Comments on: Monte Carlo simulation of the Monty Hall Problem in Ruby and Python

By: Using Programming to Teach Mathematics

Using Programming to Teach Mathematics — Fri, 29 Jul 2016 12:00:57 +0000

[…] example, the Monty Hall problem even managed to fool some seasoned mathematicians. Running a simple simulation we can quickly see that switching doors is the winning strategy (and that we’ll win roughly […]

By: Al

Al — Thu, 10 Dec 2009 13:12:07 +0000

How do you modify the script to simulate a case where there are say 2 cars and 5 goats?

By: Blinkozo

Blinkozo — Wed, 18 Nov 2009 21:25:08 +0000

Gustav, it’s not a true 50/50. Since he picked his door when there were more goats than cars, the veteran’s door is more likely to have a goat in it, His door, therefore, is tainted by previous bad odds.

The only time it’d be 50/50 is if Monty removed the goat first, then let the vet select a door. One may think this is the same thing, but remember: the door eliminated in your scenario is guaranteed NOT to be the contestant’s door. This means the contestant keeps the odds he was originally dealt (1:3), unless he switches.

By: Gustav Isaac

Gustav Isaac — Fri, 02 Oct 2009 20:08:18 +0000

Monty Hall problem
Lets suppose the contestant comes on stage and selects door # 1 in the hope that a prize is in door 1 and not a goat. Before Monty Hall does or says anything, an associate comes out and whispers something to Monty Hall. Now Monty Hall turns to the contestant and says: I just have been informed that you are a veteran. Therefore I want to better your odds and have you decide only between 2 doors. He than tells a Stage hand to remove Door 3 from the stage. ( Of course Monty knew that there was a goat behind door 3). Now he turns to the contestant and says: “There you are, 2 doors. New game. You want to switch from door # 1 to door # 2, or stay with your original choice?” Same setup as in the original problem, nothing changed except there only 2 doors now, and the contestant has now a chance of staying with his original choice or selecting the other door.
That’s now 50-50 probability, even though it’s the same problem. Which means the original problem must also be a 50-50 proposition.

By: Francesco

Francesco — Thu, 05 Mar 2009 09:26:57 +0000

What if I switch the previously selected door with itself? It’s the same as switching with the other one, right?
I’m no math-geek at all, I just want to see the light because I’m really puzzled! 🙂

By: Jérôme

Jérôme — Fri, 06 Feb 2009 11:01:56 +0000

On my desktop :

time python montecarlo.py 

real	0m7.882s
user	0m7.880s
sys	0m0.004s

time ruby montecarlo.rb   (ruby 1.9.1)

real	0m3.924s
user	0m3.904s
sys	0m0.000s

By: Ryan

Ryan — Tue, 03 Feb 2009 03:59:50 +0000

I almost said something awful just because of your comment, Kent.
Not really, just kidding.

By: Kent

Kent — Wed, 07 Jan 2009 18:13:59 +0000

What’s really amazing is the number of posts here without a single disrespectful comment. In this day of internet anonymity, what are the odds of that?

By: Daniel

Daniel — Wed, 07 Jan 2009 15:29:13 +0000

I find it easier to think of this as if you are actually switching from your single chosen door to both of the two other doors.

Since the rules are known before the game you know you’ll pick one door and then you’ll be able to switch to the other two. Since you know you’ll switch you can think of it as if you already “own” the two other doors. When Monty opens one of them, he is just helping you to open a losing door first.

It’s actually just like if you were given the choice of selecting one of the three doors or two of the three doors.

By: Monty Hall with a billion doors « Random Determinism

Monty Hall with a billion doors « Random Determinism — Wed, 07 Jan 2009 05:56:22 +0000

[…] Jump to Comments In order to provide intuition behind the solution of the Monty Hall problem, Antonio Cangiano says: If there were a billion doors, you picked one, and then Monty proceeded to open up all the […]