Generating a weighted random number

I currently have a variable that looks like this

var = 1:4

which every run through the for loop it picks a random number 1 - 4.

I need it to pick 1 half of the time and 2:4 the other half. so for the 1 x 4 matrix [1 2 3 4] i need the probabilities to be [ 0.5 0.16 0.16 0.17]

How would I write this?

Answer 1

[John D'Errico]

Use randsample:

Y = randsample(N,K,true,W) or randsample(POPULATION,K,true,W) returns a
    weighted sample, using positive weights W, taken with replacement.  W is
    often a vector of probabilities. This function does not support weighted
    sampling without replacement.

So, for your problem, use it like:

var = 1:4;
W = [0.5, 1/6, 1/6, 1/6];
X = randsample(var,1,true,W);

If you don't have the stats toolbox (where randsample seems to live) this is not difficult to do.

[Adam Vilanova-Goldstein]

thank you so much. Is the Stats toolbox "Statistics and Machine Learning Toolbox"?

[Hans Scharler]

Yes, its the Statistics and Machine Learning Toolbox.

Answer 2

[Steven Lord]

Set up a vector of probabilities.

dist = [2 1 1 1 1 1];
p = normalize(dist, 'norm', 1); % sum(abs(p)) is effectively 1

Use those probabilities as bin edges.

edges = [0, cumsum(p)];
edges(end) = 1; % Make the last bin edge exactly 1, no round-off

Create some sample data.

x = rand(1, 14000);

Use discretize to bin the elements in x into bins delimited by the edges vector.

D = discretize(x, edges);

Let's see how we did.

histogram(D)

The first bin (rolling a 1) has about twice as many hits as each of the other five (rolling 2, 3, 4, 5, or 6.)

Answer 3

[chris j]

Did a similar problem before, essentially rolling a weighted dice.

Var = 1:4;
Odds = [ 0.5 0.16 0.16 0.17]; 
Odds = cumsum(Odds./sum(Odds)); % re-write as a normalised cumulative sum.

Result = Var(find(Odds>=rand,1,'first'));
% generate a random number between 0 and 1
% use the boundaries defined by Odds to check result

%To show it works:
for J=1:10000
    Res(J) = Out(find(Odds>=rand,1,'first'));
end
hist(Res)

[Adam Vilanova-Goldstein]

Thank you so much for the help!