
 strip(put(chosen,chosen.)) || " child.";
else
 t1 = " " || strip(nkids) ||" ________________________ " ||
 strip(put(chosen,chosen.));
run;

/* then we print. Notice the options to print in plain text, shorten the 
page length and width, and remove the date and page number from the SAS output, as
well as in the proc print statement to remove the observation number and
show the line number, with a few other tricks */
options nonumber nodate ps = 60 ls = 68;
OPTIONS FORMCHAR="|----|+|---+=|-/\*";
proc print data = k2 (obs = 3) noobs label sumlabel;
by family;
var t1;
label t1 = '00'x family = "Envelope";
run;

---------------------------- Envelope=1 ----------------------------



 If there are 2 children then choose the 2nd oldest child.
 3 ________________________ 3rd oldest
 4 ________________________ 4th oldest
 5 ________________________ 5th oldest


---------------------------- Envelope=2 ----------------------------



 If there are 2 children then choose the 2nd oldest child.
 3 ________________________ oldest
 4 ________________________ oldest
 5 ________________________ 3rd oldest


---------------------------- Envelope=3 ----------------------------



 If there are 2 children then choose the 2nd oldest child.
 3 ________________________ 2nd oldest
 4 ________________________ 3rd oldest
 5 ________________________ 2nd oldest

sample(5, size = 1) 
# choose a (discrete uniform) random integer between 1 and 5

apply(matrix(2:5),1,sample,size=1) 
# choose a random integer between 1 and 2, then between 1 and 3, etc., 
# using apply() to repeat the call to sample() with different maximum number
# apply() needs a matrix or array input
# result of this is the raw data needed for one family

replicate(3,apply(matrix(2:5),1,sample,size=1))
# replicate() is in the apply() family and just repeats the 
# function n times

 [,1] [,2] [,3]
[1,] 2 1 2
[2,] 2 1 2
[3,] 2 2 2
[4,] 3 5 4

test=replicate(100000, apply(matrix(2:5), 1, sample, size=1))
apply(test, 1, summary)
 [,1] [,2] [,3] [,4]
Min. 1.0 1 1.000 1.000
1st Qu. 1.0 1 1.000 2.000
Median 1.0 2 2.000 3.000
Mean 1.5 2 2.492 3.003
3rd Qu. 2.0 3 3.000 4.000
Max. 2.0 3 4.000 5.000
# this is not so helpful-- need the count or percent for each number
# this would be the default if the data were factors, but they aren't
# check to see if we can trick summary() into treating these integers
# as if they were factors
methods(summary)
# yes, there's a summary() method for factors-- let's apply it
# there's also apply(test,1,table) which might be better, if you remember it
apply(test, 1, summary.factor)
[[1]]
 1 2 
50025 49975 

[[2]]
 1 2 3 
33329 33366 33305 

[[3]]
 1 2 3 4 
25231 25134 24849 24786 

[[4]]
 1 2 3 4 5 
19836 20068 20065 20022 20009 
# apply(test,1,table) will give similar results, if you remember it

mylist=replicate(5, apply(matrix(2:5), 1, sample, size=1))
# brief example data set

# We'll need to use some formatted values (section 1.14.12), as in SAS. 
# Here, we'll make new value labels for each number of children,
# which will make the output easier to read. We add in an envelope 
# number and wrap it all into a data frame.
df = data.frame(envelope = 1:5,
 twokids=factor(mylist[1,],1:2,labels=c("youngest","oldest")),
 threekids=factor(mylist[2,],1:3,labels=c("youngest", "middle", "oldest")),
 fourkids=factor(mylist[3,],1:4,labels=c("youngest", "second youngest", 
 "second oldest", "oldest")),
 fivekids=factor(mylist[4,],1:5,labels=c("youngest", "second youngest", 
 "middle", "second oldest", "oldest"))
)

# now we need a function to take a row of the data frame and make a single slip
# the paste() function (section 1.4.5) puts together the fixed and variable 
# content of each row, while the cat() function will print it without quotes
slip = function(kidvec) {
 cat(paste("------------- Envelope", kidvec[1], "------------------"))
 cat(paste("\nIf there are", 2:5, " children, select the", kidvec[2:5],"child"))
 cat("\n \n \n")
}

# test it on one row
slip(df[1,])

# looks good-- now we can apply() it to each row of the data frame
apply(df, 1, slip)

------------- Envelope 1 ------------------
If there are 2 children, select the youngest child 
If there are 3 children, select the youngest child 
If there are 4 children, select the second youngest child 
If there are 5 children, select the youngest child
 
 
------------- Envelope 2 ------------------
If there are 2 children, select the youngest child 
If there are 3 children, select the youngest child 
If there are 4 children, select the second oldest child 
If there are 5 children, select the middle child


------------- Envelope 3 ------------------
If there are 2 children, select the youngest child 
If there are 3 children, select the youngest child 
If there are 4 children, select the youngest child 
If there are 5 children, select the second youngest child

# and so forth

# finally, we can save the result in a file with
# capture.output()
capture.output(apply(df,1,slip), file="testslip.txt")