# Selecting a random number with some probability

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Aftab Ahmed Khan il 25 Feb 2015
Risposto: Steven Lord il 16 Giu 2022
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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### Risposta accettata

Torsten il 25 Feb 2015
x=rand;
if x<0.6
select=1;
else
select=2;
end
Best wishes
Torsten.
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Aftab Ahmed Khan il 25 Feb 2015
Thank you all for the response. Take care.

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### Più risposte (5)

pankaj singh il 10 Mar 2018
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 il 11 Giu 2020
same problem.
Jonathan Ford il 11 Lug 2021
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

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Daniel il 25 Feb 2015
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 il 17 Gen 2019
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 il 22 Mag 2019
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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Jos (10584) il 25 Feb 2015
Modificato: Jos (10584) il 25 Feb 2015
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.
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amd jafarzadeh il 10 Gen 2021
Dear Jos.
Thanks alot. It works properly.

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Steven Lord il 16 Giu 2022
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.
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Ka Mirul il 14 Nov 2017
Modificato: KSSV il 16 Giu 2022
I have create a video about generating random number in MATLAB
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