如何在我的随机函数(MATLAB)中定义概率？

Question

我试图用两个条件进行实验，这两个条件应该以一定的概率显示出来。在这种情况下，"Trial_p“应该出现在2/3的情况下，"Trial_c”应该出现在1/3的情况下。当我进行实验时，我确实得到了一个相当随机的序列，但概率并不像我所定义的那样！有人能帮忙吗？

  Trial_p = [0 4 1 2]; % trial with percept
  td1 = [0 1 0.5 2];
  Trial_c = [0 4 3]; % control trial
  td2 = [0 1 2];

  total_ntrials = 9; % 120 trials

  prob1 = 1/3*2; %0.66
  prob2 = 1/3*1; %0.33

  my_sequence = [];
  my_duration = [];

 randomized_seq = randsrc(1,total_ntrials,[1,2; prob1,prob2]); % randomize trialtypes 
  %x = rand(1)*(10-5) + 5;
   for i = 1:length(randomized_seq)
       if randomized_seq(i) == 1
           td1(1) = floor(rand(1)*(5-1)+1);    %(30-20) + 20);
           my_sequence = cat(2, my_sequence, Trial_p);
           my_duration = cat(2, my_duration, td1);
       elseif randomized_seq(i) == 2
           td2(1) = floor(rand(1)*(10-5) + 5);
           my_sequence = cat(2, my_sequence, Trial_c);
           my_duration = cat(2, my_duration, td2);
       end
   end

Answer 1

除非观测的数量很高(大数定律)，观察到的比例自然会偏离预期的概率。尝试生成大量值并检查：

>> prob1 = 2/3;
>> prob2 = 1/3;
>> randomized_seq = randsrc(1,1e6,[1,2; prob1,prob2]);
>> mean(randomized_seq==1)
>> mean(randomized_seq==2)

给出

ans =
   0.666610000000000
ans =
   0.333390000000000

如果你想要一个确切的比例，比如每种情况中的80和40：

randomized_seq = repelem([1 2], [40 80]); % generate exact number of cases
randomized_seq = randomized_seq(randperm(numel(randomized_seq))); % random shuffle