a guest Oct 22nd, 2019 71 Never
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- %Birthday problem simulation method
- numPeople = 86;
- %Let's ignore leap years
- numDays = 356;
- numTrials = 100000;
- cumTriples = zeros(numTrials, 1);
- numPeople = round(numPeople);
- for i = 1:numTrials
- birthdays = round(numDays .* rand(numPeople, 1));
- numTriples = sum(hist(birthdays, unique(birthdays)) > 2);
- cumTriples(i) = numTriples;
- percent = sum(logical(cumTriples)) / numTrials;
- % It seems as though 86 is the number of people one would expect to require
- % based on this simulation.
- % Birthday problem Poisson Approximation
- probability = 1 - exp(-(nchoosek(numPeople, 3)) / 365^2);
- fprintf('In Simulation 3 in %d people share a birthday with probability %f\n', numPeople, percent);
- fprintf('By a Poisson Approximation, 3 in %d people share a birthday with probability %f \n', numPeople, probability);
- % It seems as though 84 is the number of people one would expect to require
- % if we were going on the Poisson approximation.
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