
Game Show Problem (1990) - SanderMak
http://marilynvossavant.com/game-show-problem/
======
StefanKarpinski
I really only understood the Monty Hall Problem intuitively when I sat down to
write a simulation of it to convince someone else that the math (which had
already convinced me) was actually correct. Here's Julia code for it:

    
    
        function monty_hall(switch::Bool; n=3)
          winner, guess = rand(1:n), rand(1:n)
          reveal = setdiff(1:n,[winner,guess])[rand(1:end)]
          if switch
            guess = setdiff(1:n,[guess,reveal])[rand(1:end)]
          end
          return guess == winner
        end
    

This allows you to simulate the switch and don't switch strategies:

    
    
        julia> mean([ monty_hall(false) for _=1:100000 ])
        0.33166
    
        julia> mean([ monty_hall(true) for _=1:100000 ])
        0.6676
    

But more importantly, you can see that in the don't-switch case, the reveal
choice is dead code – it has no effect on wether the original guess was right
or not. That's the key insight: since the original guess had 1/3 chance of
being right before the reveal, it _still_ has a 1/3 change of being right
after the reveal. Before the reveal the other two choices also each had a 1/3
chance of being winners, but after the reveal their combined 2/3 chance is
concentrated entirely on the door that wasn't revealed, making that the best
option.

The code also lets us see what happens with a generalized Monty Hall problem
for more doors:

    
    
        julia> mean([ monty_hall(false, n=4) for _=1:100000 ])
        0.24862
    
        julia> mean([ monty_hall(true, n=4) for _=1:100000 ])
        0.37462
    
        julia> mean([ monty_hall(false, n=5) for _=1:100000 ])
        0.19763
    
        julia> mean([ monty_hall(true, n=5) for _=1:100000 ])
        0.2666
    

Even for higher numbers of doors, it's better to switch.

------
lutusp
What a shame that the title doesn't reveal what this is -- it's the original
appearance of the Monty Hall problem, when Marilyn Vos Savant answered the
problem correctly, only to be criticized (wrongly) by any number of well-
educate academics, like this one:

"Since you seem to enjoy coming straight to the point, I’ll do the same. You
blew it! Let me explain. If one door is shown to be a loser, that information
changes the probability of either remaining choice, neither of which has any
reason to be more likely, to 1/2\. As a professional mathematician, I’m very
concerned with the general public’s lack of mathematical skills. Please help
by confessing your error and in the future being more careful. -- Robert
Sachs, Ph.D."

Doctor Sachs was entirely wrong, as were about ten thousand other academics
who ganged up on Vos Savant in the months following the appearance of this
column.

The real outcome is that, if the contestant doesn't change doors, his
probability of winning is 1/3, but if he does change doors, his chance goes up
to 2/3\. Details:

[http://en.wikipedia.org/wiki/Monty_Hall_problem](http://en.wikipedia.org/wiki/Monty_Hall_problem)

~~~
TallboyOne
Is there a particular reason that so many people were wrong... that seems like
a glitch in the matrix almost. I'm hungover, and suck at math, and understand
after she explained it...

Even after she explained it she still received so many replies saying she was
wrong, I just am wondering what's going on here that's causing this. That's
just so bizarre.

~~~
asdfologist
From the wikipedia article: "Paul Erdős, one of the most prolific
mathematicians in history, remained unconvinced until he was shown a computer
simulation confirming the predicted result."

~~~
lutusp
Now I don't feel so bad for not getting it at first. :)

------
totallymike
I finally got it! It's because whether the host has opened a door or not
_doesn 't matter_.

You know from the start that one out of three doors has a car behind it, and
his opening one of the remaining doors doesn't change it. He just shows you
what you already know.

When you select your door, you know there is a 1/3 chance that your door has a
car behind it, and a 2/3 chance that it doesn't. The host opens another door,
and there is _still_ a 2/3 chance your door has no car. It has only been
proven that the door the host opened has no car behind it.

EDIT: To clarify, I'll rephrase. When you select your door, you know it has a
1 in 3 chance of hiding the car, and importantly, you know that there is a 2
in 3 chance that the car is _not_ behind your door.

When the host opens his door, you still know there is a 2 in 3 chance that
your door doesn't have a car behind it, but are now guaranteed that, if you
chose the wrong door in the first place, you will now choose the right door.

It's not that the host's door is eliminated from the possibilities, it's that
you can either open your door or effectively open _all the rest of the doors_
to find the car. Which would you choose?

~~~
JoeAltmaier
Correct. Intuition fools us in this situation. To illustrate, extend it to 100
doors, only 1 car. You choose 1. The host opens 98 doors with no car (remember
s/he knows where the car is so can always do this). NOW do you want to change
your door for the single door held by the host? Of course.

~~~
totallymike
Thanks :) I updated my initial comment, and noticed that you'd replied only
after I edited. After thinking about it some more it seems so intuitive.

------
tedsanders
Understanding this problem means you can differentiate between the two
scenarios below:

(1) If the host reveals an empty door at random, your odds are 1/3 and you
should switch.

(2) If the host reveals a door at random that happens to be empty, your odds
are 1/2 and it doesn't matter.

The reason that so many people get confused is that they read the words and
their minds automatically jump to scenario 2.

~~~
teddyh
“And a very small percentage of readers feel convinced that the furor is
resulting from people not realizing that the host is opening a losing door on
purpose. (But they haven’t read my mail! The great majority of people
understand the conditions perfectly.)“ — Marilyn vos Savant

~~~
tedsanders
Interesting. I guess I'm part of that small percentage! But I really do think
that's the case. Even if people say the words ("the host opens a losing
door"), they may not be processing what it means ("the host purposefully opens
a door that is known to be losing"). I wonder if I'm wrong.

------
elnate
I find it easist to understand if you increase the number of doors. Imagine
you have to pick one of ten then the host removes eight wrong answers.

~~~
pavel_lishin
Thank you! This is an incredibly intuitive explanation of this.

------
asdfologist
For those who like these sorts of problems, I find this to be another tough
one to wrap your head around:

[http://en.wikipedia.org/wiki/Boy_or_Girl_paradox](http://en.wikipedia.org/wiki/Boy_or_Girl_paradox)

------
deckar01
I think people incorrectly try to simplify the problem by combining equivalent
events without adding the probabilities of those events.

