Comments on: Monty Hall in Monte Carlo

By: Joe Gallo

Joe Gallo — Fri, 21 Jan 2011 02:11:41 +0000

The way I’ve explained it to people is to imagine 10,000 doors, and you pick one, and then the host runs through the other 9,999 except for number 5,521 (of which he says, “uh… i’m just going to skip this one…”).

Do you still think there’s a 50/50 chance you are right, after he’s opened all the doors except the one you chose and door 5521?

By: Steve Gilardi

Steve Gilardi — Sun, 28 Jun 2009 02:25:08 +0000

Whether your first choice has the prize or not is unaffected by whatever happens after. Your first choice will have the prize 1/3 of the time.

Never switch policy: The probability of getting the prize is the probability that your first choice is right: 1/3

Always switch policy: The probability of getting the prize is the probability that your first choice is wrong: 2/3

By: incanter

incanter — Tue, 23 Jun 2009 20:30:04 +0000

That’s correct, the host can’t change the door he initially placed the prize, but he can change the door he reveals based on what you choose. If you choose the door with the prize, he can choose either of the remaining doors. If you choose the door without the prize, he must choose the other door without the prize. That is the key, but since people have debated this for nearly 20 years, I’m sure we won’t convince each other :)

By: Pickle Pumpers

Pickle Pumpers — Tue, 23 Jun 2009 20:16:48 +0000

That’s completely wrong but don’t feel bad you in some pretty good company.

I’ve written a proof using a simulation instead of the flawed logic behind those charts.

Basically it comes down to the third state. At the third state you try to say that the host can now CHANGE the door he initially picked. Um, NO!

Here’s the proof with a program and source code.

http://picklepumpers.com/wordpress/?p=130