arithmetically challenged people

[Dave Platt]

In article <bvudndZ-Vc0WKUjQnZ2dnUVZ5tidnZ2d@giganews.com>,
Bill Graham <weg9@comcast.net> 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 <dplatt@radagast.org> 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!

 

[Bill Graham]

Dave Platt wrote:

In article <bvudndZ-Vc0WKUjQnZ2dnUVZ5tidnZ2d@giganews.com>,
Bill Graham <weg9@comcast.net> 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?

 

[Smitty Two]

In article <nvWdnTXm1reRkUrQnZ2dnUVZ5tOdnZ2d@giganews.com>,
"Bill Graham" <weg9@comcast.net> 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.

 

[Jamie]

Smitty Two wrote:

In article <nvWdnTXm1reRkUrQnZ2dnUVZ5tOdnZ2d@giganews.com>,
"Bill Graham" <weg9@comcast.net> 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

 

[Trevor]

"Bill Graham" <weg9@comcast.net> wrote in message
news:nvWdnTXm1reRkUrQnZ2dnUVZ5tOdnZ2d@giganews.com...

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.

 

[Don Pearce]

On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham" <weg9@comcast.net>
wrote:

Dave Platt wrote:
In article <bvudndZ-Vc0WKUjQnZ2dnUVZ5tidnZ2d@giganews.com>,
Bill Graham <weg9@comcast.net> 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

 

[John Williamson]

Don Pearce wrote:

On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham" <weg9@comcast.net
wrote:

Dave Platt wrote:
In article <bvudndZ-Vc0WKUjQnZ2dnUVZ5tidnZ2d@giganews.com>,
Bill Graham <weg9@comcast.net> 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.

 

[Trevor]

"John Williamson" <johnwilliamson@btinternet.com> wrote in message
news:93peh8Fro5U1@mid.individual.net...

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.

 

[Don Pearce]

On Sat, 21 May 2011 19:55:55 +1000, "Trevor" <trevor@home.net> wrote:

"John Williamson" <johnwilliamson@btinternet.com> wrote in message
news:93peh8Fro5U1@mid.individual.net...
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

 

[PeterD]

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.

 

[Smitty Two]

In article <ir8997$2gd3$2@pyrite.mv.net>, PeterD <peter2@hipson.net>
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?

 

[Bill Graham]

Smitty Two wrote:

In article <ir8997$2gd3$2@pyrite.mv.net>, PeterD <peter2@hipson.net
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)

 

[Trevor]

"Don Pearce" <spam@spam.com> wrote in message
news:4dd79a03.316359700@news.eternal-september.org...

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]

"Trevor" <trevor@home.net> wrote in message
news:4ddb578e$0$22471$afc38c87@news.optusnet.com.au...

"Don Pearce" <spam@spam.com> wrote in message
news:4dd79a03.316359700@news.eternal-september.org...
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.

 

[Don Pearce]

On Tue, 24 May 2011 17:00:57 +1000, "Trevor" <trevor@home.net> wrote:

"Don Pearce" <spam@spam.com> wrote in message
news:4dd79a03.316359700@news.eternal-september.org...
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

 

[Don Pearce]

On Tue, 24 May 2011 17:06:22 +1000, "Trevor" <trevor@home.net> wrote:

"Trevor" <trevor@home.net> wrote in message
news:4ddb578e$0$22471$afc38c87@news.optusnet.com.au...
"Don Pearce" <spam@spam.com> wrote in message
news:4dd79a03.316359700@news.eternal-september.org...
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

 

[Trevor]

"Don Pearce" <spam@spam.com> wrote in message
news:4ddbcd36.591611280@news.eternal-september.org...

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.

 

[Trevor]

"Don Pearce" <spam@spam.com> wrote in message
news:4ddcce1b.591839525@news.eternal-september.org...

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.