Creating a random matrix with different probabilities

Hey, I have 4 situations, and each situation has a different probability to be happened. P(0) = 0.2 P(1) = 0.46 P(2) = 0.16 P(3) = 0.18

I want to create a matrix(52,1) with [0 3] but with inequal probabilities. A= randi([0 3],52,1); With this, the probability to be have number 0,1,2,3 is 25%. Can u help me out?

Thanks in advance.

[Stephen23]

Stam Kavid's "Answer" moved here:

A= randi([0 3],52,1);
limit=4;
W=zeros(size(A));
C=zeros(size(A));
k=1;
for i=1:length(A)
C(k)=C(k)+1;
if W(k)>=limit
k=k+1
end
W(k)=W(k)+A(i)
end
W(k+1:end)=[ ];
C(k+1:end)=[ ];

I want to know how many times the A exceed limit(4), but every time that exceed the limit for example A=5 i want to start with 5-4=1 over again, if A=6 6-4=2 etc. How can i do this one? Thanks in advance

Answer 1

[Stephen23]

P = rand(52,1);
P = (P>0.2) + (P>0.66) + (P>0.82)

Note that the values used here are just cumsum([0.2,0.46,0.16]), i.e. [0.2, 0.2+0.46, 0.2+0.46+0.16]. If you wanted to automatically adjust for different input values then you could use cumsum and something like hist or discretize... but the idea is the same.

I tested this code on 1e6 iterations:

N = 1e6;
V = nan(52,N);
for k = 1:N
  P = rand(52,1);
  P = (P>0.2) + (P>0.66) + (P>0.82);
  V(:,k) = P;
end

and got a distribution that matches what you requested:

>> cnt = histc(V(:),0:3)/(N*52)
cnt =
   0.20002
   0.46002
   0.15995
   0.18000

[Stam Kavid]

Thank you very much. I appreciate your effort a lot!

Answer 2

[Steven Lord]

% Generate the vector of probabilities for each of your classes
p = [0.2 0.46 0.16 0.18];

% Cumulative probability vector
probabilityBins = cumsum([0 p]);

% Sample data
x = rand(1, 1e6);

% Bin each element of the sample data into the appropriate bin
% whose edges are in probabilityBins
D = discretize(x, probabilityBins, 0:3);

% Show the first 10 sample data points and their bins
[x(1:10); D(1:10)]

% Show that the probabilities are roughly what you'd expect
histogram(D, 'Normalization', 'probability')

% Turn on the grid to see how each bar matches its probability
yticks(sort(p))
grid on

I'd say that looks pretty good.

[Stam Kavid]

Thank you very much. You helped me a lot!