I've spent some time thinking about what factors determine exam results. It clearly depends upon the amount of effort you put into it (and the preparations) and your skill level, but also how lucky you are with the tasks you get on the exam. I think I've come up with a mathematical relationship for it

Result = Effort x Skills x Fluke

That is, the exam result is the product of effort, skills and how lucky you are. Here, all factors are non-negative, which gives us a non-negative result as well. A result of zero is a fail, anything above that is a pass.

As we can see, if any of the three factors are zero, you will fail. This is easy to explain. First, if the effort is zero, you didn't even bother showing up on the exam, so you will clearly fail. Further on, if you have zero skills, you don't even know how to read, yet alone remember anything, so no matter how much effort you put into it, you will still fail. Finally, a zero luck means maximum bad luck. An example of this could be to get the following exam:

Task 1 of 1: (100 %)
Is P = NP? Prove your answer.

We can also get some more interesting relationships out of the formula. By dividing by Effort on both sides, we get
Result / Effort = Skills x Fluke.

This tells us that, if you got a non-failing grade on the exam, while the effort approaches zero, you either have extreme skills, or you had a huge amount of luck.

We can also get Result / Skills = Effort x Fluke. So if you got a passing grade, and have zero skills, you put in impossible amounts of effort, or had impossible amounts of luck, which is just not possible, so you could not possibly pass the exam.

So, by using this formula, you should be able to develop a strategy for receiving good exam results  Good luck!

I am studying for my exam in Performance Engineering these days, and came across a nice way of estimating Pi in the chapter about Monte Carlo simulations (random number simulations) yesterday. The idea is that you use random numbers to estimate something, in this case the number Pi.

Let's take a look at the figure below. Here we have the unit circle (a circle with radius 1 and center in origo) drawn in a coordinate system. Up to the right, in the first quadrant, I have shaded a square area with sides of length 1.

As you probably should remember from school, the area of a circle is A = Pi * r^2, and this circle therefore has an area of A = Pi * 1^2 = Pi. One fourth of the circle is in the shaded area, and this part of the circle therefore has the area Pi / 4. The shaded area itself has an area of 1 * 1 = 1. Then we define R to be the ratio between the area of the shaded part of the circle and the whole shaded area. This ratio is R = (Pi / 4) / 1 = Pi / 4.

Now, let's imagine throwing darts at the shaded area at random (we do not aim for any specific area). Of all the darts hitting the shaded area, a ratio of them, corresponding to R (since there is equal probability for hitting any part of the square), should hit inside the circle. For instance, if the ratio is 0.7, and 10 darts hit the shaded area, 7 should hit inside the shaded part of the circle.

If we throw a fair amount of darts, we should be able to calculate R using the formula R = DartsInsideGreyCircle / TotalDartsInsideGreyArea. As we have from the paragraph above, R = Pi / 4, which means that Pi = R * 4. We have estimated Pi!

The JavaScript simulation below illustrates this. Click start and watch what happens! If you let it run for a while, the result will actually get quite close to the actual value of Pi.