% do the following several times:
rand(1)
% see that values returned are between 0-1

% do the following several times:
rand(1)*100
% see that values are between 0-100

% Ask about the average (50)
% How do you get a random number between 50-150 with average 100?
rand(1)*100+50

% show the following:
mean = 50;
x = zeros(1,10000);
for i=1:10000 x(i) = ceil(rand(1)*2*mean); end
% ask on the average how many random numbers are 10, 11,..?
% 10000 total numbers and 100 integer values they can take so answer is 100
% on the average

% ask do you expect that the average is the same for all numbers?  The number
% of times the random value 10 is picked is about the same as the number of
% times the random value 12 is picked?

% design an experiment to find out. (exp_1)
mean = 50;
npts = 10000;
x = zeros(1,npts);
for i=1:npts x(i) = ceil(rand(1)*2*mean); end
v = zeros(1,2*mean);
for i=1:npts v(x(i)) = v(x(i)) +1; end
plot(v);
axis([0 2*mean 0 1.2*max(v)]);

% A question about means: If I have 100 variables of mean 1, what is the mean
% of the sum of the variables?
% Try an experiment: (exp_2)
mean = 1;
npts = 100;
nexp = 100;
v = zeros(1,npts);
for i=1:nexp  x = rand(1,npts)*2*mean;  v(i) = sum(x); end
plot(v);
axis([0 nexp 0 1.2*npts*mean]);
% Answer: it is the sum of the means of the variables.

% Well maybe this does not work if the variables have different means
% try an experiment where means are increasing (exp_3)
mean = 1;
npts = 1000;
nexp = 1000;
v = zeros(1,nexp);
for i=1:nexp 
   x = zeros(1,npts);
   for j=1:npts  x(j) = rand(1)*2*j*mean;  end
   v(i) = sum(x); 
end
plot(v);
axis([0 nexp 0 0.6*npts*npts*mean]);
sum(v)/nexp  % the average

% That variance is troubling - let's define it:
% var = average of (X-mean)^2
% var of uniform distribution: (exp_4)
mean = 50;
npts = 100000;
x = zeros(1,npts);
for i=1:npts x(i) = (rand(1)*2*mean - mean)^2; end
printf('var= %f std. dev.= %f\n',sum(x)/npts, sqrt(sum(x)/npts));

% variance of the other distribution (exp_5)
mean = .001;
npts = 1000;
nexp = 1000;
x = zeros(1,nexp);
y = zeros(1,nexp);
for k=1:nexp
   for i=1:npts x(k) = x(k) + 2*i*mean; end
   y(k) = (x(k) - npts*mean)^2;
end
printf('mean= %f var= %f std. dev.= %f\n',sum(x)/nexp, sum(y)/nexp, sqrt(sum(y)/nexp));

% Normal distribution
randn(1)

% Find the mean  (exp_6)
nexp = 1000;
x = zeros(1,nexp);
for i=1:nexp x(i) = randn(1); end
printf('mean= %f variance= %f std.dev.= %f\n', sum(x)/nexp, sum((x-sum(x)/nexp)^2), sqrt(sum((x-sum(x)/nexp)^2)));

% Experiment with the standard deviation - use exp_6
s*randn(1)

% Plot (exp_7)

% Show that sum of random vars is gaussian with mean = sum of means and 
% std dev = sum of std dev. (exp_8)

%% Cells
a{1} = [1 2 3]
a{2} = [4 5]
a{3} = [6 7 8 9 10]
a
l = a{1}

%% Make rounds
% Returns a cell array where each cell is the outcome of dice rolls for a round
function rnds = rounds(n);
   point = 0;
   for i=1:n  % n rounds
      % come out roll
      roll = ceil(rand(1)*6) + ceil(rand(1)*6);
      tmp = [roll];
      if roll == 4 | roll == 5 | roll == 6 | roll == 8 | roll == 9 | roll == 10
         point = roll;
         roll = ceil(rand(1)*6) + ceil(rand(1)*6);
         tmp = [tmp roll];
         while roll ~= point & roll ~= 7
            roll = ceil(rand(1)*6) + ceil(rand(1)*6);
            tmp = [tmp roll];
         end
      end
      rnds{i} = tmp;
   end
end