Let me prefix by stating that this is not really a homework problem, just something I am curious about. I posted it here because it is probably too easy to go into the "big boy" forums . Last night my wife said she needed a 23% on her final to maintain a grade of A in her course. I asked her if the test was multiple choice, she said yes and usually the questions were 4 choice answers. I asked how many questions were usually in the exam, she said about 80. I said, "no need to study, you can probably guess randomly and have a really good chance of getting a 23% or better on the exam." Then I decided to put my theory to the test. I put together a simulator to "randomly generate" an 80 question test and then "randomly answer" the test and score it. After running about 2 million trials, the simulator says about 64% of them score >= 23%. So, I have my answer...now...how to get there with math. I scaled the problem down to a 2 question test. I assume a lower probability to score > 23%, and it seems I am correct; my simulator says 44%. This I think I can do in math: I solved this by chosing a strategy, "answer only A" (assuming a random distribution of answers on the test). Given a two question test, I need to answer either one or both questions correctly to score > 23%. So, there are 4^2 different random answer keys, and of them 7 have one or more A's in the key. So 7/(4^2) = 7/16 = .437 or 44% that my simulator says. Excellent. So...can I extend this solution method to the 80 question test? Is there an easier way to do this? If I can extend it to the 80 question test, how do I do the counting? Anyone want to help enlighten an intrigued mind.... I wish I could remember my game theory class, lol. I bet somewhere in those notes I'd find how to solve this, alas it was just too long ago and my mind is dusty. Please bring on the Pledge .