fun with the monty hall problem

[The Cheesehusker, Trade Warrior]

Fun read about this - and yes, I've had fun with this in the past

http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/

[JGibson]

On Monday, February 23, 2015 at 12:59:07 PM UTC-5, The Cheesehusker, Trade Warrior wrote:
> Fun read about this - and yes, I've had fun with this in the past
>
> http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/

I remember that column. It caused quite the uproar in my high school calculus class. Initially, my math teacher sided with the people berating vos Savant, but within a few days, she came back in and said it was correct to switch. The illustration with about 100 doors really helped me back then.

[xyzzy]

Sometimes a problem just needs to be solved with brute force, rather than mathematical elegance or cleverness (said just about every real world software engineer at one point or another).

[Ken Olson]

"xyzzy" <xyzzy...@gmail.com> wrote in message
news:b2019594-01b5-4e55...@googlegroups.com...

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

If it doesn't fit, force it. If it still doesn't fit, get a bigger hammer.

[Michael Press]

In article <b2019594-01b5-4e55...@googlegroups.com>,

I prefer the elegant solution. Monty Hall offers you the
opportunity to choose _two_ doors, the other two doors
from the one you did choose.

One of the assumptions is that MH always opens a door.
Notice that this and other assumptions are not stated.
Now suppose he does not always open a door and more often
opens a door when the contestant chooses the door with the car.

I can make assumptions that do not contradict the
problem statement but that alter the probability of
winning all over the map.

--
Michael Press

[Michael Press]

In article <rubrum-1C027E....@news.albasani.net>,

Michael Press <rub...@pacbell.net> wrote:

> In article <b2019594-01b5-4e55...@googlegroups.com>,
> xyzzy <xyzzy...@gmail.com> wrote:
>
> > On Monday, February 23, 2015 at 3:24:07 PM UTC-5, JGibson wrote:
> > > On Monday, February 23, 2015 at 12:59:07 PM UTC-5, The Cheesehusker, Trade Warrior wrote:
> > > > Fun read about this - and yes, I've had fun with this in the past
> > > >
> > > > http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
> > >
> > > I remember that column. It caused quite the uproar in my high school calculus class. Initially, my math teacher sided with the people berating vos Savant, but within a few days, she came back in and said it was correct to switch. The illustration with about 100 doors really helped me back then.
> >
> > Sometimes a problem just needs to be solved with brute force, rather than mathematical elegance or cleverness (said just about every real world software engineer at one point or another).
>
> I prefer the elegant solution. Monty Hall offers you the
> opportunity to choose _two_ doors, the other two doors
> from the one you did choose.

Then you disagree with me about this, xyzzy.

--
Michael Press

[jim brown]

On Monday, February 23, 2015 at 11:59:07 AM UTC-6, The Cheesehusker, Trade Warrior wrote:
> Fun read about this - and yes, I've had fun with this in the past
>
> http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/

This reminds of me of the old adage, figures don't lie, but liars figure. To me its a 50/50 chance of the car being behind your door. The rest is smoke and mirrors.

[The Cheesehusker, Trade Warrior]

The key to this is whether or not Monty knows where the goat is - if he does, then the odds say switch

[Futbol Phan]

The key is both that Monty knows where the goat is AND he isn't lying when he tells you the door that does not contain the car.

I think enough people have watched the show to know that Monty doesn't always rule out one of the doors and thus they treat the the information provided in this problem with skepticism.

[Michael Press]

In article <830eb399-083e-4011...@googlegroups.com>,

Your words do not specify anything.
Try to specify exactly what you intend to convey.

--
Michael Press

[xyzzy]

Attribution error?

[jim brown]

On Thursday, February 26, 2015 at 12:52:46 AM UTC-6, Michael Press wrote:
> In article <830eb399-083e-4011...@googlegroups.com>,

> wrote:
>
> > On Monday, February 23, 2015 at 11:59:07 AM UTC-6, The Cheesehusker, Trade Warrior wrote:
> > > Fun read about this - and yes, I've had fun with this in the past
> > >
> > > http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
> >
> >
> >
> > This reminds of me of the old adage, figures don't lie, but liars figure. To me its a 50/50 chance of the car being behind your door. The rest is smoke and mirrors.
>
> Your words do not specify anything.
> Try to specify exactly what you intend to convey.
>
> --
> Michael Press

I think I was clear. If there were two doors, and you chose one, you have a 50% chance at the car. Once the third door is eliminated, you are left with the above....Its that simple. The rest is a bunch of educational elites trying to justify their existence.

[The Cheesehusker, Trade Warrior]

Negatory, my good friend.

Initially, you had a 1/3 chance of the correct door - and that doesn't change. With one door now eliminated, the math sez the door you didn't choose is worth 2/3s

[Con Reeder, unhyphenated American]

You could pretty easily run a Monte Carlo simulation on it if you doubt.

I play bridge, and here is the same thing presented a different way:

http://en.wikipedia.org/wiki/Principle_of_restricted_choice

--
Experience is what allows you to recognize a mistake the second time you
make it. -- unknown

[jim brown]

No...It wasn't behind door number three....thus it is behind door number one or two ... 50/50


What makes the door you didn't choose any more special than the door you did choose? What if you chose opposite, is the current choice suddenly worth more just because you didn't choose it?

[The Cheesehusker, Trade Warrior]

What matters is how the problem is set up - the revealed door must be a nonprize winner for this to work.

We know there's a prize behind one of 3 doors, right? In our initial choice, all doors carry the same weight.

However, once one door is shown to be a nonprize, the odds shift dramatically.

Initially, we had 1/3 chance of guessing the right door - which means there was a 2/3s chance of the right door being in the other group.

And that does't change once a nonprize door is revealed - our initial guess is still valued at 1/3 and the remaining doors, of which there is now only one door.

Think of it with 100 doors - easier to grok

We choose one and the odds of that being the right door are 1% - which means there's a 99% chance of the prize being behind the other doors, right?

Now - 98 doors are opened and shown to not be the prize door - which now leaves only 2 doors - the one we choose and the other. Obviously, there's still only a 1% chance that the door we chose initially is right - which means that the remaining door has a 99% chance of being the winner.

If you prefer to use algebra - label your door X and all other doors, place them in the group of Y so that: x + Y = 1
Y is (D1+d2+d3+....)

or x + (d1+d2+d3....)=1

If d2 and onward are now 0, then d1 can only be 1-x

[xyzzy]

What are you some kind of educational elite?

Btw I agree with you that using the 100 doors example is what clarifies the answer better than anything else.

[The Cheesehusker, Trade Warrior]

We all have our crosses to bear....

> Btw I agree with you that using the 100 doors example is what clarifies the answer better than anything else.

Yep - and it all hinges on knowing the removed doors are non-prize winners - or 0s

[Con Reeder, unhyphenated American]

Because the a priori chances don't change. The fact that the car wasn't
behind one of the two doors affords a presumption that it is more likely
to be behind the other door.


--
Research is what I'm doing when I don't know what I'm doing.
-- Wernher Von Braun

[xyzzy]

In a pure statistical exercise, yes. But in the world of Let's Make a Deal, you do have to contend with the fact that Monty Hall knows where the car is and may be motivated to try to guide you one way or the other. I.e., if he knows you picked the winner but wants to induce you to switch, that's exactly what he'd do. This is definitely a case where you can overthink.

[xyzzy]

Do what real programmers do, Connie. Run a brute force simulation, build a table of all possible inputs and outcomes.

You'll see that switching yields a winner more often than not.

http://www.stayorswitch.com/

[The Cheesehusker, Trade Warrior]

man - is there nothing on the web anymore?

[xyzzy]

For those who doubt the integrity of that simulation, there's this.

http://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions

(the table at the beginning of the section is most convincing to me. The picture under it is useful but probably won't convince someone who is sure of the wrong answer)

It is an interesting logic problem and an example of the problem with overthinking. It's much easier and quicker to lay out the possible scenarions and count outcomes than to try to logically deduce a convincing answer.

[Michael Press]

In article <51686220-3d17-4c35...@googlegroups.com>,

xyzzy <xyzzy...@gmail.com> wrote:

> On Wednesday, February 25, 2015 at 2:56:48 PM UTC-5, Michael Press wrote:
> > In article <rubrum-1C027E....@news.albasani.net>,
> > Michael Press <rub...@pacbell.net> wrote:
> >
> > > In article <b2019594-01b5-4e55...@googlegroups.com>,
> > > xyzzy <xyzzy...@gmail.com> wrote:
> > >
> > > > On Monday, February 23, 2015 at 3:24:07 PM UTC-5, JGibson wrote:
> > > > > On Monday, February 23, 2015 at 12:59:07 PM UTC-5, The Cheesehusker, Trade Warrior wrote:
> > > > > > Fun read about this - and yes, I've had fun with this in the past
> > > > > >
> > > > > > http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
> > > > >
> > > > > I remember that column. It caused quite the uproar in my high school calculus class. Initially, my math teacher sided with the people berating vos Savant, but within a few days, she came back in and said it was correct to switch. The illustration with about 100 doors really helped me back then.
> > > >
> > > > Sometimes a problem just needs to be solved with brute force, rather than mathematical elegance or cleverness (said just about every real world software engineer at one point or another).
> > >
> > > I prefer the elegant solution. Monty Hall offers you the
> > > opportunity to choose _two_ doors, the other two doors
> > > from the one you did choose.
> >
> > Then you disagree with me about this, xyzzy.
>
> Attribution error?

No.

--
Michael Press

[Michael Press]

In article <2696873e-a36f-4b28...@googlegroups.com>,

No, it does not. I clarified it better than anything that has been presented here
in my initial reply to you when you spoke about elegance.

--
Michael Press

[Michael Press]

In article <5966af27-233e-4ac7...@googlegroups.com>,

jim brown <naberh...@gmail.com> wrote:

> On Thursday, February 26, 2015 at 12:52:46 AM UTC-6, Michael Press wrote:
> > In article <830eb399-083e-4011...@googlegroups.com>,
> > wrote:
> >
> > > On Monday, February 23, 2015 at 11:59:07 AM UTC-6, The Cheesehusker, Trade Warrior wrote:
> > > > Fun read about this - and yes, I've had fun with this in the past
> > > >
> > > > http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
> > >
> > >
> > >
> > > This reminds of me of the old adage, figures don't lie, but liars figure. To me its a 50/50 chance of the car being behind your door. The rest is smoke and mirrors.
> >
> > Your words do not specify anything.
> > Try to specify exactly what you intend to convey.
> >
>
>
>

> I think I was clear. If there were two doors, and you chose one, you have a 50% chance at the car. Once the third door is eliminated, you are left with the above....Its that simple. The rest is a bunch of educational elites trying to justify their existence.

I see what you are getting at, but you did not make yourself perfectly clear.
I made the matter perfectly plain and simple in my reply to xyzzy.

--
Michael Press

[xyzzy]

Basically your elegant solution is what is pictured in the second picture of the wiki link I posted.

I agree that solution is simple and elegant. However I don't think it would convince someone who is sure of the wrong answer. Only laying out all the possibilities and results can do that, IMO.

[Futbol Phan]

I did an demonstration of this with my nephew (he of the 800 Math SAT score and a disbeliever) a few Thanksgivings ago. I used a deck of cards, and the Ace of Spades was the grand prize and all other cards the goats. So I shuffle up the cards and (knowing which one is the grand prize) watch him pick the damb ace of spades. Of course once I reject 50 of the cards, leaving him with the choice of the one remaining card on the table or the one he chose initially, he opts to hold onto his card. And looks at me like I am a moran.

[Futbol Phan]

I have used an even bigger N example in my statistics course-- a winning lottery ticket. Let's say I have a room full of all sold PowerBall tickets, one of which is the winner. I ask you to pick one, and then proceed to rip up all but one of the remaining 30,000,000 tickets. Then you have to decide whether the one you picked initially is the one you want to keep, or whether you would like to switch to the one remaining ticket. Interestingly, even when presented with this scenario, there are a few holdouts who still want to continue to hold on to their ticket, despite the fact that the odds of them holding the winner are now 1 in 30 million...

[The Cheesehusker, Trade Warrior]

That's funny as hell

[Michael Press]

In article <565da8fb-38e6-419b...@googlegroups.com>,

MP -- I choose door 2.
MH -- You can have door 2 OR you can have both of doors 1 and 3.
MP -- I choose both doors 1 and 3.

--
Michael Press

[jim brown]

But you never answered my question....what makes door 2 better than door 1 (when door 3 is opened)? Why is the prize more likely to behind one randomly selected door than the other? What if you DID pick door 1...then suddenly door 2 is the better door?

Even with the 100, once you get down to two, its 50/50....one door is not better than the other...

[jim brown]

Allright....now THAT makes a little more sense to me.

[xyzzy]

That's what makes this puzzle so maddening. It isn't intuitive.

Look at Michael Press's explanation. I.e., you get to pick one of three doors. What if you got to pick two of the three doors instead? That would obviously be better odds. What's what you get for all practical purposes when one of the non-picked doors is opened and you get to switch. You get all the probability of having picked both of the other doors, with the plus that you know one of the other doors isn't it.

[jim brown]

Noone has answered the question....when there are three doors and one is opened, what makes door 2 better than door 1? What if you pick door 1, is door 2 suddenly better?


But the explanation with the lottery tickets makes perfect sense....I just don't see it transferring to a 3 door scenario without proper answers to the above questions.

[Eric Ramon]

On Thursday, February 26, 2015 at 4:05:31 PM UTC-8, jim brown wrote:

>
> But the explanation with the lottery tickets makes perfect sense....I just don't see it transferring to a 3 door scenario without proper answers to the above questions.

here's my attempt, with the disclaimer that I didn't believe it until I saw the brute force version.

You select a door knowing there's a 1 in 3 chance that you've got the right one and a 2 in 3 chance that it's one of the others.

You also know that it can't be behind both of the other doors so 1 of them is the "goat" (as they say in this example). But there's still a 2 in 3 chance that your original pick is wrong.

So when one of the doors open and it's not the big prize that doesn't affect the original odds. It's still 1 in 3 for your door and 2 in 3 that it's one of the others.

That probably doesn't help you either but it works for me.

[Michael Press]

In article <66ebeefd-0470-4146...@googlegroups.com>,

> Noone has answered the question.... when there are three doors and one is opened, what makes door 2 better than door 1?

I answered it.

> What if you pick door 1, is door 2 suddenly better?

No.

ME -- I pick door 1.
MH -- Okay. You can have door 1 OR you can have doors 2 _and_ 3.
ME -- I pick doors 2 and 3.

--
Michael Press

[Some dued]

A room is full of gnomes, each is wearing either a white hat or a black hat. They can see everyone else, but do not (and can not) know what color their own hat is. They can in no way communicate with one another.

They must enter another room one by one and line up against a wall so that when finished all black hats will be on one side, all white hats the other.

How can this be achieved?
This isn't a trick question and has a logical and simple answer, but its tough to come up with.

[The Cheesehusker, Trade Warrior]

Yes - the door you didn't pick is always better. Doesn't mean it'll win fo' sho' - just that the odds are with it.

[Eric Ramon]

the one by one is the key. The first guy is just going to stand around in the middle of the room. The second one sees ...no, wait a minute...hold on..the logic is there somewhere if I can only lasso it.

[Michael Press]

In article <aacbaa5c-10f3-409e...@googlegroups.com>,

Make it work for exactly two gnomes.

--
Michael Press

[Thomas R. Kettler]

In article <rubrum-7CF320....@news.albasani.net>,

Make it work for one gnome.
--
Remove blown from email address to reply.

[Michael Press]

In article <tkettler-B53821...@88-209-239-213.giganet.hu>,

I asked first, but since I am gracious and only spread
sweetness and light, I will answer.
The single gnome goes into the second room and stands at a wall.

--
Michael Press

[Wolfie]

"Michael Press" wrote

> The single gnome goes into the second room and stands at a wall

The second stands next to him, and

X
<--- X?O
O

- or -

X
X
<- X?O

until done.

[Michael Press]

In article <mcr7bk$rv$1...@dont-email.me>, "Wolfie" <bgbd...@gte.net>
wrote:

This is a proof of what?

--
Michael Press

[Some dued]

Wolfe got it, always go to the center between the white hats and black hats.

[Eric Ramon]

On Saturday, February 28, 2015 at 6:55:37 PM UTC-8, Some dued wrote:
> Wolfe got it, always go to the center between the white hats and black hats.

how does that do anything? The first one goes in, goes to a wall, the next one goes to the center....and? The third comes in and goes to the center. Eventually that center's going to be a bit crowded.

[xyzzy]

If you're the third one in the room and the two already in the room are wearing different colored hats, stand between them. If they are wearing the same colored hats, stand to one side of them. And so on as each person enters the room.

[Eric Ramon]

so if you see two white hats then you stand to the side. The next guy comes in. If you've got a white hat then he stands to the side of you so you know you've got a white hat. If he stands in between then you know you have a purple hat (or whatever the other color is). Thanks. That clears it up.

[Michael Press]

In article <40fcb405-fa8b-4fd7...@googlegroups.com>,

Some dued <theodo...@gmail.com> wrote:

> Wolfe got it, always go to the center between the white hats and black hats.

They are communicating, contrary to your problem statement.
Also, it does not work with 2 gnomes.

--
Michael Press

[Michael Press]

In article <d80ef159-872e-465d...@googlegroups.com>,

Consider these three situations.

1) Suppose there are exactly two gnomes.

2) Suppose there are more than two gnomes
and all wear black hats.

3) Suppose there are more than two gnomes
and all but one wear black hats
and the white hat enters the room last.

--
Michael Press

[Wolfie]

"Michael Press" wrote

> They are communicating, contrary to your problem statement.

No, they're not.

> Also, it does not work with 2 gnomes.

Of course it does.

[Wolfie]

"Michael Press" wrote

> Consider these three situations.

> 1) Suppose there are exactly two gnomes.

The two end up side-by-side. They're either: BB, BW, or WW.
And they're sorted.

> 2) Suppose there are more than two gnomes
> and all wear black hats.

Each adds themselves to an end of the line. So, for four:
B
BB
BBB
BBBB

And so on for any amount.

> 3) Suppose there are more than two gnomes
> and all but one wear black hats
> and the white hat enters the room last.

From

BBBBBBBBBB to (BBBBBBBBBBW or WBBBBBBBBBB)

Again: If everyone has the same color hat, add yourself
to one of the two ends. If there are two colors, add
yourself between the two colors (which means you're
either the new head of the black hats or the new head
of the white hats.)

[Michael Press]

In article <mcutt1$kiu$1...@dont-email.me>, "Wolfie" <bgbd...@gte.net>
wrote:

> "Michael Press" wrote
>
> > Consider these three situations.
>
> > 1) Suppose there are exactly two gnomes.
>
> The two end up side-by-side. They're either: BB, BW, or WW.
> And they're sorted.

Okay. I misunderstood the problem statement.

--
Michael Press

[Michael Press]

In article <mcusrh$gv5$1...@dont-email.me>, "Wolfie" <bgbd...@gte.net>
wrote:

> "Michael Press" wrote
>
> > They are communicating, contrary to your problem statement.
>
> No, they're not.

Yes, they are.

> > Also, it does not work with 2 gnomes.
>
> Of course it does.

I got that wrong.

--
Michael Press

[Some dued]

I typed the original problem statement on my phone, so I wasn't being overly verbose. However there is no need for communication between any two gnomes that I can see as long as they all know what they are supposed to do. Are you referring to the ability of a gnome's eyes to communicate what he sees to his brain?

[Michael Press]

In article <4e3eb7b8-b56e-4933...@googlegroups.com>,

I misread the problem. It is not on you. It is on me.
The bit about communication is not well posed but I
knew the intention.

When a gnome places himself between two gnomes
he tells the each of the two adjacent gnomes their hat color.

A better way to pose these puzzles is to speak about
a prior, agreed upon protocol, or a protocol that
all logicians will construct.

--
Michael Press

[Wolfie]

"Michael Press" wrote

> When a gnome places himself between two gnomes
> he tells the each of the two adjacent gnomes their hat color.

That's implied and not essential to the
solution.

It's not true with one, two or three gnomes.
And the last gnome doesn't know regardless
of number.

It's not true if the gnomes position themselves
facing the wall (i.e., each entering gnome can
see the gnomes already there, but the others
can not see the entering gnome or the others
against the wall.)