Why does the probability “lock in” and not change if you remove a variable but doubles for the other one?
I read it multiple times and it still doesnt make any logical sense to me.
In: Mathematics
You pick a door, let’s just say door 1.
What are the chances you guessed the door with the prize? 1 out of 3.
That means there’s a 2 out of 3 chance you guessed wrong, correct?
So now Monty reveals the prize was not behind door 2.
The chances that you guessed wrong initially *are still* 2 out of 3. By switching, you’re taking the odds that you initially guessed wrong.
Imagine it was 10 doors. You had a 90% chance of guessing the wrong door at the beginning.
Monty opens *eight* of the other doors.
The chance you guessed wrong initially is *still* 90%, and you’re being asked “Do you *really* think you’re *that* lucky to have guessed right? Or is it starting to get suspicious that I, the host of the show who knows where the prize is, haven’t opened this *one* other door?”
There doors, one of them has the prize. The host will open an empty door once you pick one. Then you can either switch to the other or go with your first choice.
When do you win if you don’t switch? When your first choice was right. Thats 1 out of 3 so the probability of you winning is ⅓.
When do you win when you switch? When your first choice was wrong, because if thats the case you can only switch to the winning door. Your first choice is wrong 2 out of 3 times so the probability of you winning with switching is ⅔.
With no switching you only win when your first choice was right (⅓) and with switching you only win when your first choice was wrong (⅔).
In a nutshell…all random chance is lost once the host opens the door without the prize. 1/3 only works on random chance, but the host has knowledge that is not random.
The easiest way to see *that* it works is literally to write out all the possibilities on paper, and count how many results in which outcome.
When you start, you have a 1/3 chance of choosing the correct door. Once a “wrong” door has been opened, your odds don’t change from 1/3. Because that wrong door would always have been opened. You’re getting new information and not reacting to it.
If you stick, your chances are still 1/3.
If you change, your odds are 2/3. Because if you decide to change no matter what, you always win if you initially picked a wrong door. You only lose if you initially pick the right door.
There’s nothing “special” about the door you initially picked. If you were right, one of the wrong doors would be opened. If you were wrong, the other wrong door would be opened.
Say there are doors 1, 2, 3. And the prize is behind door 2.
Let’s assume you never change your decision.
You pick door 1 to start with. Monty opens door 3. You stick and lose – 0/1. Let’s say you picked door 2. Monty opens door 1 and you stick and win – 1/2. Let’s say you pick door 3 and Monty opens door 1, you stick and lose – 1/3.
Now, let’s assume you always change. You pick door 1, monty opens door 3, you change to door 2 and win – 1/1. Now you pick door 2, monty opens door 1, you change and lose – 1/2. Now you pick door 3, Monty opens door 1, you change and win – 2/3.
The key to remember is it’s all even odds… But then new information is introduced – Monty doesn’t reveal the prize, ever.
Break it down to just two doors, one has poop behind it, the other has 1 million cash 50/50 odds.
You pick a door.
Monty now shows you a door which contains the poop (might be yours, might not be) – that’s the new information.
Now you know with absolute certainty which door contains the 1 million…
You see how the odds changed because of new information? Works slightly differently with more doors, but the above example should make something click for you.
There are three doors. If you picked the correct door (1/3 chance) then switching bones you. If you picked the wrong door (2/3 chance) then switching gets you the car. Since switching wins if you picked _either_ of the wrong doors, but staying only wins if you picked the _only_ correct door, switching has better odds (2/3).
The trick is that Monty always removes an empty door.
3 doors:
If you pick the prize, Monty removes an empty door. You switch and you LOSE.
If you pick the empty door, Monty removes the other empty door. You switch and you WIN the prize.
If you pick the other empty door, Monty removes the second empty door. You switch and you WIN the prize.
So switching results in a 2/3 chance to win. Not 50/50.
Here’s the part to focus on: if you are wrong, the host will **certainly** offer you the correct door. That’s it.
If you’re wrong, the host’s door will be right **for sure**. Convince yourself of this.
So, there are 3 doors. That means the chances you get it wrong are 2/3, yes? And we know, if you get it wrong, the host will certainly, definitely offer you the correct door.
So with a chance of 2/3, the host is offering you the correct door.
If you only focus on this, I think it makes perfect sense.
It might help though if you imagine there are a thousand doors. If there are a thousand doors, the chances that you get the right door are very, very low. Its 1/1000 that you chose the right door. Right?
So then, with a chance of 999/1000, **the host is offering you the right door**. So you should take the door being offered by the host, its almost certainly the correct one.
Make sense?
If you wanna understand it without the maths talk:
Imagine it with 10,000 doors. You pick one, the host opens 9,998 other doors and leaves you with yours and another one. Do you think its more likely that the host left *that one door* which is the winner, or you picked the 1 in 10,000 door that is the winner first time… Thats why you switch.
Try thinking this way; he only offers to switch if your pick was wrong so now you know it has to be in one of the other two doors.
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