Simulation of Expected Wait Time

Related previous post: Expected Wait Time given Multiple Trains

In the previous post, a simulation with 100 million trials was run on MATLAB to verify that in a scenario of two trains running independently with frequency of 10 minutes, the expected waiting time until the earlier train is 10/3 minutes. This was the code in a .m script, along with added comments for explanation:

%We will run 100 million trials, results to be stored in an array
results = zeros(1,100000000);

for i=1:100000000
    %Initiate the variable 'lowest' at 20 for now
    %It will eventually record the time to the earlier train
    %sure be less than 20
    lowest = 20;
   
    %Let both trains start at a random time t=(0,10)
    %This is the first round of the trains
    train1Start = 10*rand;
    train2Start = 10*rand;
   
    %There will be trains 10 minutes ahead, representing the second round
    another1Train = train1Start+10;
    another2Train = train2Start+10;

    %Let the passenger board the platform at a random time t=(10,20)
    passenger = 10*rand+10;
   
    %The next train may be train 1 from first round
    lowest = passenger-train1Start;
    %Or it may be train 2 from the first round
    lowest = min(lowest, passenger-train2Start);
   
    %Or it may be train 1 from the second round
    %Positive value verifies that passengers arrives become the train
    if passenger-another1Train>=0
        lowest = min(lowest, passenger-another1Train);
    end
    %Finally, the earlier train may be train 2 from the second round
    if passenger-another2Train>=0
        lowest = min(lowest, passenger-another2Train);
    end
   
    %Record the time to the earliest train into the results array
    results(i) = lowest;
end

%At last, we want to return the results
disp('Mean:')
disp(mean(results))
disp('Standard deviation:')
disp(std(results))

Upon running the script, the result comes out as a mean of 3.3334 (occasionally may get 3.3333) along with a standard deviation of 2.3572.