Selecting a random number with some probability

Question

Aftab Ahmed Khan 2015-2-25

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回答： Steven Lord 2022-6-16

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);

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Answer 1

Torsten 2015-2-25

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

Best wishes

Torsten.

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Aftab Ahmed Khan 2015-2-25

Thank you all for the response. Take care.

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Answer 2

pankaj singh 2018-3-10

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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.

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Ana Gabriela Arteaga 2020-6-11

same problem.

Jonathan Ford 2021-7-11

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

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Answer 3

Daniel 2015-2-25

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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.

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Trung Khoa Le 2019-1-17

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 2019-5-22

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".

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Answer 4

Jos (10584) 2015-2-25

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编辑：Jos (10584) 2015-2-25

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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

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amd jafarzadeh 2021-1-10

Dear Jos.

Thanks alot. It works properly.

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Answer 5

Steven Lord 2022-6-16

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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);