**This day that year :: 1989**

Mathematics had eluded my senses (both common and uncommon) and I was literally treading this pathetic path (identifying problem patterns and memorizing its solutions). It was destination “disaster”. I was traumatized and the probability of a success was merely 0.5

**This day that year :: 1990**

An apple fell. I should say it literally broke or rather rewired the entire logic circuits of my psyche. I was so glad it fell unlike Mr. Newton who thought why it fell though I felt it fell late (better late than never, right?) I began to reason, analyze and attack the problems. I should confess that I became passionate, obsessed and finally possessed by Math.

**This day this year :: 2009**

Almost 2 decades. Mathematics rules my senses today. Till day, I don’t really know how this Boolean State Transition (0-1) occurred that too in such a short span. I attribute this success primarily to Mohan sir (my tutor) and to an unseen, unknown force (let me call it God) who, for a purpose, made this happen, which I am still in pursuit of.

Last weekend, I was flipping through the pages of a book on Discrete Mathematics that I bought (thanks to one of my mentors Pai for recommending this book) recently. And I stumbled upon this chapter on probability. It was a feeling that I cannot express through mere words. It was mere bliss as I laid down the Lego Blocks of knowledge one by one and I am sure that I did come up with the most beautiful structure that Richard Johnsonbaugh would have envisioned his reader to create.

Now with this confidence (or rather an element of defiance/arrogance, if I am to confess), I decided to try my fortune on the “Monty Hall” problem. It wasn’t that late before I joined the huge majority that failed trying. My obvious answer was obviously wrong. Hey! At least I tried. I tried to reason where I went wrong and read the solution that was illustrated in detail. I could very easily understand the rationale behind the solution (which is arguably been contested by many as confusing). Thanks to my foundation on probabilistic analysis that was re-solidified by Johnsonbaugh. I would like to share my take on it with the hope that it sheds light to any confused soul.

**Monty Hall Problem**

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? (courtesy: Wikipedia)

**:: My Take on IT ::**

The key to the solution is to ascertain your best odds of nailing the prize with the choice you had made, assuming your chosen door has the prize hidden behind it, based on the 2 options that you have.

**Option 1** (*We decide to stick with our initial choice*)

Remember the “key”, I had outlined above. Based on this, we would opt for Option 1 because we believe we chose the door with the prize behind it. The probability of choosing “this” door out of the 3 choices given is quite obviously **1/3**.

**Option 2** (*We decide to change our initial choice*)

Remember the “key” again. Why does one need to change the initial choice? Because, he/she believes that the remaining closed door (leave apart the closed door that was initially chosen and the opened door without the prize) has the prize behind it.

With this assumption our initial choice (say first event in probability world) had left us with a winning probability (p1) of 2/3 (i.e. the door with the prize is one among the remaining 2 non-selected doors out of the 3 choices). Now, here comes the crucial step in this experiment. The host (unlike in Option 1 where his/her action doesn’t affect the result) is actually telling us something very important i.e. the remaining closed door (leave apart the closed door that was initially chosen and the opened door without the prize) holds the prize in this case. And we can confidently pick this door (say second event in probability world) with a winning probability (p2) of 1 (certainty). So now, as the experiment is *fair* and the events are *independent*, the probability of the entire experiment would be (p1 * p2) i.e. **2/3**.

**Conclusion**

To conclude, let’s analyze our findings on options 1 and 2.

- Assertion 1: The chance of winning with Option 1 is 1/3.
- Assertion 2: The chance of loosing with Option 1 is 2/3 (1-1/3).
- Assertion 3: The chance of winning with Option 2 is 2/3.
- Assertion 4: The chance of loosing with Option 2 is 1/3 (1-2/3).

Now, do I need to tell you what to do? Yeah! You guessed it right. The Assertions says it all!

**:: It is best to switch your choice (Option 2) to increase your odds of winning the prize ::**

Well that wasn’t that difficult, right? I guess not!

And, if you still find it confusing, please try visiting this site which helped me understand this solution to the problem.