I'm not smart enough

Use the example they gave with 100 doors.

You pick one door, and then 98 are opened to show they are not the right one. When you picked the first door there was a 1 in a hundred chance it was the right one. After they opened all the wrong doors, it is still a 1 in a 100 chance you were initially right. There is a 99 in a 100 chance the other door is the correct door. The common mistake is revaluating the problem once all the doors are open. You can't turn it into a 50/50 situation after all the other wrong doors were opened.

B

 

Play the game on this site if you still are not convinced.

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

B

 

Well, let's start from scratch. Doors A, B, and C.

Let's say you pick Door A, and Door B is the correct answer:

100% of the time, the host will open Door C.

Let's say you pick Door A, and Door C is the correct answer:

100% of the time, the host will open Door B.

Let's say you pick Door A, and Door A is the correct answer:

50% of the time, the host will open door B. 50% of the time, the host will open door C.

Two out of the three scenarios, the host is giving you the right answer. If you look at the three options above, you'll be right by switching in the first two scenarios; you'll be right by staying only in the third scenario.

I think the mathematical logic has to do with the fact that in two of the scenarios, he's not just picking a random door. You're forcing the host to knock out a particular door (since you've blocked one and the correct door is another one).

 

In 100 games never switching I won 29 times below what would be considered average or 33%.

I will now play 100 times switching every time

 

Oh yeah, I remember this from college. It's a real brain teaser, I didn't believe it at first but I eventually figured it out. Just keep thinking about it...

 

You've done what the other sites couldn't.....put it in words a dumb Pole could understand.

HA! Going into it, I agreed with your view and was set to try to prove her wrong. That didn't go so well.

 

I read the story on the webpage, and here's what I thought as I went along:

It's to confuse him. Beforehand, he thought he had 1 chance in 3 of
winning the car. (He picked Door 1.) He now knows that the car is
not behind Door 3, which contains a goat. If he stays where he is,
he has a 1 in 2 chance. If he switches to Door 2, he still has a 1
in 2 chance.

This is assuming that no one has tampered with anything. No matter
which one the contestant picks, there will always be a goat behind at
least one of the other ones. That door can be opened and the goat
shown to the contestant.

<b>(Isabel finds out she has the popular but incorrect answer. Promises
to think about
it before reading any further.)</b>

OK, thinking about it some more... Monty doesn't decide which door
he's going to open until the contestant picks one. The possibilities:
Car behind 1 - contestant picks 1 - Monty opens 2 or 3
Car behind 1 - contestant picks 2 - Monty has to open 3
Car behind 1 - contestant picks 3 - Monty has to open 2
(repeat for "Car behind 2" and "Car behind 3", but same scenario)
There is a 1 in 3 chance that the contestant has already picked right.
However, MOST likely (2 in 3), the contestant has already picked
wrong. To show him a goat, Monty only has one choice of which door to
open. This will narrow it down as to where the car is. Looks like
Marilyn was right... wish I'd seen it the first time.

 

In 100 games in which I switched every time and I won 62 times well above average of 50%

What does this mean? Who the hell knows but it successfully killed thirty minutes of my day. Good Job!

 

sorry, Major had already explained it. forgot to hit "refresh" while I was working on it