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
What if I give you
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!
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 = 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
n2 = 9 * (3/4) * (2/3).
Do you *see* a pattern? Exactly!
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)
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.
For a quiz with
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! :)
4 choices each;
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 * (k / c), for k < cis .
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!