On Tue, 17 May 2011 20:05:34 +0000, Carey Carlan wrote:

spamtrap1888 wrote in
:

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still 1/2.

The big difference: In the Monty Hall problem there is only one "coin
flip". Only one random choice is made -- the first choice of a door. In
the coin flip situation, there are five coin flips, five random
choices.

Now, in contrast, if the car and remaining goats were randomly shuffled
after each goat door was revealed, then the situation would be
different. But in the MHP problem the car does not move.

Still trying to get my head around this.

How would shuffling unknown values affect my choice? If I didn't know
before and you shuffle the choices, it's still a random choice on my
part.

The host acts as a leak of information. It might help to imagine an
alternate game, where the host does not know the contents of the doors,
and the game is void if the host reveals the car. This version puts you
back to 50/50 when the host reveals a goat, whether you switch doors or
not.

"philicorda" wrote in message
...

On Tue, 17 May 2011 20:05:34 +0000, Carey Carlan wrote:

spamtrap1888 wrote in
:

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still
1/2.

The big difference: In the Monty Hall problem there is only
one "coin
flip". Only one random choice is made -- the first choice of a
door. In
the coin flip situation, there are five coin flips, five
random
choices.

Now, in contrast, if the car and remaining goats were randomly
shuffled
after each goat door was revealed, then the situation would be
different. But in the MHP problem the car does not move.

Still trying to get my head around this.

How would shuffling unknown values affect my choice? If I
didn't know
before and you shuffle the choices, it's still a random choice
on my
part.

The host acts as a leak of information. It might help to imagine
an
alternate game, where the host does not know the contents of the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or knew
it was there makes absolutely no difference. You should still
switch doors.

David

On 05/14/2011 06:48 AM, Arny Krueger wrote:
wrote in message


Declaring that there is no connection between the two
situations is the source of the poster's error. Monty
Hall knew if the player was correct or not, and so the
player's choice of the door in the first round
influenced the selection of the goat door. The graphic
helps you understand that there are still three scenarios
once a goat door has been revealed.

I get it now.

I think, several years ago when I originally saw this, I
argued as vehemently as you, Arny. It is extremely counter-intuitive,
which goes to show intuition isn't always right!
--
Randy Yates % "Watching all the days go by...
Digital Signal Labs % Who are you and who am I?"
% 'Mission (A World Record)',
http://www.digitalsignallabs.com % *A New World Record*, ELO

In article ,
Carey Carlan wrote:

Not trying to be argumentative, but I still don't see the logic.

The visual explanations on the web make it rather easy to see, no pun
intended. I posted a link earlier, so did others.

On Tue, 17 May 2011 21:29:45 -0500, "David"
wrote:

"philicorda" wrote in message
...

On Tue, 17 May 2011 20:05:34 +0000, Carey Carlan wrote:

spamtrap1888 wrote in
:

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still
1/2.

The big difference: In the Monty Hall problem there is only
one "coin
flip". Only one random choice is made -- the first choice of a
door. In
the coin flip situation, there are five coin flips, five
random
choices.

Now, in contrast, if the car and remaining goats were randomly
shuffled
after each goat door was revealed, then the situation would be
different. But in the MHP problem the car does not move.

Still trying to get my head around this.

How would shuffling unknown values affect my choice? If I
didn't know
before and you shuffle the choices, it's still a random choice
on my
part.

The host acts as a leak of information. It might help to imagine
an
alternate game, where the host does not know the contents of the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or knew
it was there makes absolutely no difference. You should still
switch doors.

David


If the host does not know, he might quite as easily reveal the car.
You then can't win it. Do you guarantee yourself 2/3 odds by switching
then? No. If the host reveals a goat by chance, the odds do indeed
drop to 50/50.

d

"Randy Yates" wrote in message
m
On 05/14/2011 06:48 AM, Arny Krueger wrote:
wrote in message


Declaring that there is no connection between the two
situations is the source of the poster's error. Monty
Hall knew if the player was correct or not, and so the
player's choice of the door in the first round
influenced the selection of the goat door. The graphic
helps you understand that there are still three
scenarios once a goat door has been revealed.

I get it now.

I think, several years ago when I originally saw this, I
argued as vehemently as you, Arny. It is extremely
counter-intuitive, which goes to show intuition isn't
always right!

In my studies of this item, I found a statement that about 10% of
*everybody* never gets it, regardless of their intelligence or education.

That suggests to me that some people learn things about problem solving that
keep them from seeing certain solutions. The trick is to not do that, or if
you do, somehow redirect how you approach these things.

"Don Pearce" wrote in message
...
The host acts as a leak of information. It might help to imagine
an
alternate game, where the host does not know the contents of the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or
knew
it was there makes absolutely no difference. You should still
switch doors.

David


If the host does not know, he might quite as easily reveal the
car.
You then can't win it. Do you guarantee yourself 2/3 odds by
switching
then? No. If the host reveals a goat by chance, the odds do
indeed
drop to 50/50.

d
***
Sorry, I disagree. Yes the host could reveal a car if he is
unaware of the situation. If this happens, the game was defined
as void. If the host instead reveals a goat, there is no
difference whether he guessed or knew the goat was there.

David

On May 18, 6:44*am, "David" wrote:
"Don Pearce" *wrote in message

...



The host acts as a leak of information. It might help to imagine
an
alternate game, where the host does not know the contents of the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or
knew
it was there makes absolutely no difference. You should still
switch doors.

David

If the host does not know, he might quite as easily reveal the
car.
You then can't win it. Do you guarantee yourself 2/3 odds by
switching
then? No. If the host reveals a goat by chance, the odds do
indeed
drop to 50/50.

d
***
Sorry, I disagree. *Yes the host could reveal a car if he is
unaware of the situation. If this happens, the game was defined
as void. If the host instead reveals a goat, there is no
difference whether he guessed or knew the goat was there.


Let's look at the case of the ignorant host.

There are three possibilities at the start of the game. The
probability of each is 1/3

_1 2 3_
aCGG
bGCG
cGGC

Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
...............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
...............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
...............Host picks Door 3 Result Car. Contestant loses.

Of the six possible scenarios, the contestant loses four times. If the
contestant does not switch after the ignorant host opens a door, the
contestant loses four times. If we discard the times the host opens a
door with a car behind it, the contestant wins two out of four times
when he switches, and two out of four times when he doesn't switch.
Therefore, switching picks has no effect on the odds when the host
randomly opens one of the other doors.

spamtrap1888 wrote in news:88df2861-f695-449b-
:

Let's look at the case of the ignorant host.

There are three possibilities at the start of the game. The
probability of each is 1/3

_1 2 3_
aCGG
bGCG
cGGC

Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
..............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
..............Host picks Door 3 Result Car. Contestant loses.

Of the six possible scenarios, the contestant loses four times. If the
contestant does not switch after the ignorant host opens a door, the
contestant loses four times. If we discard the times the host opens a
door with a car behind it, the contestant wins two out of four times
when he switches, and two out of four times when he doesn't switch.
Therefore, switching picks has no effect on the odds when the host
randomly opens one of the other doors.

Then go back to the original where the host knows where the car is and
the contestant switches.

Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3,
loses.
...............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2,
wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins

Or the contestant doesn't switch.

Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1, wins.
...............Host picks Door 3 Result Goat. Contestant keeps Door 1,
wins.
Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses
Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses

After the Host opens the door the odds are even. Makes no difference if
the contestant changes doors or not. This is the same as there only
being two doors.

The original claim was that the odds remained 1 in 3 even after the Host
opened the door. I still don't see it.

On May 19, 7:54*am, Carey Carlan wrote:
spamtrap1888 wrote in news:88df2861-f695-449b-
:



Let's look at the case of the ignorant host.

There are three possibilities at the start of the game. The
probability of each is 1/3

_1 2 3_
aCGG
bGCG
cGGC

Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 *Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
..............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
..............Host picks Door 3 *Result Car. Contestant loses.

Of the six possible scenarios, the contestant loses four times. If the
contestant does not switch after the ignorant host opens a door, the
contestant loses four times. If we discard the times the host opens a
door with a car behind it, the contestant wins two out of four times
when he switches, and two out of four times when he doesn't switch.
Therefore, switching picks has no effect on the odds when the host
randomly opens one of the other doors.

Then go back to the original where the host knows where the car is and
the contestant switches.

Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3,
loses.
..............Host picks Door 3 *Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2,
wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins

Or the contestant doesn't switch.

Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1, wins.
..............Host picks Door 3 *Result Goat. Contestant keeps Door 1,
wins.
Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses
Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses

After the Host opens the door the odds are even. *Makes no difference if
the contestant changes doors or not. *This is the same as there only
being two doors.

The original claim was that the odds remained 1 in 3 even after the Host
opened the door. *I still don't see it.

Without switching, the contestant has a 1/3 chance of winning:
Case a: Contestant picked door with car. Host can open either door,
his choice, to reveal goat.
Case b: Contestant picked door with goat. Host must open Door 3 to
reveal goat.
Case c: Contestant picked door with goat. Host must open Door 2 to
reveal goat.

With switching, the contestant now has a 2/3 chance of winning:
Case a: Host can open either door, his choice. Contestant switches to
unopened door, loses.
Case b: Host opens Door 3, contestant switches to Door 2, wins.
Case c: Host opens Door 2, contestant switches to Door 3, wins.

On Wed, 18 May 2011 08:44:57 -0500, "David"
wrote:

"Don Pearce" wrote in message
...
The host acts as a leak of information. It might help to imagine
an
alternate game, where the host does not know the contents of the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or
knew
it was there makes absolutely no difference. You should still
switch doors.

David


If the host does not know, he might quite as easily reveal the
car.
You then can't win it. Do you guarantee yourself 2/3 odds by
switching
then? No. If the host reveals a goat by chance, the odds do
indeed
drop to 50/50.

d
***
Sorry, I disagree. Yes the host could reveal a car if he is
unaware of the situation. If this happens, the game was defined
as void. If the host instead reveals a goat, there is no
difference whether he guessed or knew the goat was there.

David



Void is not one of the permitted outcomes. Suppose the host
accidentally revealed the car - to be equivalent to the intentional
goat revelation, he would then have to say "never mind, take the car
anyway". That would leave you in the 1/3 2/3 situation. If he reveals
a goat by chance the game degenerates to the simple situation - the
host has chosen one of the three, and you get to pick between the
remaining two, always assuming that he did not pick the car.

The point of the intentional revelation is that by switching you get -
in effect - both doors, not just the one.

d

"Don Pearce" wrote in message
...

On Wed, 18 May 2011 08:44:57 -0500, "David"

wrote:

"Don Pearce" wrote in message
...
The host acts as a leak of information. It might help to
imagine
an
alternate game, where the host does not know the contents of
the
doors,
and the game is void if the host reveals the car. This version
puts you
back to 50/50 when the host reveals a goat, whether you switch
doors or
not.
***
Not true. When the host reveals a goat whether he guessed or
knew
it was there makes absolutely no difference. You should still
switch doors.

David


If the host does not know, he might quite as easily reveal the
car.
You then can't win it. Do you guarantee yourself 2/3 odds by
switching
then? No. If the host reveals a goat by chance, the odds do
indeed
drop to 50/50.

d
***
Sorry, I disagree. Yes the host could reveal a car if he is
unaware of the situation. If this happens, the game was defined
as void. If the host instead reveals a goat, there is no
difference whether he guessed or knew the goat was there.

David



Void is not one of the permitted outcomes. Suppose the host
accidentally revealed the car - to be equivalent to the
intentional
goat revelation, he would then have to say "never mind, take the
car
anyway". That would leave you in the 1/3 2/3 situation. If he
reveals
a goat by chance the game degenerates to the simple situation -
the
host has chosen one of the three, and you get to pick between
the
remaining two, always assuming that he did not pick the car.

The point of the intentional revelation is that by switching you
get -
in effect - both doors, not just the one.

d
***
Start at the beginning of this post and read all of the quoted
stuff. The initial assumption is that 'void' IS a permitted
outcome. If the void assumption is changed , I concede.
David

I don't know why people make the Monty Hall Paradox so complex. I've
explained it simply, twice.

All you have to do is understand why the initial probability of getting the
big prize is 1/3 -- and everything else falls out in a completely
straightforward manner.

On 5/19/2011 10:54 AM, Carey Carlan wrote:
wrote in news:88df2861-f695-449b-
:

Let's look at the case of the ignorant host.

There are three possibilities at the start of the game. The
probability of each is 1/3

_1 2 3_
aCGG
bGCG
cGGC

Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
..............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
..............Host picks Door 3 Result Car. Contestant loses.

Of the six possible scenarios, the contestant loses four times. If the
contestant does not switch after the ignorant host opens a door, the
contestant loses four times. If we discard the times the host opens a
door with a car behind it, the contestant wins two out of four times
when he switches, and two out of four times when he doesn't switch.
Therefore, switching picks has no effect on the odds when the host
randomly opens one of the other doors.

Then go back to the original where the host knows where the car is and
the contestant switches.

Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3,
loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2,
wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins

Or the contestant doesn't switch.

Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1, wins.
..............Host picks Door 3 Result Goat. Contestant keeps Door 1,
wins.
Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses
Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses

After the Host opens the door the odds are even. Makes no difference if
the contestant changes doors or not. This is the same as there only
being two doors.

The original claim was that the odds remained 1 in 3 even after the Host
opened the door. I still don't see it.

OK, try this: blow the game up to 100 doors. Your chance of picking
the winning door on the first try is 1 out of 100. Stick with that
choice and each time a zonk is revealed the chance of the prize being
in the remaining group increases. When you get down to two doors the
chance of the prize being behind the other door is 99 out of 100.
The chance of your first pick being correct is still 1 out of 100.

Later...
Ron Capik
--

Ron Capik wrote in news:XqSdncQ0dq_JzkjQnZ2dnUVZ_q-
:

OK, try this: blow the game up to 100 doors. Your chance of picking
the winning door on the first try is 1 out of 100. Stick with that
choice and each time a zonk is revealed the chance of the prize being
in the remaining group increases. When you get down to two doors the
chance of the prize being behind the other door is 99 out of 100.
The chance of your first pick being correct is still 1 out of 100.

That may have penetrated.

It all hinges on the host knowing in advance which door hides the prize and
revealing known (to him) bad choices. I still don't agree with the logic,
but I think I follow it.

It's called the "Monty Hall problem"? I'll continue to research.

Thank you for your patience.

Ron Capik wrote in news:XqSdncQ0dq_JzkjQnZ2dnUVZ_q-
:

OK, try this: blow the game up to 100 doors. Your chance of picking
the winning door on the first try is 1 out of 100. Stick with that
choice and each time a zonk is revealed the chance of the prize being
in the remaining group increases. When you get down to two doors the
chance of the prize being behind the other door is 99 out of 100.
The chance of your first pick being correct is still 1 out of 100.

Here's the simple answer that I can understand:

"An even simpler solution is to reason that switching loses if and only if
the player initially picks the car, which happens with probability 1/3, so
switching must win with probability 2/3"

Eureka! That finally makes sense. And yes, everyone was trying to tell me
that.

Carey Carlan wrote:
spamtrap1888 wrote in
news:88df2861-f695-449b-
:

Let's look at the case of the ignorant host.

There are three possibilities at the start of the game. The
probability of each is 1/3

_1 2 3_
aCGG
bGCG
cGGC

Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
..............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door
3, wins
..............Host picks Door 3 Result Car. Contestant loses.

Of the six possible scenarios, the contestant loses four times. If
the contestant does not switch after the ignorant host opens a door,
the contestant loses four times. If we discard the times the host
opens a door with a car behind it, the contestant wins two out of
four times when he switches, and two out of four times when he
doesn't switch. Therefore, switching picks has no effect on the odds
when the host randomly opens one of the other doors.

Then go back to the original where the host knows where the car is and
the contestant switches.

Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2,
wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins

Or the contestant doesn't switch.

Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1,
wins. ..............Host picks Door 3 Result Goat. Contestant keeps
Door 1, wins.
Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses
Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses

After the Host opens the door the odds are even. Makes no difference
if the contestant changes doors or not. This is the same as there
only being two doors.

The original claim was that the odds remained 1 in 3 even after the
Host opened the door. I still don't see it.

I claqim there are two games. In the first game, you go to the studio, pick
a door, and then go home to wait and see if they call you and tell you that
you either won or lost. Your odds are only 1/3 of winning this game. But if
you play the second game, then you go to the studio and mess around until
the host opens up a door and shown you the donkey behind it. then you can
play the game with 50-50 odds of winning. The only thing I have trouble
explaining is why, in order to play this second game with the better odds,
you have to switch doors. But, in fact, you do have to switch in order to
switch games and take advantage of the better odds.

"Bill Graham" wrote in
:

I claqim there are two games. In the first game, you go to the studio,
pick a door, and then go home to wait and see if they call you and
tell you that you either won or lost. Your odds are only 1/3 of
winning this game. But if you play the second game, then you go to the
studio and mess around until the host opens up a door and shown you
the donkey behind it. then you can play the game with 50-50 odds of
winning. The only thing I have trouble explaining is why, in order to
play this second game with the better odds, you have to switch doors.
But, in fact, you do have to switch in order to switch games and take
advantage of the better odds.

It's not just 50/50. It's 67/33 in your favor.

Here's another super-simple explanation:

As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can
keep your one door or you can have the other two doors."

In article ,
Bill Graham wrote:

I claqim there are two games. In the first game, you go to the studio, pick
a door, and then go home to wait and see if they call you and tell you that
you either won or lost. Your odds are only 1/3 of winning this game. But if
you play the second game, then you go to the studio and mess around until
the host opens up a door and shown you the donkey behind it. then you can
play the game with 50-50 odds of winning. The only thing I have trouble
explaining is why, in order to play this second game with the better odds,
you have to switch doors. But, in fact, you do have to switch in order to
switch games and take advantage of the better odds.

Yup.

The distinction between the two games is based on the amount of
information available to you at the moment you make your final decision.

In the first game, the only information you have is that every door
available to you to choose has a 1/3 chance of being correct.

In the second game, additional information is given to you by the
host, after you make your initial decision and before you make your
second one. "The prize is *not* behind that door over there."

Actually, you don't *have* to switch doors to "play the second game".
You're playing it from the moment the host gives you this extra
information. It's just that if you ignore this extra information, and
*think* you're still playing the first game (as many people do), you
are more likely than not to make a poor choice in the decision you
make in this second game.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

Dave Platt wrote:
In article ,
Bill Graham wrote:

I claqim there are two games. In the first game, you go to the
studio, pick a door, and then go home to wait and see if they call
you and tell you that you either won or lost. Your odds are only 1/3
of winning this game. But if you play the second game, then you go
to the studio and mess around until the host opens up a door and
shown you the donkey behind it. then you can play the game with
50-50 odds of winning. The only thing I have trouble explaining is
why, in order to play this second game with the better odds, you
have to switch doors. But, in fact, you do have to switch in order
to switch games and take advantage of the better odds.

Yup.

The distinction between the two games is based on the amount of
information available to you at the moment you make your final
decision.

In the first game, the only information you have is that every door
available to you to choose has a 1/3 chance of being correct.

In the second game, additional information is given to you by the
host, after you make your initial decision and before you make your
second one. "The prize is *not* behind that door over there."

Actually, you don't *have* to switch doors to "play the second game".
You're playing it from the moment the host gives you this extra
information. It's just that if you ignore this extra information, and
*think* you're still playing the first game (as many people do), you
are more likely than not to make a poor choice in the decision you
make in this second game.

Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and shows
everyone that there is a donkey behind that door. Now, will it make any
difference if the other two switch their initial picks or not? And, if they
do swap doors, with they both enjoy a 2/3 chance of winning?

In article ,
"Bill Graham" wrote:

Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and shows
everyone that there is a donkey behind that door. Now, will it make any
difference if the other two switch their initial picks or not? And, if they
do swap doors, with they both enjoy a 2/3 chance of winning?

You can't just go changing the rules of the game willy nilly. The game
is played with goats, not donkeys. Sheesh.

Smitty Two wrote:

In article ,
"Bill Graham" wrote:


Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and shows
everyone that there is a donkey behind that door. Now, will it make any
difference if the other two switch their initial picks or not? And, if they
do swap doors, with they both enjoy a 2/3 chance of winning?


You can't just go changing the rules of the game willy nilly. The game
is played with goats, not donkeys. Sheesh.
No sheep ?

Jamie

"Bill Graham" wrote in message
...
Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and
shows everyone that there is a donkey behind that door. Now, will it make
any difference if the other two switch their initial picks or not? And, if
they do swap doors, with they both enjoy a 2/3 chance of winning?

It get's really silly now when both switch doors! :-) However neither gets
100% chance of winning and one still loses. They both can't win 66% of the
time however. The problem is that the host can now only open one door if he
is not to pick one already selected, and since his door now has a 1/3 chance
of winning (rather than none as in the original game where he must always
select one that has a goat) NO new information is provided if his is
actually a goat/donkey. Therfore both contestentants now have a 50% chance
of winning whether they switch or not.
This in fact seems far more logical and thus obvious IMO.

Trevor.

On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham"
wrote:

Dave Platt wrote:
In article ,
Bill Graham wrote:

I claqim there are two games. In the first game, you go to the
studio, pick a door, and then go home to wait and see if they call
you and tell you that you either won or lost. Your odds are only 1/3
of winning this game. But if you play the second game, then you go
to the studio and mess around until the host opens up a door and
shown you the donkey behind it. then you can play the game with
50-50 odds of winning. The only thing I have trouble explaining is
why, in order to play this second game with the better odds, you
have to switch doors. But, in fact, you do have to switch in order
to switch games and take advantage of the better odds.

Yup.

The distinction between the two games is based on the amount of
information available to you at the moment you make your final
decision.

In the first game, the only information you have is that every door
available to you to choose has a 1/3 chance of being correct.

In the second game, additional information is given to you by the
host, after you make your initial decision and before you make your
second one. "The prize is *not* behind that door over there."

Actually, you don't *have* to switch doors to "play the second game".
You're playing it from the moment the host gives you this extra
information. It's just that if you ignore this extra information, and
*think* you're still playing the first game (as many people do), you
are more likely than not to make a poor choice in the decision you
make in this second game.

Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and shows
everyone that there is a donkey behind that door. Now, will it make any
difference if the other two switch their initial picks or not? And, if they
do swap doors, with they both enjoy a 2/3 chance of winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

d

Don Pearce wrote:
On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham"
wrote:

Dave Platt wrote:
In article ,
Bill Graham wrote:

I claqim there are two games. In the first game, you go to the
studio, pick a door, and then go home to wait and see if they call
you and tell you that you either won or lost. Your odds are only 1/3
of winning this game. But if you play the second game, then you go
to the studio and mess around until the host opens up a door and
shown you the donkey behind it. then you can play the game with
50-50 odds of winning. The only thing I have trouble explaining is
why, in order to play this second game with the better odds, you
have to switch doors. But, in fact, you do have to switch in order
to switch games and take advantage of the better odds.
Yup.

The distinction between the two games is based on the amount of
information available to you at the moment you make your final
decision.

In the first game, the only information you have is that every door
available to you to choose has a 1/3 chance of being correct.

In the second game, additional information is given to you by the
host, after you make your initial decision and before you make your
second one. "The prize is *not* behind that door over there."

Actually, you don't *have* to switch doors to "play the second game".
You're playing it from the moment the host gives you this extra
information. It's just that if you ignore this extra information, and
*think* you're still playing the first game (as many people do), you
are more likely than not to make a poor choice in the decision you
make in this second game.
Suppose for the moment that there are two contestants. One picks door two,
and the other picks door one. Then the moderator opens door three and shows
everyone that there is a donkey behind that door. Now, will it make any
difference if the other two switch their initial picks or not? And, if they
do swap doors, with they both enjoy a 2/3 chance of winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

--
Tciao for Now!

John.

"John Williamson" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will it
make any difference if the other two switch their initial picks or not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.

Trevor.

On Sat, 21 May 2011 19:55:55 +1000, "Trevor" wrote:


"John Williamson" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will it
make any difference if the other two switch their initial picks or not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.

Trevor.

I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.

d

On 5/20/2011 8:49 PM, Bill Graham wrote:

Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will it
make any difference if the other two switch their initial picks or not?
And, if they do swap doors, with they both enjoy a 2/3 chance of winning?

This is called the Monty Hall problem. And yes, switching does change
the probabilities of winning, and this can be proved with a simple
computer program. I use this in both my math classes when we cover
probabilities, and in my programming classes.

--
I'm never going to grow up.

In article , PeterD
wrote:

On 5/20/2011 8:49 PM, Bill Graham wrote:

Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will it
make any difference if the other two switch their initial picks or not?
And, if they do swap doors, with they both enjoy a 2/3 chance of winning?

This is called the Monty Hall problem. And yes, switching does change
the probabilities of winning, and this can be proved with a simple
computer program. I use this in both my math classes when we cover
probabilities, and in my programming classes.

Did you just wake up from a long winter's nap?

Smitty Two wrote:
In article , PeterD
wrote:

On 5/20/2011 8:49 PM, Bill Graham wrote:

Suppose for the moment that there are two contestants. One picks
door two, and the other picks door one. Then the moderator opens
door three and shows everyone that there is a donkey behind that
door. Now, will it make any difference if the other two switch
their initial picks or not? And, if they do swap doors, with they
both enjoy a 2/3 chance of winning?

This is called the Monty Hall problem. And yes, switching does change
the probabilities of winning, and this can be proved with a simple
computer program. I use this in both my math classes when we cover
probabilities, and in my programming classes.

Did you just wake up from a long winter's nap?

No, but I must have been napping when they mentioned Monty Hall. I never
heard of him or his math problems, and I have a BS in Math. (Santa Clara,
1974)

"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

trevor.

"Trevor" wrote in message
u...
"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door
three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat, the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Trevor.

On Tue, 24 May 2011 17:00:57 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

trevor.




The host doesn't pick the remaining door, he opens it. He reveals a
goat/donkey whatever. That means the two contestants have door each,
one of which has a car behind it. They now each have 50/50 odds. There
is nothing in this scenario that puts one contestant's odds higher
than the other since they both picked a door each at the start.

d

On Tue, 24 May 2011 17:06:22 +1000, "Trevor" wrote:


"Trevor" wrote in message
. au...
"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door
three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat, the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Trevor.


Of course - and that is precisely the new scenario presented. As I
said - re-read. And no, it is nothing like the original MH problem.

d

"Don Pearce" wrote in message
...
The host doesn't pick the remaining door, he opens it. He reveals a
goat/donkey whatever. That means the two contestants have door each,
one of which has a car behind it.

ONLY if it is not behind the one the host already opened. Since he no longer
has a choice of doors, he must have a 1/3 chance of showing the car.

They now each have 50/50 odds. There
is nothing in this scenario that puts one contestant's odds higher
than the other since they both picked a door each at the start.

Right, where did I say otherwise, IF the host has not already shown the car?
The whole point is that the game is now no longer like the Monty Hall
scenario in any way.

Trevor.

"Don Pearce" wrote in message
...
NOPE, IF the host picks the remaining door he now has a 1/3 chance of
the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat,
the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Of course - and that is precisely the new scenario presented. As I
said - re-read. And no, it is nothing like the original MH problem.


Why do I need to re-read, that is exactly what I said!!!!!!!!!!!!!!!!!
What the hell are you objecting to??????????????????

Trevor.