"Hello World!"

Yes, its a tradition, and a good one at that! Whether you like it or not, randomness rules our lives. The degree of randomness may vary, but the inevitable fact remains. This is a blog post about how I predicted the score I would get on a quiz. Rather, the score that I was *supposed* to get if I chose the answers randomly.

I had taken up a course on Coursera. I must say, I *probably* learnt quite a bit of probability in the course. It so happened that I was running behind schedule by a large margin for a quiz that was due on a Sunday. I had no clue what the course content was, let alone what the questions meant. And the clock was ticking! Remember? randomness controls us. Then I thought, "Why not apply what I had learnt? Lets see if I can predict the score when I answer questions randomly.".

Here's how this particular quiz was structured: There were 9 questions with 4 choices each. You get 3 attempts at the quiz. The questions, nor the choices change after each attempt. The order of the choices might change though.

Lets get predicting!

**Attempt 1:**

Given a single question, `4`

choices and 1 correct answer, what is the probability that you'll pick the right choice if you're asked to choose randomly? Well, thats simple. Its `(1/4)`

.

What if I give you `2`

questions, `4`

choices each and 1 correct answer, what then? `(2/4)`

, isn't it?

What if I give you `3`

questions? You'll get `3/4`

questions right.

If you had `9`

questions, you'll get `9/4`

right. Hold on to that thought!

**Attempt 2**

The questions and choices will not change in this attempt, remember? Good. Now that you've got `9/4`

answers already right, you can just carry them over to this attempt. In this attempt, you'll have to answer `9 - 9/4 = 9 * (3/4)`

questions right. For the sake of brevity, lets call this `n1`

. So, `n1 = 9 * (3/4)`

.

In this attempt, lets do something smart. For the `n1`

questions that you got wrong, you know what you had chosen earlier. The same choice cannot be the answer, obviously. Now, your sample space has reduced to `3`

, instead of `4`

. If you choose the answers randomly for these `n1`

questions, how many will you get right? `n1 / 3`

, right!

The remaining questions are `n2 = n1 - (n1/3) = n1 * (2/3)`

.

Lets substitute for `n1`

: `n2 = 9 * (3/4) * (2/3)`

.

Do you *see* a pattern? Exactly!

**Attempt 3:**

The same rules apply here: same questions, same choices. For the remaining `n2`

questions, you know the 2 choices that are wrong. Your sample space is further reduced to 2. Following the same process, you'd get `n2 / 2`

questions right.

The remining questions are,

`n3 = n2 - n2 / 2`

`= n2 * (1/2)`

`= 9 * (3/4) * (2/3) * (1/2)`

`= 9/4`

.

By the end of the third attempt, you would've answered `9 - (9/4) = 9 * (3/4)`

questions right. Thats a whopping 75%! Thats great, for someone who doesnt even know what the questions are. In this case, 75% turns out to be **6.75** questions.

I put this theory to the test. And guess, what.. **I got 7 questions right out of 9!**

But that cant always be that accurate. There *must* be a method to the madness. There must be a *distribution*.

**Generalization:**

For a quiz with `n`

questions, `c`

choices and `k`

attempts, the number of remaining right answers is given by:

`n * (k / c), for k < c`

Wondering how I got this formula? Drop me an email, lets discuss! :)

`9`

questions; `4`

choices each; `3`

attempts; `1000`

experiments. In each experiment, I noted down the number of correct answers. Here's what the distribution of the number of correct answers looks like. Feel free to change the parameters and run simulations.

`n`

:
`c`

:
`k`

:
`n * (k / c), for k < c`

is **.**

As you can see from the plot, the mean of the distribution is centered around our theoritically determined value. I dont know how to theoritically compute the variance of the plot thought. Any thoughts?

Initially, I had written an **R** script to simulate the experiment. Here's the code. I know, its a bit clumsy. I'm sure there are much better ways to simulate this quiz. Do let me know if you know any of them. I hope this blog post inspired you to look into the subtle everyday randomness. Until next time.. adios!