If you want to know why, continue reading below...
Firstly, for those who don't know what am talking about, Maina's Cash Cow is a challenge that runs on a weekdays radio progamme popularly know as Maina and King'ang'i in the morning, on Classic FM, a popular radio station in major Kenyan towns and their environs, especially Nairobi.
Here is how it basically runs:
- Maina, the presenter, invites radio listeners to call the station and attempt their luck in counting his cash; only one lucky caller gets his call picked up, he will listene the sound of a Note Counting Machine and all he has to do is guesstimate the correct number of notes.
- After the lucky caller guesses the number of notes, say 51, Maina goes ahead and gives the callewr two more possibilities. e.g 'What if I told you it's either 51, 52 or 53?', (should you still stick to your original answer? at this point it does not matter, change to whatever of the 3 you want, it was a guess anyway)
- After the caller settles on one of the three numbers, Maina discloses that one of them is not correct. e.g The user settled on 51, and Maina goes like, 'What if I told you it's either 51 or 53?' (i.e rules out 52).
Apparently, the correct probability solution will predict that SWITCHING from your original answer to whatever remaining option (i.e from 51 to 53 in this example) doubles your chances, from 33% to 67%. Well, it did not make sense to me too at first, I initially worked out that it's just two options (51, 53) and thus the probabilities should be 50% vs 50%. I will not delve into the probability theories that explain this, instead, I will direct you to a simplified python snippet that you can study and also execute to see the results for yourself.
I implemented a simple computer algorithm to simulate the scenario and demonstrate why it is always NOT advisable to stick to your original answer as the caller tackling Maina's Cash Cow challenge. The simulation simply runs 10,000 challenges for each of two guys, one who always sticks to the original answer, and the other always switches. The simulation clearly demonstrated those who stick to their original answer only win ~ 33% of the time while those who switch their answers won ~ 67% of the time. The code is here.
Some of the assumptions made are that:
- Maina knows the correct number of notes even before the caller begins guessing.
- Maina is going to be honest
- Of course it does not have to be real notes, it could be computer generated sound based on a predetermined number of notes or a real Note Counter, the math still stands.
Note: This is a classic probability puzzle known as the Monty Hall Problem.
