pull down to refresh
I get the concept but I am struck on the idea that odds go from 2/3 to 1/2 once you eliminate a door.
The combined probability of two doors is always 2/3
We start with
1/3 + 1/3 + 1/3 = 1
We eliminate a door
1/3 + 0 + 2/3 = 1
The key is your initial selection probability never changes
Yes but then the combined probability of my door and the one already opened is also 2/3 so if my door plus opened and the door I didn’t pick plus the opened one both are 2/3 why is it beneficial to switch?
The combined probability of your door and the door already open is 1/3 + 0
The probability of the open door is 0
If you switch p is 2/3
If you stay p is 1/3
I get what you are saying but still doesn’t seem right.
The probability of each door has to add up to 1 or 3/3
Eliminating one of the goat 🐐 doors doesn’t change the original probabilities
Original probabilities are 1/3 🏆 and 2/3 goat
Let me start over and be more concise
Imagine there are 100 doors
You pick 1 door
Probability is 1 percent
Monty opens 98 doors with goats
There is a 99 percent chance it’s not the door you selected
3 doors 🚪, one prize 🏆
I pick door 🚪 2 which has p = 1/3 where p is probability
That means my chance of getting nothing is 2/3
The combined probability of the two doors 🚪 I didn’t select is 2/3
Now Monty Hall opens door 🚪 3 which has no prize ; this means p = 0 for door 🚪 3
That means door 🚪 1 has p = 2/3
The combined probability of door 1 and door 3 is always 2/3
Since p for door 🚪 3 is 0 then p for door 🚪 1 is 2/3
Monty Hall asks me do I want to stay or choose door 🚪 1
I tell Monty Hall I want to switch , give me door 🚪 1
I double my probability by switching
Monty did us a favor by showing us one empty door
Hope this is clear. It’s easier to explain with pen 🖊️ and paper 📝