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
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.
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