Here's the exact text of the original question, posed to Marilyn in her column in parade magazine:

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

http://www.fortunecity.com/victorian/vangogh/111/9.htm

The stipulation that the host knows what's behind the doors, and always opens a door with a goat is a given for this problem.

In fact, the additional information that it was 'Monty Hall' came later... which messed the whole thing up because they interviewed him, and he stated that he sometimes opened the prize door right away... and then everyone forgot the stipulations of the original question

I actually remember when this happened, because my college statistics professor was *consumed* with proving Marilyn wrong, and ended up conceding she was correct only after writing a program to do 10,000 permutations that bore her true

Trek

And that statement, while better than how it is presented the vast majority of the time, still isn't sufficient to make the answer unambiguous. There is still the question of Monty's motivations which could make it either adviseable to switch (up to 2/3 odds of winning) or to stay (up to 100% chance of winning).

An interesting follow-up on Marilyn. Later she tackled a restricted version of the problem that I presented at Spooky math problem (the restriction being that the two envelopes hold money, one has twice the money of the other) and correctly analyzed an argument for whether you should always switch. But she incorrectly analyzed whether you could do better than even odds. I know a couple of probability theorists who pointed out her mistake to her, but she never admitted to her mistake.

Make of that what you will.

And that statement, while better than how it is presented the vast majority of the time, still isn't sufficient to make the answer unambiguous. There is still the question of Monty's motivations which could make it either adviseable to switch (up to 2/3 odds of winning) or to stay (up to 100% chance of winning).

I might be misunderstanding the issue then.

The question, as I understood it, is that I find myself on a game show, and I've picked a door... the host has just opened a door to reveal a goat, and then asked me if I want to change my choice to the other door.

So, in order to make my decision on this *one* event, I have to test it... and in order to test it, I must set up *identical* events, devoid of the host's motivation, to test whether I should change in this one instance.

It would be different if the question was 'in general, what should I do if I don't know the host's motivations', but the question states that the host opened a door with a goat... therefore my examination of the question should assume that fact

I don't know if this is coming across correct in text or not, so I apologize if it isn't...

Put another way, I'm setting up a model to that mimics the original question, which assumes the host opens a goat door

The possibility that the host *might* not offer the choice... or might reveal the prize first... etc... don't enter into the answering of this question, because the question supposes that the opening of the goat door has already happened, and now you have to make a decision based on the probabilities in play.

Trek

I'm not sure of the relevance of this. Stats was never my strong suit but...

If the following simulation bears any resemblance to reality (of the re-stated version of the problem), then you only get a greater chance of winning if you always switch your choice after the host opens the first door on the goat.

If at that time you make another choice--either, whether to switch or not, or a random choice between the two remaining doors (which amounts to the same thing?)--then your odds of success remain at 33.33%.

#! perl -slw
use strict;
use List::Util qw[ shuffle ];

our $TRIALS ||= 1_000_000;
my $DIVISOR = $TRIALS / 100;

my( $stick, $switch, $newChoice );

for my $try ( 1 .. $TRIALS ) {
      ## Randomly hide the prizes behind the doors.
    my @doors = shuffle 'car', 'goat', 'goat';

      ## Pick a door
    my $choice = int rand 3;

      ## The host opens a door that I didn't pick
      ## and that isn't the car
    my $opened = grep{ $_ != $choice and $doors[ $_ ] ne 'car' } 0 .. 
+2;
    
      ## Count my wins if I stick or switch
    $doors[ $choice ] eq 'car' ? $stick++ : $switch++;
    
        ## Make a new choice from the remaining two
    my $new = ( grep{ $_ != $opened } 0 .. 2 ) [ rand 2 ];

      ## And if I make a completely new random choice.
    $doors[ $new ] eq 'car' and $newChoice++;
}

printf "Odds of
    Choose again: %.3f
    Win if you don't switch: %.3f
    Win if you do switch: %.3f\n",
    $newChoice / $DIVISOR, $stick / $DIVISOR, $switch / $DIVISOR;

__END__
P:\test>test
Odds of
    Choose again: 33.331
    Win if you don't switch: 33.286
    Win if you do switch: 66.714

P:\test>test
Odds of
    Choose again: 33.370
    Win if you don't switch: 33.353
    Win if you do switch: 66.647
[download]

Assuming that this simulation isn't at fault, I'd love to see a (preferably layman's terms) explaination for why a predetermined strategy (always switch) would have such an effect on the odds of success?

---

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algorithm, algorithm on the code side." - tachyon

I've never seen the Monty Hall problem presented with that stipulation explicit.

Update: strike off the cuff response... By the way (after a quick read and next to no analysis) the spooky math problem looks more like a divide-by-an-infinite-number problem. The probability that you pick a number between two numbers in an infinite continuum is undefined.

Please read Re^7: Marilyn Vos Savant's Monty Hall problem carefully. You'll find that it is not enough to say how much knowledge Mr. Hall has - to make the answer unambiguous you also need to specify how Mr. Hall will use that knowledge.

On the spooky math problem, there is no divide by an infinite number problem, but there is a lot of subtle behaviour with infinity lurking around. I strongly recommend against coming to any opinion without thinking it through very carefully. It took me about a week to really understand what was happening when I first ran into the problem. (It didn't help that Laurie let me get myself well and truly convinced that there was no way that you could beat 50% before he gave me the answer and told me to figure out for myself why it worked...)

Given any pair of numbers in the envelopes, and given the distribution from which you pick a random number, your probability of winning is well-defined and will be bigger than 50%. Depending on those three factors, the actual odds can be anywhere between (but not including) 50% and 100%. But you are not provided with sufficient information to figure out what the odds are.

Please read Re^4: Marilyn Vos Savant's Monty Hall problem carefully, especially where I wrote

the host knows and will not open the door with the car (emphasis mine)

The knowledge Mr. Hall has and how he uses it is unambiguous. He always opens a door, he never opens the door with the car. and he always offers a switch. What's more important than the information he has is the information he provides by using what he knows in a predictable way.

I fully understand your previous node, and actually we are in agreement about the necessary information to calculate the probablilities. We just differ on what the original problem statement is and/or what assumptions can be safely made for this specific problem.

Re the spooky math problem: what I wrote were my initial thoughts when I read that root node. I'll reserve judgement and not argue either side until I investigate further (depends on life, kids, the missus etc. if I ever get to it). From what I've seen, if you are this sure you are right then you probably are. I've seen stranger things with my own eyes, so I won't rule out any possibility.

People seem to be arguing with tilly about this, but this has been my experience too. I have seen this problem published at least a dozen times; I have never seen the assumptions (e.g. that Monty knows what is behind the doors, etc.) stated clearly before.