Today I have chosen quite a common problem that uses both interesting and helpful reasoning called the Monty Hall Problem
it begins in a game show with three doors, 2 of those doors contain goats (to make it simpler, I will just call it “dud”) and one contains a sports car (for clarity, I will just call it “win”).
After announcing the rules, the host lets you pick one of the doors; therefore, you have a 1 in 3 chance of picking “win”
But now the host opens another door always containing a “dud” and gives you the option of switching to the remaining door or sticking with your original choice
Now it may seem like you have a 1 in 2 chance either way and in a way – your right but not in the way you would think, let’s look at this in a different way.
In the second stage, these are the three possible options
You chose the “win” and both other options are the “dud”
You chose the “dud” and one of the other options is a “dud” but the other is a “win”
You chose the “dud” and one of the other options is a “win” but the other is a “dud”
Out of these options, only in one do you choose “win” meaning both other options would be a “dud” and by switching you would lose, in both other options the host eliminates the “dud” and by switching you would win.
So going back to the 1 in 2 quiery, there are in fact two different options but the “win” being behind the unchosen door occurs more frequently (2/3)
Now a TWIST
What would happen if your drunk friend ran up to the doors in the second phase and randomly opened one of the two other doors (instead of the host), it turns out that he opens a door with a “dud” behind it and the host decides to carry on, should you switch?
It turns out, this completely ruins your advantage and ill explain why
For the sake of this point imagine 18 people do this separately and the arrangement stays the same every time of 1 and second are “duds” while 3 is a “win” (unknown to them), so 6 people chose the first door, 6 people choose the second and 6 people choose the third.
However, this time, half of the people who chose the first and second door will have their friend open the “dud” door (these are the only ones we will pay attention to) while all of the people on the third door will see a “dud” since those are the only other options
Altogether, 6 people will see a goat on the first 2 doors while all six people will see a goat
From that information we no longer can tell wich door is better to choose since you are one of those 12 people and you have a 1/2 chance of choosing either door!
Weird isnt it
I hope you’ve learnt something from this or at least been intrigued!
And with that I bid you farewell!
