This video will explain the controversial Monty Hall Problem once and for all!
The Monty Hall Problem is a brain teaser loosely based on the American TV game show "Let's Make a Deal" and named after its host Monty Hall.
The problem goes like this:
Suppose you're on a game show and you're given the choice of three doors where behind one door is a car and behind the others, goats. So you pick a door without opening it and the host opens one of the doors containing goats. The host then asks "Do you want to (1) switch or (2) stay with the door of your choice?".
Assuming you like the idea of winning a car more than a goat, which of the 2 choices will give the higher chance for the better outcome?
*** ANSWER ***
You should always switch doors!
The chance of winning the car is 2/3 when you switch to the other door, while it's 1/3 when you stay with your first choice.
But how is this possible?
There's 2 doors! Shouldn't the chance of getting a goat or car be 50%?
It's pretty much a coin toss at this point right?
What difference does it make if the host opens one of the doors that contain a goat?
These are all great questions! And this is exactly why the Monty Hall problem is so controversial and tricky.
*** EXPLANATION ***
Allow me to explain how the answer came about...
First we have 3 doors. The chance of picking the door with a car behind it is 1 in 3 and a goat is 2 in 3 (Since there are 2 goats and 1/3 + 1/3 = 2/3).
Once the host reveals one of the goats, you are left with 2 doors. So, why is it not 50%?
The confusion arises when people see 2 doors and immediately assume a 50% chance each door is a goat or a car irregardless of the host's reveal.
The part that most people miss is that your chances of picking correctly before and after the reveal remains the same (1/3).
Ask yourself, why would your initial choice now increase to 1/2 or 50% because of the reveal? You did not do anything to change the outcome. So there should be no change. However... the gameshow gives you a chance at redemption by giving you a chance at switching to the only other door and removing the only other outcome.
Mathematically, if your chance of claiming the car is 1/3 with one option then the ONLY other option is 2/3 because the probability that the car is in one of the doors is 100%. (1 - 1/3 = 2/3).
Well.. that was my best shot at a logical proof.
Still not convinced? Then I have a more convincing proof coming up!
Ok! Now imagine instead of 3 doors there's 1 million doors, where there's 1 car and 999,999 goats.
You choose a random door.
What are the chances you chose the car correctly the first time? It's 1 in a million.
The host then opens all but 2 doors concealing the car and the last goat.
After the reveal, do you still believe switching makes no difference?
Switching doors in the 1M door scenario now gives you a 999,999/1M chance to win that car!
Now going back to the 3 door problem, you initially have a 1/3 chance of choosing correctly and because of the reveal you are now given a choice to choose the option that gives you a 2/3 chance of winning. It's just much harder to see in the 3 door scenario since it's close to the 2 door scenario.
With this in mind, switching doors becomes a no brainer!
Thanks for watching!
Hope this clears up the Monty Hall Problem once and for all.
Hit like, Subscribe or Comment if you now understand the Monty Hall problem, disagree or if you just like clicking on random things to warm up your APM :)
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