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 have three doors: P (prize), E1 (Empty) and E2 (empty again). It’s important to keep track of the difference between E1 and E2. So now we draw up a table of all possible scenarios, and mark it up with the odds of you and monty choosing each option, from which we can calculate the probability for what the third door has inside
|You|Probability|Monty|Probability|Remaining Door|Probability|
|:-|:-|:-|:-|:-|:-|
|P|1/3|E1|1/2|E2|1/6|
|P|1/3|E2|1/2|E1|1/6|
|E1|1/3|E2|1/1|P|1/3|
|E2|1/3|E1|1/1|P|1/3|
There’s four possible scenarios, two of which (with total probability 1/3) are bad, and two of which (total probability 2/3) that are good.
Do the same situation with 100 doors.
You pick one. The host opens 98.
Odds You got it right the first guess or it’s behind the other door?
Now do it with 97. Same thing.
96, 95, etc.
All the way down to 3.
If we we play with a sports car and two goats..
If your initial plan is to swap doors, it means you’d have to pick a goat on the first try (67% chance), as you’ll then be moved to the sports car. If you picked the car (33%), you will then be moved to a random goat.
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