thisnumbersdisconnected
New Member
No, there really are only two.That is because there are three curtains.
No -- and I do fully understand what you've been saying. You're just not grasping the reality of the game, but allowing a door that is never in the equation to become a distraction for you.If you would just bear with me for a bit and answer my yes or no question I think you'll see why there is never a 50/50 chance for any single curtain. Question: Before any curtain is opened do you agree that if curtains #2 and #3 are in area "B" that there is a 2/3rds chance that the car is in area "B"? Yes or no?
First, let me deal once again with your altered version. In your scenario, Area "A" is one choice with one door in it. Area "B" is one choice, even though it has two doors in it. That is two choices -- a 50/50 chance of winning the car.
Now, to the real game scenario. Just as with your altered scenario, there are never more than two doors in the running. You need to understand how the game works. Standing on the floor in front of you when you're next to Monte Hall, there appear to be three choices -- but only apparently.
Now, let me explain why your scenario is the same choice Monty gives me, and why both represent a 50/50 chance of winning the car. From the very start, Monty is going to eliminate one of the non-prize doors. Before you ever make a choice, one of those doors is going to be tossed. He always does that. You know he always does that. No matter which door you pick, before you see where the car is, one of the non-prize doors will be removed from the equation.
Therefore, it is never in the equation to begin with. There is not a one-in-three chance of anything. There is a one-in-two chance. The third door is an illusion, a "false flag," a distraction -- whatever you want to call it. Let me illustrate.
Here are the three doors:
X X X
You pick Door #3:
X X X
Monty removes Door #1:
X X
Now, the same scenario, only Monty removes Door #2 this time:
X X X
X X X
X X
I can do the same thing with you picking another door to show you that Monty is going to remove one of the other doors and while you might get this ...
X X
... that's still the same thing as either of the other two scenarios.
In all the scenarios, how many doors are left? Answer: 2
Was Monty ever going to remove the door that has the prize behind it? Answer: No
Which door, therefore, has the prize been behind all along? Answer: Either your door, or the other one.
Was the door Monty removed ever a factor? Answer: No.
Why was the door Monty removed never a factor? Answer: Because it did not have the car behind it, and you did not pick it.
No matter where the car is, Monty is always going to take one door out of the equation before revealing the prize. Also, before revealing the prize, he is going to barrage you with enticements to trade your door for cash, another prize, or the other door. He ignores the third door, never offers it to you, never speaks of it except to eliminate it.
It doesn't matter if you already have the door with the car, or the remaining door has the car, he's going to do the same thing: Take away one door, entice you to change, and regardless of your decision, reveal the door hiding the car. The fact he removes one door before doing anything else means that door was never in the equation. Mathematically, you had a 50/50 chance from the very beginning. However, because intuition trumps probability, it is nonetheless prudent to trade your door for the remaining door.
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