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'[OT] The Monty Hall problem'
2007\12\04@182303 by

This was cited as an example in a discussion about intuition

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

I still find it counter-intuitive, and if you haven't come across
it before, takes some figuring out

Probably there's an electronic use for this statistical cleverness

Only had a quick look at this, but the logic seems flawed to me.

Under Solution, option 1 is in fact two options:
1a. Player picks car. Host revealr Goat A. Switching loses.
1b. Player picks car. Host reveals Goat B. Switching loses.
2. Player picks Goat A. Host must reveal Goat B. Switching wins.
3. Player picks Goat B. Host must reveal Goat A. Switching wins.

Thus the chances of winning by switching are 50/50.

Anyone agree?

John

Jinx <joecolquittclear.net.nz> wrote:
This was cited as an example in a discussion about intuition

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

I still find it counter-intuitive, and if you haven't come across
it before, takes some figuring out

Probably there's an electronic use for this statistical cleverness

> http://en.wikipedia.org/wiki/Monty_Hall_problem
>
>   Thus the chances of winning by switching are 50/50.
>
>   Anyone agree?

To be honest, I still don't fully understand. I've never seen the
show so perhaps there's something I'm unaware of (the article
says that the past can't be ignored and talks about host behaviour
and I haven't grasped the relevance of that). If you have a choice
to make between two doors, then the odds must be 50:50

> Only had a quick look at this, but the logic seems flawed to me.
>
>   Under Solution, option 1 is in fact two options:
>   1a. Player picks car. Host revealr Goat A. Switching loses.
>   1b. Player picks car. Host reveals Goat B. Switching loses.
>   2. Player picks Goat A. Host must reveal Goat B. Switching wins.
>   3. Player picks Goat B. Host must reveal Goat A. Switching wins.
>
>   Thus the chances of winning by switching are 50/50.
>
>   Anyone agree?

Not me. The total chance in 1a + 1b together is 1/3, 2 is 1/3, and 3 is
the remaining 1/3. So the chance of losing by switching is 1/3, the
chance of winning is 2/3.

Another way to see it (as also explained in the wiki): you pick a door.
now if, after that, you were (always!) offered to pick the other two
doors together, instead of your one door, what would you do? That is the
essentially the same option you get when the host shows one goat and
allows you to pick the other door.

Wouter van Ooijen

-- -------------------------------------------
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> If you have a choice to make between two doors, then the odds
> must be 50:50

As a catch-all statement that is (of course) nonsense. If you have a
choice between two doors, but know there is no car, your odds are 0:0.
If there are two cars, the odds are 1:1. If I am honest and I tell you
where the car is your odds are 1:0 (or 0:1). So you realy must do some
thinking, applying gut-feeling won't work. Of course it won't, this case
gets so much attention *because* gut-feeling 'logic' does not work here.

Wouter van Ooijen

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To my mind, provided the host acts the same whether or not the correct door
was initially selected,  then there are only 2 possible outcomes. The odds
are then 50-50.

But I seem to remember from a stats course I did that the maths can proove
otherwise.

RP

On 05/12/2007, wouter van ooijen <woutervoti.nl> wrote:
{Quote hidden}

> -
You have a 1:3 chance of picking the car and a 2:3 chance of picking a
goat on the first pick.

If you pick a car on the first try and switch you loose because you can
only switch to a goat.

If you pick a goat on the first try and switch you always win.  One
thing that is left out when you try to make math logic out of it is that
the game show host knows where the car is.  If you pick a goat, then he
opens up the other door for the goat.  That leaves you only the car to
switch to.

> {Original Message removed}
> To my mind, provided the host acts the same whether or not
> the correct door was initially selected,  then there are only
> 2 possible outcomes. The odds are then 50-50.

Only when those two outcomes are equally likely. When I throw a ball in
the general direction of a basket, there are two possible outcomes: a
hit or a no-hit. Do you realy think those odds are 50-50?

This example is of course for US. Europeans: I am a member of the Dutch
national football team, and I take a penalty kick. Down under: I kick a
sheep in the direction of a - ^%\$#*&(*, no goals here, and the sheep has
run away. Never mind.

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
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>Only had a quick look at this, but the logic seems flawed to me.
>
>  Under Solution, option 1 is in fact two options:
...
>  Thus the chances of winning by switching are 50/50.
>
>  Anyone agree?

I agree.
On Wed, Dec 05, 2007 at 02:12:01AM -0500, J FLETCHER wrote:
> Only had a quick look at this, but the logic seems flawed to me.

It's not.

{Quote hidden}

I don't. Reason being that in either case a goat is revealed.

You initial pick has a 1/3 chance of winning, the other two doors have a
2/3 chance of winning.

The host helps you by eliminating one of the other two doors, which has to
be a goat (either one). It's the door elimination that shifts the odds
because you initially had 3 doors not 2.

BAJ
{Quote hidden}

> -
On Wed, Dec 05, 2007 at 02:55:40AM -0500, Jinx wrote:
> > en.wikipedia.org/wiki/Monty_Hall_problem
> >
> >   Thus the chances of winning by switching are 50/50.
> >
> >   Anyone agree?
>
> To be honest, I still don't fully understand. I've never seen the
> show so perhaps there's something I'm unaware of (the article
> says that the past can't be ignored and talks about host behaviour
> and I haven't grasped the relevance of that).

The relevance is that the host will never ever reveal the car, only a goat.

BTW I used to watch the show as a kid.

> If you have a choice
> to make between two doors, then the odds must be 50:50

But it's not. It's a choice between 3 doors and the host is proving to you
that one of the doors is not the winner.

Let's put it another way. You have three doors and one is a winner. I give
you the option of picking 1 door, or the option of picking 2 doors, which
do you pick?

The answer is pretty obvious that you have a better chance to win if you
pick 2 doors, because it's twice as likely that 1 of them is a winner.

Now the Monte Hall problem is the same situation except that choice occurs
midway through the game.

1. Pick one of three doors. That's the same as picking 1 door in the
scenario above.

2. One of the two doors the hosts keeps is revealed. It must be a loser.

3. Now you get your option to pick 1 or 2 doors again. Staying pat with the
initial door means you only pick 1 door. However, switching means picking 2
doors with a known loser. But it's still 2 doors!

The logic is right. Switching has a better chance of winning.

BAJ
On Wed, Dec 05, 2007 at 04:59:42AM -0500, Richard Prosser wrote:
> To my mind, provided the host acts the same whether or not the correct door
> was initially selected,  then there are only 2 possible outcomes.

The host does not act the same. That's where the probability shift occurs.

The host knows where the car is. So he will never reveal the car. That
means that his behavior changes depending on if the car is selected
initially or not.

1. If you pick the car, then he shows you 1 of the 2 goats.
2. However, if you pick a goat, he cannot show you the car! So he must
eliminate the other goat. And that's all the difference.

>The odds  are then 50-50.

It's 67/33 for switching. See my 1 door, 2 door picking posts elsewhere for
a clearer explanation.

BAJ
>The host does not act the same. That's where the probability shift occurs.
>
>The host knows where the car is. So he will never reveal the car. That
>means that his behavior changes depending on if the car is selected
>initially or not.

Is that really correct? Surely the host always picks a known non-prize door,
and gives the contestant a second choice, whether or not they picked the
correct one?

Alan B. Pearce wrote:

>>The host knows where the car is. So he will never reveal the car. That
>>means that his behavior changes depending on if the car is selected
>>initially or not.
>
> Is that really correct? Surely the host always picks a known non-prize door,
> and gives the contestant a second choice, whether or not they picked the
> correct one?

This were the rules, if I understood correctly.

--
Ciao, Dario

> -----Original Message-----
> From: piclist-bouncesmit.edu [piclist-bouncesmit.edu] On
Behalf
> Of Alan B. Pearce
> Sent: 05 December 2007 13:23
> To: Microcontroller discussion list - Public.
> Subject: Re: [OT] The Monty Hall problem
>
> >The host does not act the same. That's where the probability shift
> occurs.
> >
> >The host knows where the car is. So he will never reveal the car.
That
> >means that his behavior changes depending on if the car is selected
> >initially or not.
>
> Is that really correct? Surely the host always picks a known non-prize
> door,
> and gives the contestant a second choice, whether or not they picked
the
> correct one?

That's my understanding as well, the hosts behaviour is always the same.
After the contestant selects a door, the host will always open one of
the other doors with a goat behind it.  The behaviour is the same
whether or not the contestant picked the car on his first choice.

Regards

Mike

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>> Is that really correct? Surely the host always picks a known non-prize
>> door,
>> and gives the contestant a second choice, whether or not they picked
>> the correct one?
>
>That's my understanding as well, the hosts behaviour is always the same.
>After the contestant selects a door, the host will always open one of
>the other doors with a goat behind it.  The behaviour is the same
>whether or not the contestant picked the car on his first choice.

So then the first choice is irrelevant, and it comes down to 50:50 chance
that you pick the right one.

2007\12\05@094453 by
Alan B. Pearce wrote:
>>> Is that really correct? Surely the host always picks a known non-prize
>>> door,
>>> and gives the contestant a second choice, whether or not they picked
>>> the correct one?
>>>
>> That's my understanding as well, the hosts behaviour is always the same.
>> After the contestant selects a door, the host will always open one of
>> the other doors with a goat behind it.  The behaviour is the same
>> whether or not the contestant picked the car on his first choice.
>>
>
> So then the first choice is irrelevant, and it comes down to 50:50 chance
> that you pick the right one.
>
>

Seems like a type of a frame-of-reference question to me, thus both
answers (1:1, 2:3) are correct.

From the perspective of the original three closed doors: 2:3
From the perspective of the remaining two closed doors: 1:1

But I'm probably wrong.
Daniel

> -----Original Message-----
> From: piclist-bouncesmit.edu [piclist-bouncesmit.edu] On
Behalf
> Of Alan B. Pearce
> Sent: 05 December 2007 14:18
> To: Microcontroller discussion list - Public.
> Subject: Re: [OT] The Monty Hall problem
>
> >> Is that really correct? Surely the host always picks a known
non-prize
> >> door,
> >> and gives the contestant a second choice, whether or not they
picked
> >> the correct one?
> >
> >That's my understanding as well, the hosts behaviour is always the
same.
> >After the contestant selects a door, the host will always open one of
> >the other doors with a goat behind it.  The behaviour is the same
> >whether or not the contestant picked the car on his first choice.
>
> So then the first choice is irrelevant, and it comes down to 50:50
chance
> that you pick the right one.

That was my gut feeling, and it seems totally logical.  Does it make a
difference if the host opened a door before or after the contestant
makes his/her first choice?

However, I found a simulator that does appear to show that switching
significantly improves you chances (run say 1000 rounds and note that
switching gives you a 2/3 chance of winning, whilst not switching gives
you a 2/3 chance of losing).  I hate probabilities.

http://www.userpages.de/monty_hall_problem/

Regards

Mike

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Put another way:

If you switch doors you win if you initially picked one of the two goats. 2/3
chance of winning the car.

If you don't switch doors you win if you pick the car. 1/3 chance of winning
the car.

To me this feels more like a logical thing than a statistical thing and the use
for it may be in some sorting algorithm perhaps.

/Ruben

{Quote hidden}

==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
rubenpp.sbbs.se
==============================
> >> Is that really correct? Surely the host always picks a known non-prize
> >> door,
> >> and gives the contestant a second choice, whether or not they picked
> >> the correct one?
> >
> >That's my understanding as well, the hosts behaviour is always the same.
> >After the contestant selects a door, the host will always open one of
> >the other doors with a goat behind it.  The behaviour is the same
> >whether or not the contestant picked the car on his first choice.
>
> So then the first choice is irrelevant, and it comes down to 50:50 chance
> that you pick the right one.
>

No, you always win if you initially pick a goat and then switch doors (since
the host has to pick the other goat). 2/3 chance of winning.

/Ruben
==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
rubenpp.sbbs.se
==============================
It is your first choice along with "always switch" or "never switch"
after being shown a goat that you have to consider.

If you always switch, then you'll always win if you pick a goat first.
That gives you 2:3 to win.

Oh, I just remembered a little movie about it.

http://www.5min.com/Video/The-Monty-Hall-Game-Show-Problem-932173

> {Original Message removed}

> >That's my understanding as well, the hosts behaviour is always the
> >same. After the contestant selects a door, the host will always open
> >one of the other doors with a goat behind it.  The behaviour is the
> >same whether or not the contestant picked the car on his
> first choice.
>
> So then the first choice is irrelevant, and it comes down to
> 50:50 chance that you pick the right one.

No, because the two alternatives are not equal. The door you first
selected was a 1:3 door. The door you can switch to is a 2:3 door (it
must be, because your original door was a 1:3).

Wouter van Ooijen

-- -------------------------------------------
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consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu

> That was my gut feeling, and it seems totally logical.  Does
> it make a difference if the host opened a door before or
> after the contestant makes his/her first choice?

of course it does. if the host opens before the first choice and reveals
a goat the game is reduced to two doors, one goat and one car.

> I hate probabilities.

I always loved probability stuff. The only math you don't have to learn
anything for, just think :)

Wouter van Ooijen

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> Seems like a type of a frame-of-reference question to me, thus both
> answers (1:1, 2:3) are correct.

statistics and probability don't care about frames, but they do care

>  From the perspective of the original three closed doors: 2:3

yes

>  From the perspective of the remaining two closed doors: 1:1

no

> But I'm probably wrong.

yes

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
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On Wed, Dec 05, 2007 at 08:23:15AM -0500, Alan B. Pearce wrote:
> >The host does not act the same. That's where the probability shift occurs.
> >
> >The host knows where the car is. So he will never reveal the car. That
> >means that his behavior changes depending on if the car is selected
> >initially or not.
>
> Is that really correct?

Yes.

> Surely the host always picks a known non-prize door,

Yes.

> and gives the contestant a second choice, whether or not they picked the
> correct one?

Bingo! But since the host is prohibited from revealing the prize, the
advantage goes to the contestant.

BAJ
On Wed, Dec 05, 2007 at 09:18:02AM -0500, Alan B. Pearce wrote:
> >> Is that really correct? Surely the host always picks a known non-prize
> >> door,
> >> and gives the contestant a second choice, whether or not they picked
> >> the correct one?
> >
> >That's my understanding as well, the hosts behaviour is always the same.
> >After the contestant selects a door, the host will always open one of
> >the other doors with a goat behind it.  The behaviour is the same
> >whether or not the contestant picked the car on his first choice.
>
> So then the first choice is irrelevant, and it comes down to 50:50 chance
> that you pick the right one.

No. Let me repeat my assertion: would you have a better chance of winning
if you got to pick 2 doors instead of 1? Of course you would.

The host showing the goat regardless gives you the opportunity to pick 2
doors instead of one. But to do so you HAVE TO SWITCH! If you stand pat,
then you've only chosen 1 door throughout the process. Switching gives you
an opportunity to choose 2 doors.

It's frustrating because it's counterintuitive. But the advantage occurs
because the host must always show a goat. If he could show the car, then
you'd be right.

BAJ
> The host showing the goat regardless gives you the opportunity to pick 2
> doors instead of one. But to do so you HAVE TO SWITCH! If you stand pat,
> then you've only chosen 1 door throughout the process. Switching gives you
> an opportunity to choose 2 doors.

Maybe a technical nitpick here, but you don't have to switch, you have
to reselect randomly.
wouter van ooijen <woutervoti.nl> wrote:
> > >That's my understanding as well, the hosts behaviour is always the
> > >same. After the contestant selects a door, the host will always open
> > >one of the other doors with a goat behind it.  The behaviour is the
> > >same whether or not the contestant picked the car on his
> > first choice.
> >
> > So then the first choice is irrelevant, and it comes down to
> > 50:50 chance that you pick the right one.
>
> No, because the two alternatives are not equal. The door you first
> selected was a 1:3 door. The door you can switch to is a 2:3 door (it
> must be, because your original door was a 1:3).

I don't know why people have so much trouble with this.

You know up front that the host is going to reveal one of the "goat"
doors, taking it out of play. Therefore, it's really a two-door game
from the outset, and you have a 50% chance of picking the correct door.

The only "mystery" is that you don't know /which/ two doors will end up
being in play at the time you make your first choice, only that the door
you pick will be one of them. Not a big deal.

There's /no/ advantage in switching.

-- Dave Tweed
I think people get it wrong and think there is no advantage to switching
because they think it is all about probabilities, and it's not.  I once
told it to a friend who started writing down mathematical equations to
figure it out.  Her equations didn't reflect the fact that the "host"
alters the game for you or against you, depending on if you switch or
not.

> {Original Message removed}

> -----Original Message-----
> From: piclist-bouncesmit.edu [piclist-bouncesmit.edu] On Behalf
> Of Ruben Jönsson
> Sent: 05 December 2007 14:51
> To: Microcontroller discussion list - Public.
> Subject: Re: [OT] The Monty Hall problem
>
> Put another way:
>
> If you switch doors you win if you initially picked one of the two goats.
> 2/3
> chance of winning the car.
>
> If you don't switch doors you win if you pick the car. 1/3 chance of
> winning
> the car.

Thanks, that's by far the best description I have seen. It clicked in my head as soon as I read it!

Mike

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So...if the contestant doesn't switch doors after the host reveals Goat A,
then the host exposes Goat B behind the door the contestant didn't switch
to, the contestant still has a 1-in-3 chance of there being a car behind the
remaining door?  Monty's studio audiences must've felt pretty silly with all
their premature applause (at least two times out of three).

But why argue about it?  Mathematicians can prove that 1=0.  We're techies!
Simulate, observe, record, _then_ argue about one another's methods. ;)

RR

----- Original Message -----
From: "Michael Rigby-Jones" <Michael.Rigby-Jonesbookham.com>
To: "Microcontroller discussion list - Public." <piclistmit.edu>
Sent: Wednesday, December 05, 2007 5:51 AM
Subject: RE: [OT] The Monty Hall problem

>
>
>> {Original Message removed}
> > The host showing the goat regardless gives you the
> opportunity to pick
> > 2 doors instead of one. But to do so you HAVE TO SWITCH! If
> you stand
> > pat, then you've only chosen 1 door throughout the process.
> Switching
> > gives you an opportunity to choose 2 doors.
>
>
> Maybe a technical nitpick here, but you don't have to switch,
> you have to reselect randomly.

would not be nitpicking if it were true. (but it is not).

Wouter van Ooijen

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> I think people get it wrong and think there is no advantage
> to switching because they think it is all about
> probabilities, and it's not.

That's a word play, but IMO it *is* about probabilities. But as any
branch of maths they must be used correctly.

Wouter van Ooijen

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> You know up front that the host is going to reveal one of the
> "goat" doors, taking it out of play. Therefore, it's really a
> two-door game from the outset, and you have a 50% chance of
> picking the correct door.
> There's /no/ advantage in switching.
> -- Dave Tweed

You are the one behind the quiz in CCI? I hope you are kidding!

Alternatively: let's play this game a few times (I mean a very large
few). You'll be the quiz host, I'll be the guest. For every car I choose
you pay me 45, for every goat I choose I pay you 55. If you are right, I
will go broke. If I am right, I will be rich. This was they way such
issues were solved in the early days. The false ideas were weeded out by
financial darwinism-avant-la-lettre :)

Wouter van Ooijen

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> So...if the contestant doesn't switch doors after the host
> reveals Goat A,
> then the host exposes Goat B behind the door the contestant
> didn't switch
> to, the contestant still has a 1-in-3 chance of there being a
> car behind the
> remaining door?

No, did anyone say so?

> Mathematicians can prove that 1=0.

We all know that 1 is not an lvalue, so 1=0 is not a valid expression.

> We're techies!

for some value of 'we'

Wouter van Ooijen

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Michael Rigby-Jones wrote:
{Quote hidden}

Right, since everyone now apparently understands why you 'must' switch,
and I thought I understood too. I decided to write a program to test the
probabilities. As Rob Robson said: "We're techies! Simulate, observe,
record, _then_ argue about one another's methods." The program I wrote
tests each possible outcome....

Well, the results are surprising (to me... because I thought I
understoold), and well, here they are:

First, the legend....
\$ -> user's chosen door that has a car behind it
! -> user's chosen door with a goat behind it.
o -> the open door.
X -> switched door with a car.
. -> switched door with nothing.

Then the results...

\$o.  ->   Switcher score : 0  Sticker score : 1
\$.o  ->   Switcher score : 0  Sticker score : 2
!Xo  ->   Switcher score : 1  Sticker score : 2
!oX  ->   Switcher score : 2  Sticker score : 2
X!o  ->   Switcher score : 3  Sticker score : 2
o\$.  ->   Switcher score : 3  Sticker score : 3
.\$o  ->   Switcher score : 3  Sticker score : 4
o!X  ->   Switcher score : 4  Sticker score : 4
Xo!  ->   Switcher score : 5  Sticker score : 4
oX!  ->   Switcher score : 6  Sticker score : 4
o.\$  ->   Switcher score : 6  Sticker score : 5
.o\$  ->   Switcher score : 6  Sticker score : 6

But, then, I thought, well, there appear to be only 12 possible
outcomes, but do they have equal probibility of happening?

Rolf

and here's the program.

==================================
#!/usr/bin/perl -w

use strict;

sub printscen (\$\$\$) {
my \$chosen = shift;
my \$open = shift;
my \$car = shift;
my \$msg = "";
my \$pos;
foreach \$pos ( 1 .. 3 ) {
if (\$open == \$pos) {
\$msg .= "o";
} elsif (\$chosen == \$pos) {
\$msg .= \$car == \$pos ? "\\$" : "X";
} else {
\$msg .= \$car == \$pos ? "!" : ".";
}
}
print "\$msg  -> ";
}

my \$switchscore = 0;
my \$stayscore = 0;

my \$car;
foreach \$car (1 .. 3) {
my \$choose;
foreach \$choose (1 .. 3) {
my \$open;
foreach \$open (1 .. 3) {
# Cant't open door if it is the one the contestant chooses.
next if \$open == \$choose;
# Can't open door if car is behind it...
next if \$open == \$car;
if (\$car == \$choose) {
\$stayscore++;
} else {
\$switchscore++;
}
printscen (\$choose, \$open, \$car);
print "  Switcher score : \$switchscore  Sticker score : \$stayscore\n";
}
}
}

Wouter *van Nistelrooy ?
*

On 05/12/2007, wouter van ooijen <woutervoti.nl> wrote:
>
>
> This example is of course for US. Europeans: I am a member of the Dutch
> national football team, and I take a penalty kick.
>
On 06/12/2007, Ruben Jönsson <rubenpp.sbbs.se> wrote:

{Quote hidden}

Very good point - has clarified the whole mess for me.
> Thanks,
> Richard
>
>
>
>
> >
> > Maybe a technical nitpick here, but you don't have to switch,
> > you have to reselect randomly.
>
> would not be nitpicking if it were true. (but it is not).

Hmm.. I saw this exact problem described somewhere, and they were very
much adimant about that point, that if you didn't reselect, your odds
of picking the car were still one in three, but if you RANDOMLY
re-selected after the first goat was shown, your odds changed to 50/50
(which is self-evident).  There was a connection made to Bayesian
math, but it's been a few years and I don't remember where it was.
It works the other way round, too.

Let's say you are playing a shell game.  There are three shells and one
pea under one shell.  You mix them and let someone pick a shell.  He has
1:3 chance of picking the shell with the pea under it and winning.

But you're a good guy and so when he picks a shell you lift another
shell with no pea.  That leaves two shells on the table, the shell he
picked and the other shell.  (There is no switching here.)

So you say to the guy, "See, you have 50:50 chance of winning.  It's
this one or that one.  Place your bet."

Does he really have a 50:50 chance of winning as you promised?

(If so, why does he lose 66% of the time? (-:  )

> {Original Message removed}
It (now) looks to me like :-
If you never change - Odds are 1 in 3
If you always change - Odds are 2 in 3
If you randomly change - odds are half way between 1/3 & 2/3 = 1/2.

RP

On 06/12/2007, David VanHorn <microbrixgmail.com> wrote:
{Quote hidden}

> -
Terse and mathy..

www.speech.sri.com/people/anand/771/html/node6.html
Hi Jinx:

I'm with you, but for me it's even simpler.

At the time of the first pick the problem is "there are three doors and one of
them is a winner and the other two are losers, so the odds of a winner are 1 in 3.

At the time of the second pick the problem is "there are two doors and one of
them is a winner and the other one is a looser, so the odds of a winner are 1 in 2.

There is no memory when it comes to gambling, i.e. what's happened in the past
has no impact on the current odds.  For example suppose someone who was not in
the room for the first pick is brought into the game after one of the doors has
been opened.  They only see the second problem i.e. they are to choose one of
two doors and only one door has the prize.  The newcomer's odds are 1/2 the
same as the original contestant at that point.

--
Have Fun,

Brooke Clarke
http://www.PRC68.com
http://www.precisionclock.com
http://www.prc68.com/I/WebCam2.shtml 24/7 Sky-Weather-Astronomy Cam
> > This example is of course for US. Europeans: I am a member of the
> > Dutch national football team, and I take a penalty kick.

> Wouter *van Nistelrooy ?

(this was already way OT...)

(if you don't recognise the order of the lines: I hate top posting...)

We have a nice ironic song here (variation on Dylans "life after
death"), the punchline is "don't worry, there is life after death". One
of the couplets is "what (bad) could happen to Seedorf when he kicked
from 11 meters"

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

http://www.xs4all.nl/~werksman/liedjes/leven_na_de_dood.html

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu

> At the time of the first pick the problem is "there are three
> doors and one of
> them is a winner and the other two are losers, so the odds of
> a winner are 1 in 3.

correct, but only because the winners are distributed evenly among those
three doors.

> At the time of the second pick the problem is "there are two
> doors and one of
> them is a winner and the other one is a looser, so the odds
> of a winner are 1 in 2.

not correct, because the winners are not distributed evenly. How can
they? Door one was a 1:3, it can't suddenly change to a 1:2 beacuse the
all-konwing host has shown you which of the other two is a loser.
Showing that conveys no new information: you knew (at least) one of the
others was a loser. Now if the host has opened both other doors, that
would have conveyed new info, and hence changed the odds.

> There is no memory when it comes to gambling, i.e. what's
> happened in the past
> has no impact on the current odds.

For a fresh experiment, true. But this is not.

> For example suppose
> someone who was not in
> the room for the first pick is brought into the game after
> one of the doors has
> been opened.  They only see the second problem i.e. they are
> to choose one of
> two doors and only one door has the prize.  The newcomer's
> odds are 1/2 the
> same as the original contestant at that point.

clever, but still wrong. the newcomer who does not know which door was
first chosen indeed has a 1:2 chance, as your intuition says, and
exactly like the original guest when he chooses at random between his
original door and the alternative (someone has already shown the
calculation for this case). but a newcomer who knows the story and can
see which door was choosen by the guest is in the same position as the
guest, and should choose the other door, for a 2:3 chance.

> Have Fun,

I do, it's like debugging students code. I am always surprised by the
clever bugs they make for me :)

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu

Interesting way of evaluating it. The problem is that you have assigned two
possibilities for each time they pick they car on their first go, when in
reality there is only one - the only difference is that Monty has a choice
of doors to pick in that situation (which doubles the possibilities in you
scenario, but not in reality).

The correct matrix is as follows:
\$ -> user's chosen door that has a car behind it
! -> user's chosen door with a goat behind it.
* -> non-chosen door with a car
. -> non-chosen door with a goat

\$..   ->   Switcher score : 0  Sticker score : 1
!*.    ->   Switcher score : 1  Sticker score : 1
!.*    ->   Switcher score : 2  Sticker score : 1
*!.    ->   Switcher score : 3  Sticker score : 1
.\$.   ->   Switcher score : 3  Sticker score : 2
.!*    ->   Switcher score : 4  Sticker score : 2
*.!    ->   Switcher score : 5  Sticker score : 2
.*!    ->   Switcher score : 6  Sticker score : 2
..\$   ->   Switcher score : 6  Sticker score : 3

You can essentially ignore what Monty does, as if you pick a goat he has
only one choice, if you pick the car it doesn't matter what he does. He
doesn't add to the possibilities.

On Dec 5, 2007 7:19 PM, Rolf <learrrogers.com> wrote:

> Michael Rigby-Jones wrote:
> >
> >> {Original Message removed}
wouter van ooijen <woutervoti.nl> wrote:
> I wrote:
> > You know up front that the host is going to reveal one of the
> > "goat" doors, taking it out of play. Therefore, it's really a
> > two-door game from the outset, and you have a 50% chance of
> > picking the correct door.
> > There's /no/ advantage in switching.
> > -- Dave Tweed
>
> You are the one behind the quiz in CCI?

Yes. *

> I hope you are kidding!

Yes, sorry, brain fart. Too many different things going on today. :-(

Richard Prosser <rhprossergmail.com> wrote:
> It (now) looks to me like :-
> If you never change - Odds are 1 in 3
> If you always change - Odds are 2 in 3
> If you randomly change - odds are half way between 1/3 & 2/3 = 1/2.

I managed to confuse myself with the last case. When I began to think
about how you would set up a Monte Carlo simulation of the game, I
realized my mistake.

See, now I know why so many people have trouble with this!

If you pick randomly whether or not to swich, the odds associated
with choices prior to that don't matter any more.

If you always (or never) switch, then only the odds associated with the
first pick matter.

Another way to think about it is that if you pick a goat to begin with,
the host's hands are tied -- he has no choice, he *must* show you the
other goat, which means the car is behind the remaining door. This
happens in two cases out of three.

-- Dave Tweed

* Actually, I'm the editor of the "Engineering Quotient" column, but
I'm also the primary contributor, because I get very little in the
way of outside contributions of material. After a hiatus of about
2.5 years, the column is coming back on a bimonthly basis starting
with the January 2008 issue. If anyone wants to send me suitable
puzzles or engineering problems, you will get credit in print, and
you'll have my eternal gratitude, as well.
>Maybe a technical nitpick here, but you don't have
>to switch, you have to reselect randomly.

That is my thought too, i.e. you have to make a selection of 1 of 2 doors.
You may select the same door as previously, or you may select the other one.

It would be interesting to see if the coin flip to switch or not increases
or decreases your rate of winning in a simulation.

--
James.

{Original Message removed}
On Dec 6, 2007 1:00 PM, James Newton <jamesnewtonmassmind.org> wrote:
> It would be interesting to see if the coin flip to switch or not increases
> or decreases your rate of winning in a simulation.

In the discussion of this problem that I saw, it was crucial to going
from 1/3 to 1/2.

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