Question

THe following code generates random numbers with a given probability of success:

         n=[randi([0 1],1,8) ones(1,8)];
         n= n(randperm(10));

If the above lines are repeated random (unpridictable) values will be generated:

This is n for the first run:

 n =     1     0     1     0     0     1     1     1     1     1

This is n for the second run:

 n =     1     1     1     0     1     1     1     1     1     1

How can I make the generator picking the number of already selected as failure(0) with higher probability?

That is the probability person 2, 3 and 4 in the second round has more probability of loosing. This does not mean that they have to fail.

The entries 1 to 10 are 10 different user outputs.

ok let us say always a max of 30% of entries will be 0. In every time the above is executed. However this is done randomly. So a max of any 3 of the 10 can be zero.

I do not want the probability of success to change.Just to control which one is zero.

To clarify further What I want: If 3 will be chosen “randomly” to be zero then let the previously chosen three have higher probability to be picked and not be picked.

Answer 1

Here’s a solution with the following logic:

1. Generate up to 3 failures, assign them randomly
2. Decide how many failures are in step 2
3. Give the previous failures a higher chance to fail again

Note that I assume that it is equally likely to have 0,1,2, or 3 failures.

nRuns = 5;
allRuns = zeros(nRuns,10); %# stores success and failure for all runs

%# (1) generate from 0 to 3 failures (each outcome is equally likely)
nFailures = randi([0 3],1);
tmp = randperm(10);

firstRun = tmp > nFailures
allRuns(1,:) = firstRun;

%# (2) decide how many failures occur in the 2nd run (each outcome is equally likely)
for iRun = 2:nRuns
%# as the loop has been added later
%# I use "2" to indicate any next run
nFailures2 = randi([0 3],1);

%# (3) give previous failures a higher chance to fail again
failIdx = find(~allRuns(iRun-1,:));
successIdx = find(allRuns(iRun-1,:));
%# 5x higher chance of failing for previous failures
failCandidates = [repmat(failIdx,1,5),successIdx];

failCandidatesRand = failCandidates(randperm(length(failCandidates)));

%# if you have R2012a or later
failCandidatesRand = unique(failCandidatesRand ,'stable');
toFail = failCandidatesRand (1:nFailures2);

%# alternatively, if you have R2011b or earlier
toFail = zeros(nFailures2,1);
toFail(1) = failCandidatesRand(1);
ii = 2;
kk = 2;
while ii < (nFailures2+1)
if ~any(toFail==failCandidatesRand(kk));
toFail(ii) = failCandidatesRand(kk);
ii = ii + 1;
end
kk = kk + 1;
end

%# write failures
nextRun= true(1,10);
nextRun(toFail) = false

allRuns(iRun,:) = nextRun;
end