arithmetically challenged people
William Sommerwerck
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!
--
"We already know the answers -- we just haven't asked the right
questions." -- Edwin Land
ne...@jecarter.us
<grizzle...@comcast.net> 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!
It's all about extending the "suspense".
There's also the possibility that the viewers might not have been able
to understand why she didn't roll again ;-)
John
polymod
"William Sommerwerck" <grizzle...@comcast.net> wrote in message
news:iqetrg$ek4$1...@dont-email.me...
> 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!
Even Jackie Gleason got nervous on The $64,000 Question.
I'd probably would have done the same ;)
Poly
Kevin Krell
> As I'm working, "Let's Make a Deal" is on. A woman has four rolls of a die
> to get 10 points or more. She gets 8 on the first two rolls. So... does she
> or the host say "No need to roll again?" No! She actually rolls the die!
>
roll again, at least once more (and possibly twice if #3 is a one), as
the 8 is still less than the 10 you're saying she requires.
Sylvia Else
But she's bound to get at least 1 on each subsequent roll, so no matter
what happens she wins. Accordingly, as the OP indicated, there's no need
to bother.
Sylvia
Trevor
"Sylvia Else" <syl...@not.here.invalid> wrote in message
news:931bps...@mid.individual.net...
Perhaps his dice have a zero on one side :-)
Trevor.
Skybuck Flying
"William Sommerwerck" <grizzle...@comcast.net> wrote in message
news:iqetrg$ek4$1...@dont-email.me...
> to get 10 points or more. She gets 8 on the first two rolls. So... does
> she
> or the host say "No need to roll again?" No! She actually rolls the die!
Maybe she believed the more points she would get the bigger the prize ! ;)
=D
There is a little tale about a woman being greedy ! ;) =D
Bye,
Skybuck.
petrus bitbyter
"William Sommerwerck" <grizzle...@comcast.net> schreef in bericht
news:iqetrg$ek4$1...@dont-email.me...
Depends on the rules you did not mention. She still does not have 10 points
even though she sure will have them when she rolls one or two times more.
Nevertheless I want to know why the person is a woman. I suppose there's
another answer that will make the difference. :)
petrus bitbyter
PeterD
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.
--
I'm never going to grow up.
Sylvia Else
Rules?
Sylvia.
PeterD
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.
Bill Graham
> 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.
Bill Graham
No, you are not mathematically challenged. You are absolutely correct. So,
there are at least two of us who can read.
Bill Graham
Oh! - I get it now. She's rolling to obtain a number on the die between 1
and 6 inclusive. This wasn't made clear in the original statement. I didn't
know what she was rolling for. If its just a number on the face of the die,
then sure. She has already won the game, so there is no reason to
continue....
Bill Graham
One possibility. Or she might have to match some number from 1 to 6 that
someone else pulled out of a hat, or there are many other scenearios that
might be in the game. If all she has to do is get the number that comes up
on the die, then she has already won after two rolls.
Soundhaspriority
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
John Williamson
I've not seen the show, as I'm on the East of the Atlantic. The fact is
that a win is not inevitable with the score given. The di(c)e *could*
fall off the table on one or more subsequent rolls, for no score, unless
it's a sealed die shaker. I'd agree it's almost vanishingly unlikely....
As pointed out by another poster, the show rules (which may be known
only to the contestants and the show crew) might state that rolling must
continue until the needed score is reached or exceeded.
--
Tciao for Now!
John.
John Williamson
Yes, by about 50%, and that fact has caused a *lot* of argument and
discussion in another group that I frequent.
Michael A. Terrell
http://en.wikipedia.org/wiki/The_$64,000_Question
--
You can't fix stupid. You can't even put a Band-Aid™ on it, because it's
Teflon coated.
William Sommerwerck
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.
Soundhaspriority
"William Sommerwerck" <grizzle...@comcast.net> wrote in message
news:iqhhrp$1en$1...@dont-email.me...
Aw, John, you spilled the beans ;)
Bob Morein
(310) 237-6511
Sylvia Else
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" <we...@comcast.net> wrote in message
news:w5SdncQfzLWPqVHQ...@giganews.com...
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
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
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
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
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
> 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
wrote:
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
spamtrap1888
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.
Smitty Two
"William Sommerwerck" <grizzle...@comcast.net> wrote:
Here is a link to a good visual representation:
Carey Carlan
news:4dcd3b49....@news.eternal-september.org:
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.
Don Pearce
wrote:
Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.
d
spamtrap1888
> In article <iqj8k9$oi...@dont-email.me>,
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
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
There you go! I knew there was some good reason why my instinct told me to
switch doors, and there it is.
Arny Krueger
>> 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
> "William Sommerwerck" <grizzledgee...@comcast.net> wrote in
> messagenews: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.
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.
Bill Graham
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
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
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 :-#)#
--
(Please post followups or tech enquiries to the newsgroup)
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
Call (604)872-5757 or Fax 872-2010 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."
Plai...@yawhoo.com
wrote:
>"William Sommerwerck" <grizzle...@comcast.net> wrote in
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.
PlainBill
Plai...@yawhoo.com
wrote:
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
William Sommerwerck
news:h66dnfLBnLbDGFDQ...@giganews.com...
> "William Sommerwerck" <grizzle...@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... ;-)
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 bollocks.
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 bollocks over your eyes. Think it through
carefully, and pretty soon you'll /understand/.
William Sommerwerck
> 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.
William Sommerwerck
> 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.
David
news:iqk4b0$cop$1...@dont-email.me...
***
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
Don Pearce
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
Dave Platt
Don Pearce <sp...@spam.com> 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 <dpl...@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!
spamtrap1888
> In article <4dcd9c92.240121...@news.eternal-september.org>,
>
> Don Pearce <s...@spam.com> 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.
Trevor
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.
Trevor
Which is complete bollocks 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.
Trevor
"David" <som...@somewhere.com> wrote in message
news:iqk5jt$jn1$1...@dont-email.me...
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.
Trevor
<Plai...@yawhoo.com> wrote in message
news:fe3rs6lb23a42j9oi...@4ax.com...
>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.
Trevor
> 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.
Trevor
"John Robertson" <sp...@flippers.com> wrote in message
news:tbydnY7yjqn4ElDQ...@giganews.com...
> 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.
Trevor
"Don Pearce" <sp...@spam.com> wrote in message
news:4dcd9c92....@news.eternal-september.org...
> 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.
Trevor
"Dave Platt" <dpl...@radagast.org> wrote in message
news:jbk0a8-...@radagast.org...
> In article <4dcd9c92....@news.eternal-september.org>,
> Don Pearce <sp...@spam.com> 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.
Trevor
"spamtrap1888" <spamtr...@gmail.com> wrote in message
news:5f3fe288-fefb-4981...@z15g2000prn.googlegroups.com...
>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.
Don Pearce
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
William Sommerwerck
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.
Arny Krueger
> 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.
William Sommerwerck
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?
Don Pearce
Some will never get it, no matter how it is explained.
d
Arny Krueger
news:4dce3407$0$22473$afc3...@news.optusnet.com.au
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?
William Sommerwerck
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.
David
news:4dce2d98$0$2443$afc3...@news.optusnet.com.au...
<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
Bill Graham
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.
William Sommerwerck
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.
Carey Carlan
news:4dcd5ee1....@news.eternal-september.org:
> On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gul...@hotmail.com>
> wrote:
>
>>sp...@spam.com (Don Pearce) wrote in
>>news:4dcd3b49....@news.eternal-september.org:
>>
>>> On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
>>> <ar...@hotpop.com> wrote:
>>>
>>>>"Bill Graham" <we...@comcast.net> wrote in message
>>>>news:t_ydnZKHN4u_QlHQ...@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.
Don Pearce
wrote:
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
spamtrap1888
> s...@spam.com (Don Pearce) wrote innews:4dcd5ee1....@news.eternal-september.org:
>
>
>
> > On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulf...@hotmail.com>
> > wrote:
>
> >>s...@spam.com (Don Pearce) wrote in
> >>news:4dcd3b49....@news.eternal-september.org:
>
> >>> On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
> >>> <ar...@hotpop.com> wrote:
>
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.
Bill Graham
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.
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.
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.
Trevor
Or interpreting it from a view of TV game show reality rather than a purely
statistical basis.
Trevor.
Trevor
> 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.
Don Pearce
What has this to do with the question?
d
Trevor
> 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.
Trevor
>>(First you MUST assume the coin is untampered with, you cannot assume the
>>same for a TV game show however!)
>
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.
Carey Carlan
@news.eternal-september.org:
> That isn't the final situation. I will take this a step at a time.
>
> There are three doors - one with a car, two with goats
>
> I choose one. I have a 1 in 3 chance of being right
>
> That means there is a 2 in 3 chance of the car being in the other two
>
> I know for a fact that at least one of the other two is a goat.
>
> That does not change the odds - it is still 2 in 3 that the car is in
> one of those
>
> The host shows me one of the two - one he knows to contain a goat.
>
> This is not new information, I knew there was a goat there, I still
> know there was a goat there.
>
> The odds are still 2 in 3 that the car is in one of those two doors.
Stop there.
No, I didn't know there was a goat THERE. I knew there was a goat
behind at least one of the door besides the one I chose, but I didn't
know which one. Now a variable is removed from the equation.
Revealing a goat behind a door doesn't change the odds?
Of course it does.
Otherwise, revealing the car behind a door also wouldn't change the
odds.
Once the host has revealed a goat, then there's an even chance that the
car is behind one of the two remaining doors--and I have no information
either way (unless you're counting the psychological factors) that the
door I chose is or is not the correct one.
Not trying to be argumentative, but I still don't see the logic.
Carey Carlan
news:c5220357-4ddd-43ac...@z15g2000prn.googlegroups.com:
>> Just as in flipping coins.
>> Getting 5 heads in a row is 1/32.
>> But getting the 5th head after already getting 4 is still 1/2.
>
> The big difference: In the Monty Hall problem there is only one "coin
> flip". Only one random choice is made -- the first choice of a door.
> In the coin flip situation, there are five coin flips, five random
> choices.
>
> Now, in contrast, if the car and remaining goats were randomly
> shuffled after each goat door was revealed, then the situation would
> be different. But in the MHP problem the car does not move.
Still trying to get my head around this.
How would shuffling unknown values affect my choice? If I didn't know
before and you shuffle the choices, it's still a random choice on my part.
Don Pearce
wrote:
>sp...@spam.com (Don Pearce) wrote in news:4dd0c3d7.381246241
Sorry, but if you don't get it by now, you simply aren't going to.
Give up and try something else.
d
philicorda
The host acts as a leak of information. It might help to imagine an
alternate game, where the host does not know the contents of the doors,
and the game is void if the host reveals the car. This version puts you
back to 50/50 when the host reveals a goat, whether you switch doors or
not.
David
news:EtEAp.608$Ky....@newsfe24.ams2...
***
Not true. When the host reveals a goat whether he guessed or knew
it was there makes absolutely no difference. You should still
switch doors.
David
Randy Yates
I think, several years ago when I originally saw this, I
argued as vehemently as you, Arny. It is extremely counter-intuitive,
which goes to show intuition isn't always right!
--
Randy Yates % "Watching all the days go by...
Digital Signal Labs % Who are you and who am I?"
mailto://ya...@ieee.org % 'Mission (A World Record)',
http://www.digitalsignallabs.com % *A New World Record*, ELO
Smitty Two
Carey Carlan <gul...@hotmail.com> wrote:
> Not trying to be argumentative, but I still don't see the logic.
The visual explanations on the web make it rather easy to see, no pun
intended. I posted a link earlier, so did others.
Don Pearce
wrote:
If the host does not know, he might quite as easily reveal the car.
You then can't win it. Do you guarantee yourself 2/3 odds by switching
then? No. If the host reveals a goat by chance, the odds do indeed
drop to 50/50.
d
Arny Krueger
news:B9ydnXmRQo_4r07Q...@supernews.com
> On 05/14/2011 06:48 AM, Arny Krueger wrote:
>> "spamtrap1888"<spamtr...@gmail.com> wrote in message
>> news:dd7c10fd-38c7-43e2...@35g2000prp.googlegroups.com
>>
>>> Declaring that there is no connection between the two
>>> situations is the source of the poster's error. Monty
>>> Hall knew if the player was correct or not, and so the
>>> player's choice of the door in the first round
>>> influenced the selection of the goat door. The graphic
>>> helps you understand that there are still three
>>> scenarios once a goat door has been revealed.
>>
>> I get it now.
>
> I think, several years ago when I originally saw this, I
> argued as vehemently as you, Arny. It is extremely
> counter-intuitive, which goes to show intuition isn't
> always right!
In my studies of this item, I found a statement that about 10% of
*everybody* never gets it, regardless of their intelligence or education.
That suggests to me that some people learn things about problem solving that
keep them from seeing certain solutions. The trick is to not do that, or if
you do, somehow redirect how you approach these things.
David
news:4dd3548d...@news.eternal-september.org...
>The host acts as a leak of information. It might help to imagine
>an
>alternate game, where the host does not know the contents of the
>doors,
>and the game is void if the host reveals the car. This version
>puts you
>back to 50/50 when the host reveals a goat, whether you switch
>doors or
>not.
>***
>Not true. When the host reveals a goat whether he guessed or
>knew
>it was there makes absolutely no difference. You should still
>switch doors.
>
>David
>
If the host does not know, he might quite as easily reveal the
car.
You then can't win it. Do you guarantee yourself 2/3 odds by
switching
then? No. If the host reveals a goat by chance, the odds do
indeed
drop to 50/50.
d
***
Sorry, I disagree. Yes the host could reveal a car if he is
unaware of the situation. If this happens, the game was defined
as void. If the host instead reveals a goat, there is no
difference whether he guessed or knew the goat was there.
David
spamtrap1888
Let's look at the case of the ignorant host.
There are three possibilities at the start of the game. The
probability of each is 1/3
_1 2 3_
aCGG
bGCG
cGGC
Let us say door 1 represents the contestant's pick. The host can pick
either door 2 or door 3
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door
3, loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 2. Result Car. Contestant loses
..............Host picks Door 3. Result Goat. Contestant switches to
Door 2, wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
..............Host picks Door 3 Result Car. Contestant loses.
Of the six possible scenarios, the contestant loses four times. If the
contestant does not switch after the ignorant host opens a door, the
contestant loses four times. If we discard the times the host opens a
door with a car behind it, the contestant wins two out of four times
when he switches, and two out of four times when he doesn't switch.
Therefore, switching picks has no effect on the odds when the host
randomly opens one of the other doors.
Carey Carlan
a558-e3d...@34g2000pru.googlegroups.com:
Then go back to the original where the host knows where the car is and
the contestant switches.
Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3,
loses.
..............Host picks Door 3 Result Goat. Contestant switches to
Door 2, loses.
Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2,
wins
Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3,
wins
Or the contestant doesn't switch.
Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1, wins.
..............Host picks Door 3 Result Goat. Contestant keeps Door 1,
wins.
Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses
Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses
After the Host opens the door the odds are even. Makes no difference if
the contestant changes doors or not. This is the same as there only
being two doors.
The original claim was that the odds remained 1 in 3 even after the Host
opened the door. I still don't see it.
spamtrap1888
> spamtrap1888 <spamtrap1...@gmail.com> wrote in news:88df2861-f695-449b-
> a558-e3d119b1c...@34g2000pru.googlegroups.com:
Without switching, the contestant has a 1/3 chance of winning:
Case a: Contestant picked door with car. Host can open either door,
his choice, to reveal goat.
Case b: Contestant picked door with goat. Host must open Door 3 to
reveal goat.
Case c: Contestant picked door with goat. Host must open Door 2 to
reveal goat.
With switching, the contestant now has a 2/3 chance of winning:
Case a: Host can open either door, his choice. Contestant switches to
unopened door, loses.
Case b: Host opens Door 3, contestant switches to Door 2, wins.
Case c: Host opens Door 2, contestant switches to Door 3, wins.
Don Pearce
wrote:
Void is not one of the permitted outcomes. Suppose the host
accidentally revealed the car - to be equivalent to the intentional
goat revelation, he would then have to say "never mind, take the car
anyway". That would leave you in the 1/3 2/3 situation. If he reveals
a goat by chance the game degenerates to the simple situation - the
host has chosen one of the three, and you get to pick between the
remaining two, always assuming that he did not pick the car.
The point of the intentional revelation is that by switching you get -
in effect - both doors, not just the one.
d
David
***
Start at the beginning of this post and read all of the quoted
stuff. The initial assumption is that 'void' IS a permitted
outcome. If the void assumption is changed , I concede.
David
William Sommerwerck
explained it simply, twice.
All you have to do is understand why the initial probability of getting the
big prize is 1/3 -- and everything else falls out in a completely
straightforward manner.
Ron Capik
OK, try this: blow the game up to 100 doors. Your chance of picking
the winning door on the first try is 1 out of 100. Stick with that
choice and each time a zonk is revealed the chance of the prize being
in the remaining group increases. When you get down to two doors the
chance of the prize being behind the other door is 99 out of 100.
The chance of your first pick being correct is still 1 out of 100.
Later...
Ron Capik
--
Carey Carlan
dn...@giganews.com:
> OK, try this: blow the game up to 100 doors. Your chance of picking
> the winning door on the first try is 1 out of 100. Stick with that
> choice and each time a zonk is revealed the chance of the prize being
> in the remaining group increases. When you get down to two doors the
> chance of the prize being behind the other door is 99 out of 100.
> The chance of your first pick being correct is still 1 out of 100.
That may have penetrated.
It all hinges on the host knowing in advance which door hides the prize and
revealing known (to him) bad choices. I still don't agree with the logic,
but I think I follow it.
It's called the "Monty Hall problem"? I'll continue to research.
Thank you for your patience.
Carey Carlan
> OK, try this: blow the game up to 100 doors. Your chance of picking
> the winning door on the first try is 1 out of 100. Stick with that
> choice and each time a zonk is revealed the chance of the prize being
> in the remaining group increases. When you get down to two doors the
> chance of the prize being behind the other door is 99 out of 100.
> The chance of your first pick being correct is still 1 out of 100.
Here's the simple answer that I can understand:
"An even simpler solution is to reason that switching loses if and only if
the player initially picks the car, which happens with probability 1/3, so
switching must win with probability 2/3"
Eureka! That finally makes sense. And yes, everyone was trying to tell me
that.
Bill Graham
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.
Carey Carlan
news:bvudndZ-Vc0WKUjQ...@giganews.com:
> I claqim there are two games. In the first game, you go to the studio,
> pick a door, and then go home to wait and see if they call you and
> tell you that you either won or lost. Your odds are only 1/3 of
> winning this game. But if you play the second game, then you go to the
> studio and mess around until the host opens up a door and shown you
> the donkey behind it. then you can play the game with 50-50 odds of
> winning. The only thing I have trouble explaining is why, in order to
> play this second game with the better odds, you have to switch doors.
> But, in fact, you do have to switch in order to switch games and take
> advantage of the better odds.
It's not just 50/50. It's 67/33 in your favor.
Here's another super-simple explanation:
As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can
keep your one door or you can have the other two doors."