Well, it turns out that they both are equivalent, if you have each matching answer used once. This can be proven quite easy with the binomial distribution. This means that a 5 option matching is equivalent to 5 question multiple choice of 5 options.
function MultipleGuessExams % Simulating a multiple choice of 100 questions, 5 options, each weighted % equally. n=5; p=100; w = ones(p,1); MultipleChoice(n,p,w); % Now trying the same thing with true false choice n=2; p=100; w = ones(p,1); MultipleChoice(n,p,w); % Now looking at matching n=5; p=1000; Matching(n,p,w); % The mutliple choice follows the bionomal disbitribution. If there are % p=100 questions and n=5 options, the distributin looks like the % following. n=5; p=100; x=1:100; y = binopdf(x,p,1/n); plot(x,y); ylabel('% Achieving Score'); xlabel('Score');
Simulating a matching exam of p questions, n options per question, no repeats
function score = Matching(n,p,w) score = 0; for i=1:p question = randperm(n); % 1 to n inclusive pertubation matrix answer = randperm(n); % 1 to n inclusive pertuation matrix % Grading correct = question-answer; [loc ~] = find(correct==0); score = score + sum(loc); end score = score/(n*p); fprintf(1,'%i matching questions with %i options: %f\n',p,n,score);
1000 matching questions with 5 options: 0.202600
Simulating a multiple choice exam of p questions, n options per question
function score = MultipleChoice(n,p,w) score = 0; for i=1:p % Choising an answer for the question answer = randi([1 n],1); % Picking a guess guess = randi([1 n],1); % Guessed correctly if guess == answer score = score + w(i); end end % Normalizing the score by the total possible score score = score/sum(w); fprintf(1,'%i multiple questions with %i options: %f\n',p,n,score);
100 multiple questions with 5 options: 0.270000
100 multiple questions with 2 options: 0.480000
end
Congrats to DeSoto (women's), Tanasi (open), and Reckon (Masters) on making Nationals! Congratulations to 'Nooga by Nature as well for getting 4th in a three region bid
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