arithmetically challenged people

[Soundhaspriority]

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in message
news:iqhhrp$1en$1@dont-email.me...

The "Let's Make a Deal" paradox has been argued about for decades.

The correct answer is that changing your selection is statistically likely
to result in getting the "good" prize 2/3 of the time.

The simplest explanation is that the contestant chooses a curtain with a
bad
prize 2/'3 of the time, and the host always reveals one of the bad prizes
behind a different curtain. Ergo, 2/3 of the time the good prize is behind
the unchosen/unopened curtain, and you should switch.

QED.

Aw, John, you spilled the beans

Bob Morein
(310) 237-6511

 

[Sylvia Else]

On 13/05/2011 3:14 AM, PeterD wrote:

On 5/12/2011 10:52 AM, Sylvia Else wrote:
On 12/05/2011 10:29 PM, PeterD wrote:
On 5/12/2011 2:07 AM, Sylvia Else wrote:
On 12/05/2011 4:03 PM, Kevin Krell wrote:
On 5/11/2011 2:07 PM, William Sommerwerck wrote:
As I'm working, "Let's Make a Deal" is on. A woman has four rolls
of a
die
to get 10 points or more. She gets 8 on the first two rolls. So...
does she
or the host say "No need to roll again?" No! She actually rolls the
die!

OK, guess I'm mathematically challenged, then. Of course she has to
roll
again, at least once more (and possibly twice if #3 is a one), as
the 8
is still less than the 10 you're saying she requires.

But she's bound to get at least 1 on each subsequent roll, so no matter
what happens she wins. Accordingly, as the OP indicated, there's no
need
to bother.

Sylvia

You need to understand the rules of game TV to understand why they did
what they did. It was not an option to 'give it to her', she was
required to roll.


Rules?

Sylvia.

Yes, the federal government has rules in place for game shows, in effect
since the early 60s, following a number of scandals where contestants
were given 'special' treatment.

Any details? I note (from promos and 'news' items, not from watching
them) that contestants in Biggest Slob and Master Burgerflipper seem to
come back after being eliminated.

Sylvia.

 

[Trevor]

"Bill Graham" <weg9@comcast.net> wrote in message
news:w5SdncQfzLWPqVHQnZ2dnUVZ5gudnZ2d@giganews.com...

William Sommerwerck wrote:
As I'm working, "Let's Make a Deal" is on. A woman has four rolls of
a die to get 10 points or more. She gets 8 on the first two rolls.
So... does she or the host say "No need to roll again?" No! She
actually rolls the die!

Of course she rolls the die. She has to, because she needs to get ten, and
she only has eight. So, by the rules of the game she has to keep rolling
until she gets the ten she needs, or runs out of rolls. Getting a good
start is not the same as winning.

I think you miss the point, in this case it's "Not possibly being able to
lose (unless the dice are fixed!) is not the same as winning".
They MUST show her getting 10 so the "arithmetically challenged people"
mentioned in the header can understand what's gong on. Besides they probably
had more air time to fill!

Trevor.

 

[Trevor]

"Bill Graham" <weg9@comcast.net> wrote in message
news:xZidnX2kQ6hHqVHQnZ2dnUVZ5g6dnZ2d@giganews.com...

Kevin Krell wrote:
On 5/11/2011 2:07 PM, William Sommerwerck wrote:
As I'm working, "Let's Make a Deal" is on. A woman has four rolls of
a die to get 10 points or more. She gets 8 on the first two rolls.
So... does she or the host say "No need to roll again?" No! She
actually rolls the die!
OK, guess I'm mathematically challenged, then. Of course she has to
roll again, at least once more (and possibly twice if #3 is a one), as
the 8 is still less than the 10 you're saying she requires.

No, you are not mathematically challenged. You are absolutely correct. So,
there are at least two of us who can read.

Or perhaps don't realise the lowest number on a Die(ce) is one. So she
hasn't actually won yet, but she simply can't lose.

Trevor.

 

[Bill Graham]

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.

 

[Bill Graham]

Trevor wrote:

"Bill Graham" <weg9@comcast.net> wrote in message
news:w5SdncQfzLWPqVHQnZ2dnUVZ5gudnZ2d@giganews.com...
William Sommerwerck wrote:
As I'm working, "Let's Make a Deal" is on. A woman has four rolls of
a die to get 10 points or more. She gets 8 on the first two rolls.
So... does she or the host say "No need to roll again?" No! She
actually rolls the die!

Of course she rolls the die. She has to, because she needs to get
ten, and she only has eight. So, by the rules of the game she has to
keep rolling until she gets the ten she needs, or runs out of rolls.
Getting a good start is not the same as winning.

I think you miss the point, in this case it's "Not possibly being
able to lose (unless the dice are fixed!) is not the same as winning".
They MUST show her getting 10 so the "arithmetically challenged
people" mentioned in the header can understand what's gong on.
Besides they probably had more air time to fill!

Trevor.

Well, if they want to compare her score to many others who roll the die four
times, then she should be required to roll the die four times. But this
depends on the show and its moderators.

 

[Arny Krueger]

"Bill Graham" <weg9@comcast.net> 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.

 

[William Sommerwerck]

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. 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.

 

[Don Pearce]

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger" <arnyk@hotpop.com>
wrote:

"Bill Graham" <weg9@comcast.net> 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.

d

 

[Smitty Two]

In article <iqj8k9$oi5$1@dont-email.me>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

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. 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.

Here is a link to a good visual representation:

http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall.html

 

[Carey Carlan]

spam@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger" <arnyk@hotpop.com
wrote:

"Bill Graham" <weg9@comcast.net> 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.

 

[spamtrap1888]

On May 11, 11:07 pm, Sylvia Else <syl...@not.here.invalid> wrote:

On 12/05/2011 4:03 PM, Kevin Krell wrote:

On 5/11/2011 2:07 PM, William Sommerwerck wrote:
As I'm working, "Let's Make a Deal" is on. A woman has four rolls of a
die
to get 10 points or more. She gets 8 on the first two rolls. So...
does she
or the host say "No need to roll again?" No! She actually rolls the die!

OK, guess I'm mathematically challenged, then. Of course she has to roll
again, at least once more (and possibly twice if #3 is a one), as the 8
is still less than the 10 you're saying she requires.

But she's bound to get at least 1 on each subsequent roll, so no matter
what happens she wins. Accordingly, as the OP indicated, there's no need
to bother.

Sylvia

Rolling again takes less time than explaining, and less time by the
production staff, later on, to deal with the communiques sent by
puzzled viewers.

 

[Don Pearce]

On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulfjoe@hotmail.com>
wrote:

spam@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger" <arnyk@hotpop.com
wrote:

"Bill Graham" <weg9@comcast.net> 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.

d

 

[Bill Graham]

Arny Krueger wrote:

"Bill Graham" <weg9@comcast.net> 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.

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?

 

[Bill Graham]

William Sommerwerck wrote:

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. 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.

There you go! I knew there was some good reason why my instinct told me to
switch doors, and there it is.

 

[Arny Krueger]

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in
message news:iqj8k9$oi5$1@dont-email.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.

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 bollocks.

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.

 

[spamtrap1888]

On May 13, 7:58 am, Smitty Two <prestwh...@earthlink.net> wrote:

In article <iqj8k9$oi...@dont-email.me>,
 "William Sommerwerck" <grizzledgee...@comcast.net> wrote:

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. 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.

Here is a link to a good visual representation:

http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall.html

I learned something today -- thanks!

For people who learn inductively, try this demo:

http://www.curiouser.co.uk/monty/montygame.htm

Does not work properly on Firefox, use IE.

 

[Bill Graham]

Arny Krueger wrote:

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in
message news:iqj8k9$oi5$1@dont-email.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.

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 bollocks.

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........

 

[Bill Graham]

Bill Graham wrote:

Arny Krueger wrote:
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in
message news:iqj8k9$oi5$1@dont-email.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.

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 bollocks.

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........

Another way to look at it is, you are allowed to play the game twice. If you
only play it once, your chance of winning is only 1/3 and you can't play
again. But if you accept a loss in the first game, then they will let you
play again with a 50% chance of winning. So, you are better off by accepting
a loss in the first game, and then getting to play the game again with a 50%
chance of winning in the second game.

 

[John Robertson]

Bill Graham wrote:

Arny Krueger wrote:
"Bill Graham" <weg9@comcast.net> 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.

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?

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

And as Spamtrap said this is a great site to see the results - and other
than you have to run it under IE it will show you how it does benefit
you to change doors. Run the iteration a few hundred times - first on
keep the door and the other on change the door.

http://www.curiouser.co.uk/monty/montygame.htm

This is a variation of the three cups/shells hiding something shuffle
carney game...

John :-#)#
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