We all would have come across the reality shows and games which say pick your reward by choosing a door. Haven’t you? The math behind choosing a door is famously called the Monty Hall problem named after a Canadian game show host, producer and a philanthropist Monty Hall. He was widely known for being the host of the game show ‘Let’s make a deal’. And here is the math behind guessing what’s behind the door.
The host asks you to choose a door from three. One of them has a treasure and other two has nothing hidden. Let’s name the doors A, B and C. You don’t know what’s behind the doors unless you open them. Let’s say you choose A. So, what’s the probability that the door you have chosen has the treasure.
Insight into the problem
Let’s go by the old school way of finding the probability. There are three doors and one has the reward. So, the probability for the door to have the reward is 1/3. The probability is the same for all the three doors.
Now, the host opens one of the remaining two doors and finds nothing hidden. So, the treasure must be in one of the remaining two doors. The host asks you a question, “Do you want to switch the door?” What would be your answer? Switch or not switch? Before your answer, let’s find what math says.
Before opening the door, the probability for reward hidden behind a door is 1/3. The probability is same for the other two doors too. Now, after opening a door what would be the probability of having reward behind them. Most of your answers would be 50. That’s how you think probability works here, right? But the answer is wrong. It’s not fifty. Then, what is it? More than fifty or less?
If I say it is 66% for the door you have chosen to not have any reward, would you believe it? I understand, you bet that I’m crazy. I’m sure I’m not. So, your question must be how is it 66 then. Let me explain it to you.
First, when no doors are chosen, all the doors had the same probability of having a treasure. When one of the doors are opened, the treasure must be in either of the two doors. There are only two options now. And the probability must be 50-50. That’s what you think, right? Try some simulation games before you go forward.
But the probability doesn’t work this way here. It’s more like, where is the treasure behind the door you have chosen or behind the other doors. It’s either in what you have chosen or in the others. First, every door had the same probability of having a reward. When you look at it this way after choosing the door, the probability would have become 1/3 for the door you have chosen and 2/3 for the doors you left. You sum up the probabilities of all the other doors and make it one.
When one of the unchosen doors is opened, the probability of the treasure hidden behind the unchosen doors remains the same i.e. 2/3. But now, the number of unchosen doors has been reduced to 1. So, the probability 2/3 totally goes to the door that’s left unopened and unchosen. But, the probability for the door you have chosen still remains the same i.e. 1/3.
What if it is 4 & more?
It’s the same for any number of doors. If you have 10 doors totally and you choose 1 and open an other one, the probability of treasure in the door you have chosen would be 10/100 and for the other doors it would be 11.25/100. As the number of unopened unchosen doors decreases, the probability of the treasure in them increases.
Your choice is a random door out of 10 doors. But, the other is the one who beat 8 other doors. Now, would you switch the door or not? If you follow math then the best idea would be to switch your choice. My choice would be to switch.
When more information is added, you have to re-evaluate the choices. The fatal flaw of the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after he filters the other doors. Choose your choice wisely.
-Keerthana Vengatesan
Author
