Selecting a random number with some probability

Hello Everyone, I am using this one line of code to generate a single value either to be 1 or 2 with equal probability but my question is that how can i select the value to be 1 with 60% probability and the value to be 2 with 40% probability ? Thank you.

select=randi(2,1,1);

Answer 1

[Torsten]

x=rand;
if x<0.6
  select=1;
else
  select=2;
end

Best wishes

Torsten.

[Aftab Ahmed Khan]

Thank you all for the response. Take care.

Answer 2

[pankaj singh]

The simplest technique is to use inbuilt Matlab function 'randscr'.

Suppose you want to generate M by N matrix of W, X, Y, and Z with probabilities i,j,k, and l. Then use

out = randsrc(M,N,[W,X,Y,Z;i,j,k,l]);   % i+j+k+l = 1;

In your case, as you want a single value to be generated, your M x N = 1 x 1 matrix; the values are 1 with 60% probability (i.e. 0.6) and 2 with 40% (i.e. 0.4) probability, therefore use this;

out = randsrc(1,1,[1,2;0.6,0.4]);

Note that the above is just an example. You can create any matrix size with any number of values. The sum of probabilities must be equal to 1.

[Carles Alfageme]

Hi,

I am trying to use this function: randsrc(M,N,[W,X,Y,Z;i,j,k,l]); in a Matlab Function block in Simulink.

The problem is that when i simulate my simulink model, this error message appears:

"Function 'randsrc' not supported for code generation.

Function 'MATLAB Function3' (#87.24.74), line 3, column 5: "randsrc(1,1,[0.6,0.7,0.8,0.9;0.25,0.25,0.25,0.25])" Launch diagnostic report.

Component:MATLAB Function | Category:Coder error"

I think it is because this function comes from Communication Toolbox, but I am not sure.

Do you have any idea or suggestion?

[Ana Gabriela Arteaga]

same problem.

[Jonathan Ford]

I'm not sure what is causing your error, but you could try writing your own randsrc function, something like this:

function X = myrandsrc(M, N, A)
    X = reshape(A(1,sum(A(2,:) < rand(M*N,1)*ones(1,size(A,2)),2)+1),M,N);
end

Then:

X = myrandsrc(4,5,[1 2 3 4; 0.4 0.7 0.9 1])

will return something like:

X =

     1     2     3     2     2
     2     1     4     3     2
     3     2     2     3     2
     2     2     2     3     2

Note that for this implementation, you need to use the cumulative probability distribution in the second row of A, so with the above call you will get ~40% of 1, 30% of 2, 20% of 3 and 10% of 4.

Below is the function separated out into multiple lines, to better explain how it works:

function X = myrandsrc(M, N, A)
    % the number of elements to chose from
    sz = size(A,2);
    
    % generate some random numbers
    r = rand(M*N,1)*ones(1,sz);
    
    % determine the correct elements
    r = sum(A(2,:) < r,2)+1;
    
    % select the correct elements
    X = A(1,r);
    
    % Reshape into M x N matrix
    X = reshape(X,M,N);
end

Answer 3

[Daniel]

You can use rand, which gives uniform distribution and look if the number is below or above 0.6.

if(0.6 <= rand()){
   select = 1;
} else {
   select = 2;
}

That should give 60/40 chances. There are more elegant ways to do that though.

[Trung Khoa Le]

Could you please give me a bit explanation why this way makes sense or some documentation that I can read to gain some intuition? Thanks

[Luciano Anastassiou]

Hi Trung Khoa Le,

The intuition is simply that "rand" generates a random number between 0 and 1. Then when you apply the "if" constraint, you are telling the system to only give out the result "select = 1" when that random number is below 0.6. Otherwise it will give out "select = 2".

If you repeated this 1000s of times, it would give out "select = 1" 60% of the time, because 60% of those random numbers between 0 and 1 will be below 0.6, and the other 40% of the time it will give out "select = 2".

Answer 4

[Jos (10584)]

For two values it is simple

VAL = [10 20]   % 2 values
P = .8          % probabbility of selecting first value
Ndraw = 20      % number of draws
R = rand(Ndraw,1) < P 
SEL = VAL(2 - R) % use as index into VAL

For more complicated cases you might be interested inTake a look at my RANDP function, which picks random values with relative probabilities.

http://www.mathworks.com/matlabcentral/fileexchange/8891

[amd jafarzadeh]

Dear Jos.

Thanks alot. It works properly.

Answer 5

[Steven Lord]

Another approach is to use the discretize function to discretize a uniform random number between 0 and 1 as generated by the rand function. Because edges is [0, 0.6, 1] any values in uniform that are in the range [0, 0.6) will be mapped to 1 in oneOrTwo and any values in uniform in the range [0.6, 1] will be mapped to 2.

probabilities = [0.6 0.4];
edges = cumsum([0 probabilities])

edges = 1×3

0 0.6000 1.0000

uniform = rand(1, 1e4);
oneOrTwo = discretize(uniform, edges);

We can check using histogram that the generated numbers have the right distribution (or close to it.)

histogram(oneOrTwo, 'Normalization', 'probability')
% Draw lines at the desired probabilities
yline(probabilities, 'r:')

Those look to be in pretty good agreement with the desired probabilities to me.

Answer 6

[Ka Mirul]

I have create a video about generating random number in MATLAB

https://www.youtube.com/watch?v=MyH3-ONYL_k

Hope that will help you