%  [birth_02b.m]< [birth_02.m]    2014.11.14 

% For different values of k, k people pick a number from $\{1, \dots  n\}$.
% Find the LARGEST value of n so that:
% The probability  that {\bf  AT LEAST 2} pick the same number
% is greater than tolerance. This is the same as q <= (1 - tolerance)
% where q = Prob( all different birthdays)

tolerance = input('what is the tolerance? ');

% enter range as vector
range = input('what is the range. e.g. [5: 5: 100] ?');

store = []; % start with an empty storage vector

for k = range;
  
   n = k;  % for n < k-1, prob at least 2 pick same  = 1 (pigeon hole principle)
   q = 0;  % when n = k the probability is virtually 0
    
      while q < 1 - tolerance
          n = n + 1;
          a = [n: -1 : (n-k+1)];
          b = (1/n)*(a);
         q  = prod(b);
     endwhile

   store = [store; k (n-1)];    % concatenate new values

endfor

disp('tolerance = '), disp(tolerance)
disp(store)