A method of choosing a line randomly from a file:

 Without reading the file more than once
 When substantial parts of the file cannot be held in memory
 Without knowing how many lines are in the file
Is to:

 keep the first line of the file as a possible choice, then
 Read the second line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/2.
 Read the third line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/3.
 ...
 Read the Nth line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/N
 Return the computed possible choice when no further lines exist in the file.


Task

 Create a function/method/routine called one_of_n that given n, the number of actual lines in a file, follows the algorithm above to return an integer - the line number of the line chosen from the file. 
The number returned can vary, randomly, in each run.
 Use one_of_n in a simulation to find what woud be the chosen line of a 10 line file simulated 1,000,000 times.
 Print and show how many times each of the 10 lines is chosen as a rough measure of how well the algorithm works.


Note: You may choose a smaller number of repetitions if necessary, but mention this up-front.

Note: This is a specific version of a Reservoir Sampling algorithm: https://en.wikipedia.org/wiki/Reservoir_sampling

fcn one_of_n(lines){ # lines is any iterable
#if 0  // iterative
   choice:=Void;
   foreach i,line in ([0..].zip(lines)){
      if((0).random(i+1)==0) choice=line;
   }
   return(choice);
#else  // functional
   [0..].zip(lines).pump(Ref(Void).set,fcn([(n,line)])
      { if((0).random(n+1)==0) line else Void.Skip }).value
#endif
}
 
fcn one_of_n_test(n=10, trials=0d1_000_000){
   bins:=n.pump(List(),0); // List(0,0,0...)
   if(n){ foreach i in (trials){ bins[one_of_n((n).walker())]+=1 } }
   return(bins);
}
 
println(one_of_n_test());