Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
Sorry, but if you don't get it by now, you simply aren't going to.spam@spam.com (Don Pearce) wrote in news:4dd0c3d7.381246241
@news.eternal-september.org:
That isn't the final situation. I will take this a step at a time.
There are three doors - one with a car, two with goats
I choose one. I have a 1 in 3 chance of being right
That means there is a 2 in 3 chance of the car being in the other two
I know for a fact that at least one of the other two is a goat.
That does not change the odds - it is still 2 in 3 that the car is in
one of those
The host shows me one of the two - one he knows to contain a goat.
This is not new information, I knew there was a goat there, I still
know there was a goat there.
The odds are still 2 in 3 that the car is in one of those two doors.
Stop there.
No, I didn't know there was a goat THERE. I knew there was a goat
behind at least one of the door besides the one I chose, but I didn't
know which one. Now a variable is removed from the equation.
Revealing a goat behind a door doesn't change the odds?
Of course it does.
Otherwise, revealing the car behind a door also wouldn't change the
odds.
Once the host has revealed a goat, then there's an even chance that the
car is behind one of the two remaining doors--and I have no information
either way (unless you're counting the psychological factors) that the
door I chose is or is not the correct one.
Not trying to be argumentative, but I still don't see the logic.
But now those 2 in 3 odds have been concentrated into the one
remaining door of the two, which I will open because that is better
than the 1 in 3 chance of it being my first choice.
d
Still trying to get my head around this.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.
The host acts as a leak of information. It might help to imagine anspamtrap1888 <spamtrap1888@gmail.com> wrote in
news:c5220357-4ddd-43ac-935c-fc50796b2c9d@z15g2000prn.googlegroups.com:
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 imaginespamtrap1888 <spamtrap1888@gmail.com> wrote in
news:c5220357-4ddd-43ac-935c-fc50796b2c9d@z15g2000prn.googlegroups.com:
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.
I think, several years ago when I originally saw this, I"spamtrap1888"<spamtrap1888@gmail.com> wrote in message
news:dd7c10fd-38c7-43e2-9db6-5c4e25f4a946@35g2000prp.googlegroups.com
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.
The visual explanations on the web make it rather easy to see, no punNot trying to be argumentative, but I still don't see the logic.
If the host does not know, he might quite as easily reveal the car."philicorda" wrote in message
news:EtEAp.608$Ky.286@newsfe24.ams2...
On Tue, 17 May 2011 20:05:34 +0000, Carey Carlan wrote:
spamtrap1888 <spamtrap1888@gmail.com> wrote in
news:c5220357-4ddd-43ac-935c-fc50796b2c9d@z15g2000prn.googlegroups.com:
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
In my studies of this item, I found a statement that about 10% ofOn 05/14/2011 06:48 AM, Arny Krueger wrote:
"spamtrap1888"<spamtrap1888@gmail.com> wrote in message
news:dd7c10fd-38c7-43e2-9db6-5c4e25f4a946@35g2000prp.googlegroups.com
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!
If the host does not know, he might quite as easily reveal theThe 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
Let's look at the case of the ignorant host."Don Pearce" wrote in message
news:4dd3548d.36447387@news.eternal-september.org...
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.
Then go back to the original where the host knows where the car is andLet'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.
Void is not one of the permitted outcomes. Suppose the host"Don Pearce" wrote in message
news:4dd3548d.36447387@news.eternal-september.org...
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
Start at the beginning of this post and read all of the quoted"Don Pearce" wrote in message
news:4dd3548d.36447387@news.eternal-september.org...
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
***
OK, try this: blow the game up to 100 doors. Your chance of pickingspamtrap1888<spamtrap1888@gmail.com> wrote in news:88df2861-f695-449b-
a558-e3d119b1cb3b@34g2000pru.googlegroups.com:
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:spamtrap1888 <spamtrap1...@gmail.com> wrote in news:88df2861-f695-449b-
a558-e3d119b1c...@34g2000pru.googlegroups.com:
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.
That may have penetrated.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: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.
I claqim there are two games. In the first game, you go to the studio, pickspamtrap1888 <spamtrap1888@gmail.com> wrote in
news:88df2861-f695-449b-
a558-e3d119b1cb3b@34g2000pru.googlegroups.com:
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.
It's not just 50/50. It's 67/33 in your favor.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.