Possibility that two students answer the same on a 25 question multiple choice test I am trying to get some approximation of the possibility that two students would answer $25$ questions in a row exactly the same.  There are 4 possible responses for each question.  I calculate the random possibility as $4^{25}=1.13\times10^{15}$. Since, however, I assume some amount of familiarity with the material, I divided that by $4$.  I imagine, however, that this is not a precise method.  If I assume an average score of $60%$, is there a more precise model to determine the probability?
 A: While you can come up with a mathematical answer. It would only show how unlikely the possibility of the two students getting the same answers assuming they were to randomly guess. In reality, it is not as unlikely as you would think. For example some questions might have a really deceptive choice and both students pick that choice. Their was a trig test in my Algebra 2 Trig class where two students both failed the test with a score of a 43 and had the same exact answers as each other. Does this prove they cheated? No, they both had their calculators in radians instead of degrees and answered every question correctly. Yes, while mathmatically you can calculate the probability of two students getting the same answers with an average score of 60, but I would advise against it. Just because they have the same answers doesn't mean they cheated...If thats what you are trying to prove.
A: Simply assuming the average score is not enough. For example, take a look at the following test:


*

*Question $1$: What is $1+1$?

*Question $2$: Prove conclusively that $P=NP$.


This test will have an average score of $50\%$, and the number of students that answer the same on all questions will be equal to the number of students taking the test.
On the other hand, you can ask a multichoice question about differential equations to a set of $6$ year old children. In that case, you can more or less assume that the answers are randomly distributed.
A: Edit: We can calculate the probability if we assume that each question has a 60% likelihood of being correctly answered by either student.
For each question, the odds that the students will both choose the correct answer is $.60 \times .60=.36$. In this case, their answers will coincide. The odds that one student chooses the correct answer and one doesn't is $2 \times .60 \times .40 = .48$. In this case, their answers will not coincide. The odds that neither will choose the correct answer is $.40 \times .40 = 0.16$, in which case (assuming their choice is random) the odds that they will have the same guess is $1/3 = .333$. So, the odds per question that the guesses will coincide is $.36\times1 + .40\times0 +.16*.333 = .413$. 
Now we want to know the odds that they guess the same for 25 questions; this is $.413^{25} = 2.5 \times 10^{-10}$.
