3.1 The approach

The approach is to ask the question: if choices are made independently, what is the likelihood of a student selecting a combination that is the same as that of another student?

Let us consider student 1 and student 3. The question is, then, what are the chances that student 3 has the same answers as student 1? To answer this question, we have to narrow down to questions where student 1 and student 3 answer identically. In this case, questions 1, 3, 4, 5, 6, 7 and 8. Then, for each question, we take the choice of student 3 out, and compute the chances of selecting the same choice as student A just out of the conditional probabilities:

• Q1: Besides student 3, only student 1 (out of three students) chooses ‘A’, the probability is .
• Q3: Everyone chooses ‘E’, so the probability is 1.
• Q4: Besides student 3, only student 1 (out of three students) chooses ’A’, the probabillity is again.
• Q5: Besides student 3, two other students choose ‘B’, so the probability is .
• Q6:
• Q7:
• Q8: 1

You can consider the “chance” that students 1 and 3 pick the same answer in question 2 as 1 (100 percent) because they did not choose the same answer!

Then, we multiply all the probabilities. In this case, the product is 0.016460905. This means that if student 1 and student 3 work independently, the chances that they end up with the same answers in questions 1, 3, 4, 5, 6, 7 and 8 is 1.646 percent.

Is this a low probability? It is all relative. Let us consider the probability of student 2 and student 4 selecting the same choices. The questions where they answer the same are questions 2, 3, 4, 8. The probabilities are , 1, and 1. As a result, the product of the probabilities is 0.2222. This means there is a 22.2 percent chance that student 2 and student 4 end up with the same answers by chance.

As a class of 4, the number of probabilities to compute is , which is 6. This is because there are four choices for the first student, and three for the second. However, the probability is symmetric because we only consider questions where two students answer the same way, therefore the product is divided by 2. In this case, the six combinations are:

• students 1 and 2
• students 1 and 3
• students 1 and 4
• students 2 and 3
• students 2 and 4
• students 3 and 4

Once you have computed all the probabilities, sort them and see whether the probabilities are evenly distributed. Pay attention to the combinations with low probabilities.