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
It’s sometimes easier with problems like this to actually run through all the possible scenarios. Luckily with 3 doors there are only 9
1) prize behind 1, you choose 1. Monty opens either 2 or 3, doesn’t matter, correct coarse of action (COA) **stay**
2) prize behind 1, you choose 2. Monty opens 3, correct COA **change**
3) prize behind 1, you choose 3. Monty opens 2, correct COA **change**
And so on for the others. Of the 9 possible scenarios you’ll find that in 3 of them the best choice is to stay, and in 6 the best choice is to change.
The reason *why* it’s not 50:50 is because Monty is acting on outside information, he changes the rules. If Monty did not know where the prize was then 1/3 of the time you would guess right at the beginning, 1/3 of the time you’d want to change to the other door, and 1/3 of the time *Monty would accidentally open the prize door after your first guess*. That doesn’t happen because he knows where the prize is and intentionally avoids it. The result is that you still guess right 1/3 of the time, but in BOTH of the times you guess wrong he tells you where the prize is.
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