another puzzler
 

On Fri, 13 May 2011 12:19:45 -0700, "Bill Graham"
wrote:

Arny Krueger wrote:
"Bill Graham" wrote in message

Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a chance
to guess again, there is a 50% chance the car is behind door #1 and a
50% chance the car if behind door #2. Change your choice or not, you
have a 50% chance of being right.

But when you first entered the arena, you only had a 1/3 chance of winning.
How does that chance change halfway through the game, and why would it
matter whether you changed doors or not?
Exactly!!! In effect it is a new game. You can choose the same door,
or you can choose the other door. The car is behind one of them.
50-50.

PlainBill

another puzzler
 

"Arny Krueger" wrote in message
...
"William Sommerwerck" wrote in
message

After you know there is a goat behind door #3 and are
given a chance to guess again, there is a 50% chance the
car is behind door #1 and a 50% chance the car if behind
door #2. Change your choice or not, you have a 50%
chance of being right.

This is not correct. I explained it in a previous post.

You seriously think I didn't read your alleged explanation?
You've been known to be wrong before... ;-)

And so have you.


Because you will have initially selected the wrong door
2/3 of the time (right?) it follows that 2/3 of the time
the good prize will be behind one of the two other doors.
The host will /always/ select a door with a goat,
therefore, you should switch, because there's a 2/3
chance the other door will have the good prize.

That is sheerist ********.

It is, in fact, the correct explanation. It is simple and easily understood
(which is something of an acheivement for me).

You are ignoring the fact that the host KNOWS what is behind each door. His
choice of which door to open is not random.

Everybody has "blind spots". We carry "mental baggage" with us that keeps us
from accepting certain things that are demonstrably true. I've slowly
discarded mine over the years on occasions when I was shown the error of my
thinking.

No one is trying to pull your ******** over your eyes. Think it through
carefully, and pretty soon you'll /understand/.

another puzzler
 

But by not switching doors, you are ignoring the new information that the
prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching
doors, you are including the new information that the prize has to be
behind
one of the other two doors, and your new chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

another puzzler
 

Well said. Now if the host only offered the opportunity to chose a
different door if you had chosen the car, changing would be a bad
idea. As it is, the odds are now 1 of 2, rather than 1 of 3.

No, the new probability is 2/3.

another puzzler
 

"William Sommerwerck" wrote in message
...

But by not switching doors, you are ignoring the new
information that the
prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right.
By
switching
doors, you are including the new information that the prize has
to be
behind
one of the other two doors, and your new chance of winning is
50%.

No, it doesn't. Your new chance of winning is 2/3.

***
This is similar to another puzzle. A couple has two children.
What is the probability that the second is a boy? The couple then
volunteers that they are not both girls. Now what is the
probability the second is a boy?

The first case is 1/2. The second case is 2/3.

David

another puzzler
 

On Fri, 13 May 2011 13:11:54 -0700, wrote:

On Fri, 13 May 2011 12:19:45 -0700, "Bill Graham"
wrote:

Arny Krueger wrote:
"Bill Graham" wrote in message

Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a chance
to guess again, there is a 50% chance the car is behind door #1 and a
50% chance the car if behind door #2. Change your choice or not, you
have a 50% chance of being right.

But when you first entered the arena, you only had a 1/3 chance of winning.
How does that chance change halfway through the game, and why would it
matter whether you changed doors or not?
Exactly!!! In effect it is a new game. You can choose the same door,
or you can choose the other door. The car is behind one of them.
50-50.

No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

d

another puzzler
 

In article ,
Don Pearce wrote:

or you can choose the other door. The car is behind one of them.
50-50.

No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap.

Thank you, Don! Describing the problem in that way is without a doubt
the clearest explanation of the "paradox" I have ever read.


--
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!

another puzzler
 

On May 13, 2:27*pm, (Dave Platt) wrote:
In article ,

Don Pearce wrote:
or you can choose the other door. *The car is behind one of them.
50-50.

No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap.

Thank you, Don! *Describing the problem in that way is without a doubt
the clearest explanation of the "paradox" I have ever read.


This explanation (by subtraction) from the wikipedia article struck
me:

"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 (Carlton
2005)."

The player picks a door and has a 1/3 chance of being right. This
chance does not change when a losing door is revealed, so the only
remaining choice gives you a 2/3 chance.

another puzzler
 

"Arny Krueger" wrote in message
...
"William Sommerwerck" wrote in
message

After you know there is a goat behind door #3 and are
given a chance to guess again, there is a 50% chance the
car is behind door #1 and a 50% chance the car if behind
door #2. Change your choice or not, you have a 50%
chance of being right.

This is not correct. I explained it in a previous post.

You seriously think I didn't read your alleged explanation?

You've been known to be wrong before... ;-)

Like this...

Because you will have initially selected the wrong door
2/3 of the time (right?) it follows that 2/3 of the time
the good prize will be behind one of the two other doors.
The host will /always/ select a door with a goat,
therefore, you should switch, because there's a 2/3
chance the other door will have the good prize.

That is sheerist ********.

Your first mistake is assuming that there is a connection between your 2
guesses. In fact you have been given two different and disconnected games
to play.

Other than the fact that the car and 1 goat are carries-over from the
first game, there is no connection. If they brought in another car and
another goat, then the odds during the second game would be the same.

When you play the second game your odds of winning have improved to 1/2.
You have 1 chances out of 2, no more, no less to win when there are 2
opportunities.

Pick whichever door you will, unless you can smell the goat! ;-)

It would appear to me that the real purpose of this thread is to test the
gullibility of people.


No, YOU are simply arguing straight statistical chance, whereas TV game
shows, are always manipulated for dramatic effect. Another good example is
the quiz master who usually accepts a correct answer immediately, but often
gives a chance to change that answer if wrong. Obviously if given a chance
to switch your answer you should do so, since it is more likely your answer
would already have been accepted if correct. Whether its 66% of the time is
totally unproven, but anyone who watches these game shows knows it is NOT a
50:50 chance whenever a TV host, producer, and TV network are involved!

Trevor.

another puzzler
 

"Bill Graham" wrote in message
...
Arny Krueger wrote:
"William Sommerwerck" wrote in
message

After you know there is a goat behind door #3 and are
given a chance to guess again, there is a 50% chance the
car is behind door #1 and a 50% chance the car if behind
door #2. Change your choice or not, you have a 50%
chance of being right.

This is not correct. I explained it in a previous post.

You seriously think I didn't read your alleged explanation?

You've been known to be wrong before... ;-)

Like this...

Because you will have initially selected the wrong door
2/3 of the time (right?) it follows that 2/3 of the time
the good prize will be behind one of the two other doors.
The host will /always/ select a door with a goat,
therefore, you should switch, because there's a 2/3
chance the other door will have the good prize.

That is sheerist ********.

Your first mistake is assuming that there is a connection between
your 2 guesses. In fact you have been given two different and
disconnected games to play.

Other than the fact that the car and 1 goat are carries-over from the
first game, there is no connection. If they brought in another car
and another goat, then the odds during the second game would be the
same.
When you play the second game your odds of winning have improved to
1/2. You have 1 chances out of 2, no more, no less to win when there
are 2 opportunities.

Pick whichever door you will, unless you can smell the goat! ;-)

It would appear to me that the real purpose of this thread is to test
the gullibility of people.

But by not switching doors, you are ignoring the new information that the
prize has to be behind one of the other two doors.... You are sticking
with your original guess that had only a 1/3 chance of being right. By
switching doors, you are including the new information that the prize has
to be behind one of the other two doors, and your new chance of winning is
50%

IOW, lets suppose that you picked door #1 and then left the game, went
home, and waited by the phone to find out whether you won or not. There is
only a 1/3 chance of your getting the lucky call.

But by staying on board, and switching your guess to door #2, you are
taking advantage of the "new game" that has a 50% chance of
success........


Which is complete ******** because that has already been done for you once
the first door is proven NOT to be the main prize. Whether you switch or
not, statistically you now have a 50:50 chance. The ONLY reason to switch is
because the game host is more often than not giving you a chance to get it
right. IF nobody actually had any idea where the prize was, there would be
no advantage in switching at all, but then the first door they opened would
be the main prize 33% of the time, and as any game viewer knows, that
*never* happens.

Trevor.

another puzzler
 

"David" wrote in message
...
"William Sommerwerck" wrote in message ...

But by not switching doors, you are ignoring the new information that the
prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching
doors, you are including the new information that the prize has to be
behind
one of the other two doors, and your new chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

***
This is similar to another puzzle. A couple has two children. What is the
probability that the second is a boy? The couple then volunteers that they
are not both girls. Now what is the probability the second is a boy?

The first case is 1/2. The second case is 2/3.


Wrong, on a purely statistical basis the first case is 50:50, BB, BG, GB, or
GG. Two out of four meet the criteria.
The second case is 50:50 Boy or Girl, One out of two meets the criteria.

However IF you know the average family statistics for your Country/town, you
can change those odds because you have more data. *If* the number of two
children families with 2 boys Vs 2 girls is known, one simply substitutes
the known data. It will probably be still close to 50:50 however in most
areas AFAIK.

Trevor.

another puzzler
 

wrote in message
...
As it is, the odds are NOW 1 of 2, rather than 1 of 3.

Right, whether you switch or not! *IF* the host didn't actually know where
the car was and always offered the choice to switch. But then the car would
be revealed on the first door 33% of the time, which hardly ever happens, if
ever!

Trevor.

another puzzler
 

"William Sommerwerck" wrote in message
...
You are ignoring the fact that the host KNOWS what is behind each door.
His
choice of which door to open is not random.

Bingo! But still makes the 2/3 claim pure conjecture. Somewhere between 1/2
and 2/3 yes. They ARE known to also use reverse logic sometimes after all!

Trevor.

another puzzler
 

"John Robertson" wrote in message
...
This is a variation of the three cups/shells hiding something shuffle
carney game...

Rubbish, everyone knows the pea is in the carney's hand NOT under ANY of the
three shells!
The TV games are rarely THAT rigged, just rigged a bit for dramatic effect.

Trevor.

another puzzler
 

"Don Pearce" wrote in message
...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not, ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

another puzzler
 

"Dave Platt" wrote in message
...
In article ,
Don Pearce wrote:

or you can choose the other door. The car is behind one of them.
50-50.

No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap.

Thank you, Don! Describing the problem in that way is without a doubt
the clearest explanation of the "paradox" I have ever read.

And WHY exactly would you choose the already revealed incorrect door for the
second chance??? (unless you are a complete moron)
There are only two remaining possible correct door choices whether you
switch or not!

Trevor.

another puzzler
 

"spamtrap1888" wrote in message
...
The player picks a door and has a 1/3 chance of being right. This
chance does not change when a losing door is revealed, so the only
remaining choice gives you a 2/3 chance.

Which totally ignores the fact that the only reason the first door is opened
is because the host already knows it is incorrect. This is NOT a purely
statistical game of chance, the host can manipulate the odds either way, and
regularly do.

Trevor.

another puzzler
 

On Sat, 14 May 2011 17:39:22 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not, ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics. And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

d

another puzzler
 

I find it interesting that almost everyone who "agrees" with me is quite
wrong.

They are interpreting the problem and its explanation in terms of what they
would like the situation to be, rather than looking at it from a strictly
mathematical basis.

another puzzler
 

"Don Pearce" wrote in message
...
On Sat, 14 May 2011 17:39:22 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not,
ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics. And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.


Here's the simplest-possible correct explanation...

2/3 of the time, your initial pick is wrong. The host will then show you the
"goat" prize (the other being the good prize). Ergo, switching will get you
the good prize 2/3 of the time. 1/3 of the time you'll lose the good prize.
This is obviously better than sticking with the initial choice (which is
right only 1/3 of the time).

How much simpler does it need to be, to be comprehensible?

another puzzler
 

"spamtrap1888" 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.

another puzzler
 

On Sat, 14 May 2011 03:46:26 -0700, "William Sommerwerck"
wrote:

"Don Pearce" wrote in message
...
On Sat, 14 May 2011 17:39:22 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not,
ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics. And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.


Here's the simplest-possible correct explanation...

2/3 of the time, your initial pick is wrong. The host will then show you the
"goat" prize (the other being the good prize). Ergo, switching will get you
the good prize 2/3 of the time. 1/3 of the time you'll lose the good prize.
This is obviously better than sticking with the initial choice (which is
right only 1/3 of the time).

How much simpler does it need to be, to be comprehensible?


Some will never get it, no matter how it is explained.

d

another puzzler
 

"Trevor" wrote in message
u
"spamtrap1888" wrote in message
...
The player picks a door and has a 1/3 chance of being
right. This chance does not change when a losing door is
revealed, so the only remaining choice gives you a 2/3
chance.

Which totally ignores the fact that the only reason the
first door is opened is because the host already knows it
is incorrect. This is NOT a purely statistical game of
chance, the host can manipulate the odds either way, and
regularly do.

There are two iron rules that dictate which door the host opens.

(1) There can't be a car behind it

(2) It can't be the door the contestant picked.

If the contestant picks a door with no car, then there is only one other
door the host can choose. The host has no choice and can't affect the
outcome.

If the contestant chooses a door with a car, then the host can choose either
of two two doors that have no car, but which one he chooses doesn't seem to
affect the outcome. His effect on the odds comes from the fact that he
revealed one of the two doors with no car behind it.

How can the host manipulate the odds?

another puzzler
 

"Arny Krueger" wrote in message
...
"Trevor" wrote in message
u
"spamtrap1888" wrote in message

...
The player picks a door and has a 1/3 chance of being
right. This chance does not change when a losing door is
revealed, so the only remaining choice gives you a 2/3
chance.

Which totally ignores the fact that the only reason the
first door is opened is because the host already knows it
is incorrect. This is NOT a purely statistical game of
chance, the host can manipulate the odds either way, and
regularly do.

There are two iron rules that dictate which door the host opens.

(1) There can't be a car behind it

(2) It can't be the door the contestant picked.

If the contestant picks a door with no car, then there is only one other
door the host can choose. The host has no choice and can't affect the
outcome.

If the contestant chooses a door with a car, then the host can choose
either
of two two doors that have no car, but which one he chooses doesn't seem
to
affect the outcome. His effect on the odds comes from the fact that he
revealed one of the two doors with no car behind it.

How can the host manipulate the odds?

Correct. The host has no effect on the odds.

Part of the confusion occurs because people confuse permutations and
combinations. In the situation where the contestant has chosen the good
prize, the two bad prizes form a combination, not a permutation.

another puzzler
 

"Trevor" wrote in message
...


"David" wrote in message
...
"William Sommerwerck" wrote in message
...

But by not switching doors, you are ignoring the new
information that the
prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right.
By
switching
doors, you are including the new information that the prize
has to be
behind
one of the other two doors, and your new chance of winning is
50%.

No, it doesn't. Your new chance of winning is 2/3.

***
This is similar to another puzzle. A couple has two children.
What is the probability that the second is a boy? The couple
then volunteers that they are not both girls. Now what is the
probability the second is a boy?

The first case is 1/2. The second case is 2/3.


Wrong, on a purely statistical basis the first case is 50:50, BB,
BG, GB, or
GG. Two out of four meet the criteria.
The second case is 50:50 Boy or Girl, One out of two meets the
criteria.

snip
Trevor.

The second case is: BB, BG, GB. The couple told you the GG case
does not exist.
Get it now? The goat problem has similar probability outcome
changes.

David

another puzzler
 

William Sommerwerck wrote:
But by not switching doors, you are ignoring the new information
that the prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching doors, you are including the new information that the
prize has to be behind one of the other two doors, and your new
chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

No. You now know that the prize is not behind door $3, so your chance of
winning in the, "second game" is 50-50. But you had to buy yourself this
chance at the second game. You did this by switching doors.

another puzzler
 

"Bill Graham" wrote in message
...
William Sommerwerck wrote:
But by not switching doors, you are ignoring the new information
that the prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching doors, you are including the new information that the
prize has to be behind one of the other two doors, and your new
chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

No. You now know that the prize is not behind door $3, so your chance of
winning in the, "second game" is 50-50. But you had to buy yourself this
chance at the second game. You did this by switching doors.

I know it's unkind to tell people who agree with you that they're wrong,
but... you're wrong. You really need to think this through carefully.

another puzzler
 

another puzzler
 

On Sun, 15 May 2011 12:14:44 GMT, Carey Carlan
wrote:

(Don Pearce) wrote in
:

On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan
wrote:

(Don Pearce) wrote in
:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
wrote:

"Bill Graham" wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganew s.com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a
chance to guess again, there is a 50% chance the car is behind door
#1 and a 50% chance the car if behind door #2. Change your choice
or not, you have a 50% chance of being right.


Lets make it ten doors. You pick one, and get a one in ten chance of
being right. That means that the chances are 90% that the car is
behind one of the 9 doors you did not pick. You know for certain
that at least eight of those doors conceal a goat, so when eight
goats are revealed, you have no new information. The chances are 90%
that the car is behind one of the nine - only now there is only one
remaining to open.

One vital fact here is that the person doing the revealing knows the
contents of the doors and chooses to reveal only goats. Had he been
guessing too, and just happened to reveal only goats, then yes, you
would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the situation.
How do the preceding 8 steps affect the final step?

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.

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

another puzzler
 

On May 15, 5:14*am, Carey Carlan wrote:
(Don Pearce) wrote :



On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan
wrote:

(Don Pearce) wrote in
:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
wrote:

"Bill Graham" wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganews .com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 *and are given a
chance to guess again, there is a 50% chance the car is behind door
#1 and a 50% chance the car if behind door #2. *Change your choice
or not, you have a 50% chance of being right.

Lets make it ten doors. You pick one, and get a one in ten chance of
being right. That means that the chances are 90% that the car is
behind one of the 9 doors you did not pick. You know for certain
that at least eight of those doors conceal a goat, so when eight
goats are revealed, you have no new information. The chances are 90%
that the car is behind one of the nine - only now there is only one
remaining to open.

One vital fact here is that the person doing the revealing knows the
contents of the doors and chooses to reveal only goats. Had he been
guessing too, and just happened to reveal only goats, then yes, you
would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the situation.
How do the preceding 8 steps affect the final step?

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.

another puzzler
 

spamtrap1888 wrote:
On May 15, 5:14 am, Carey Carlan wrote:
(Don Pearce) wrote
:



On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan
wrote:

(Don Pearce) wrote in
:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
wrote:

"Bill Graham" wrote in message

Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous
problem. I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a
chance to guess again, there is a 50% chance the car is behind
door #1 and a 50% chance the car if behind door #2. Change your
choice or not, you have a 50% chance of being right.

Lets make it ten doors. You pick one, and get a one in ten chance
of being right. That means that the chances are 90% that the car
is behind one of the 9 doors you did not pick. You know for
certain that at least eight of those doors conceal a goat, so
when eight goats are revealed, you have no new information. The
chances are 90% that the car is behind one of the nine - only now
there is only one remaining to open.

One vital fact here is that the person doing the revealing knows
the contents of the doors and chooses to reveal only goats. Had
he been guessing too, and just happened to reveal only goats,
then yes, you would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the
situation.
How do the preceding 8 steps affect the final step?

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 really interesting thing is that, even if the car does not move,
conditional probability theory says the odds have changed, and you should
switch doors.

another puzzler
 

"Don Pearce" wrote in message
...
This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics.

Right!

And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

And understanding that games of pure chance have NO memory for any previous
actions. It's just like the old question, what are the odds of tossing a
coin 10 heads in a row? If you toss 9 heads in a row, what are the odds of
tossing a 10th?
(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

Trevor.

another puzzler
 

"William Sommerwerck" wrote in message
...
I find it interesting that almost everyone who "agrees" with me is quite
wrong.

They are interpreting the problem and its explanation in terms of what
they
would like the situation to be, rather than looking at it from a strictly
mathematical basis.

Or interpreting it from a view of TV game show reality rather than a purely
statistical basis.

Trevor.

another puzzler
 

"William Sommerwerck" wrote in message
...

How much simpler does it need to be, to be comprehensible?

Nobody said it wasn't comprehensible. But it simply ignores the fact that
game shows are NOT pure chance.


Trevor.

another puzzler
 

On Mon, 16 May 2011 15:08:08 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics.

Right!

And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

And understanding that games of pure chance have NO memory for any previous
actions. It's just like the old question, what are the odds of tossing a
coin 10 heads in a row? If you toss 9 heads in a row, what are the odds of
tossing a 10th?
(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

Trevor.


What has this to do with the question?

d

another puzzler
 

"David" wrote in message
...
The second case is: BB, BG, GB. The couple told you the GG case does not
exist.

Oops you are quite correct, there are still 3 possibilities.

Trevor.

another puzzler
 

"Don Pearce" wrote in message
...
(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

What has this to do with the question?

Nothing I guess for the specific case in question. As was pointed out there
is only one possible set of events if a switch is offered for a one of 3
game of chance. I admit to confusing this with other TV games where the host
can and does influence the outcome.

Trevor.

another puzzler
 

another puzzler
 

On Tue, 17 May 2011 20:01:06 GMT, Carey Carlan
wrote:

(Don Pearce) wrote in news:4dd0c3d7.381246241
:

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

Sorry, but if you don't get it by now, you simply aren't going to.
Give up and try something else.

d

another puzzler
 

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.