Another riddle - from "The Curious Incident Of The Dog In The Nighttime"
wotsayyou
You are on a quiz show, and are presented with three doors. Two doors have
goats behind them and one has a car. Your objecive is to pick the car and
win it.
You choose a door, and the quiz show host opens one of the remaining doors
to reveal a goat (the host knows what is behind each of the doors).
The host then offers you the chance to either stick with your original
choice, or to choose the other remaining unopened door.
To have the most chance of winning the car, should you change your choice,
or stick, or does it make no difference ?
stePH
I'd say it makes no difference; having eliminated one of the "goat"
doors, there's a 50% chance of getting the correct door and no way to
tell whether you've picked the right one already or not.
stePH
--
"A lion will exert himself to the utmost, even when entering the tiger's
den to throw baby rabbits off a cliff!" -- Moroboshi Ataru
Bev Vincent
"wotsayyou" <jonho...@hotmail.com> wrote in message
news:GhcHc.226$fh5...@newsfe2-win.ntli.net...
You should switch. If you do, you have a 66.67% chance of winning.
--
Bev Vincent
www.BevVincent.com
The Road to the Dark Tower, NAL (Penguin), October 2004
Algomeysa2
Changing your choice wins 2/3rds of the time, the intuitive answer that it's
50/50 is incorrect.
The thing to keep in mind is that the host ALWAYS reveals a goat, thus
changing the odds.
The way to think of this is, imagine there are 1000 doors. You pick one,
and you have a 1 in 1000 chance of being right, 999 in 1000 of being wrong.
But if you were allowed to group pick ALL the doors but the one you
originally picked, you'd then have a 999 in 1000 chance of being right.
so if our game show host then, one after the other, opens up 998 of those
999 doors, revealing goats, until just one is left, if you switched your
guess you'd be right 999 out of 1000 times. If you stuck with your original
choice, you'd only be right 1 in 1000.
To see a simulation of this:
http://planettom.home.mindspring.com/vossavant.htm
Dr. Randall Flagg
> "wotsayyou" <jonho...@hotmail.com> wrote in message
> news:GhcHc.226$fh5...@newsfe2-win.ntli.net...
>> Aka "The Drawing Of The Two Goats And A Car" ;O)
>>
>> You are on a quiz show, and are presented with three doors. Two
>> doors have goats behind them and one has a car. Your objecive is to
>> pick the car and win it.
>>
>> You choose a door, and the quiz show host opens one of the remaining
>> doors to reveal a goat (the host knows what is behind each of the
>> doors).
>>
>> The host then offers you the chance to either stick with your
>> original choice, or to choose the other remaining unopened door.
>>
>> To have the most chance of winning the car, should you change your
>> choice, or stick, or does it make no difference ?
>
> You should switch. If you do, you have a 66.67% chance of winning.
Perhaps on the road to the Dark Tower, where east is west and west is east,
but not here . . .
Each door is equally likely to contain the car; no door is favored over
another. You have a 1/3 chance of picking the car, but that doesn't really
matter. The host reduces it to two doors. She's always going to give you the
door with the car and a door with a goat, no matter which door you initially
choose. By the nature of the game, your odds OF WINNING are 50% from the
start. This is true no matter how many door there are. With a million doors,
the host is going to eliminate all but 2, leaving your odds of winning at
50%.
Suppose you don't pick a door until after the host removes a goat. Then you
pick. Each door is still equally likely. The likelihood of the doors doesn't
change simply because of WHEN you picked a door.
I'm assuming your logic goes something like this:
There's a 2/3 chance that the door the host picked plus the door you picked
contain goats. Therefore, the door you didn't pick has the car (2/3).
The same logic can be used to justify sticking as that used to justify
switching, though:
There's 2/3 chance that the door the host picked plus the door you DIDN'T
pick contain goats. Therefore, the door you DID pick has the car (2/3).
Algomeysa2
"Dr. Randall Flagg" <cujogro...@yahooo.com> wrote in message
news:2l5crrF...@uni-berlin.de...
No, you're simply wrong about this one. See my other post.
Bev Vincent
"Algomeysa2" <Algomeys...@mindspring.comNOPESPAM> wrote in message
news:1bfHc.12214$yy1....@newsread2.news.atl.earthlink.net...
There's a live demo and several different attempts to explain this one here:
http://www.cut-the-knot.org/hall.shtml
wotsayyou
"Algomeysa2" <Algomeys...@mindspring.comNOPESPAM> wrote in message
Yup, 2/3.
Damn. I was convinced it was 1/2 for ages. Couldn't get my head round it.
You are 2/3 likely to choose a goat at the beginning though, therefore if
you change you are 2/3 likely to get a car.
Algomeysa2
"Bev Vincent" <MaxD...@hotmail.com> wrote in message
news:ETfHc.55$yo4....@monger.newsread.com...
> There's a live demo and several different attempts to explain this one
here:
>
> http://www.cut-the-knot.org/hall.shtml
also, one here:
http://planettom.home.mindspring.com/vossavant.htm
Dr. Randall Flagg
I saw your other post, and it was . . . well . . . simply wrong. It doesn't
matter how many doors there are (as long as there are at least three. There
could be 3, a hundred, a thousand, a million, a trillion, whatever.) You're
saying the odds change because of your selection. Sorry, but the odds don't
change simply because of which door you selected (or when you selected).
Would you agree that if there were only two doors to begin with, that the
odds for each door are equal? If the host gave you a choice of two doors,
call them A and B, isn't A equally as likely to reward you with the prize as
B?
Now, let's add that, before you arrived, there were 1000 doors. I chose one
and the host removed 998 bad doors, leaving doors A and B. The choice to you
is the same. It doesn't matter to you which one I chose; you have a choice
of two doors, one of which has the car.
I have the same choice.
The number of doors doesn't matter; it's all window dressing. The host is
just using some showmanship to give you a choice of two doors. How the doors
were arrived at doesn't matter. The choice to you is to pick one of two
doors.
Your logic, applied to each door results in the same--choose the 'other'
door. With doors A and B remaining, your logic applied to A:
A had 1/3 chance of being the car, thus 2/3 chance of being a goat, so I
better pick B.
B had 1/3 chacne of being the car, thus 2/3 chance of being a goat, so I
better pick A.
Both are the same. You can't apply the logic to just one door and ignore the
other.
wotsayyou
It is totally unintuitive, I have to say, but it does work out as 2/3 to get
a car if you change.
It is illustratred quite well in " The Curious Incident...":
You are asked to choose a
door
¦
/ ¦ \
/ ¦ \
/ ¦ \
/ ¦
\
/ ¦
\
You choose You choose You choose
a door with a a door with a a
door with a
goat behind it goat behind it
car behind it
¦ ¦
¦
/ \ / \
/ \
/ \ / \
/ \
/ \ / \
/ \
/ \ /
\ / \
/ \ ¦
¦ ¦ \
stick change stick change
stick change
¦ ¦ ¦
¦ ¦ ¦
You You You You You
You
Get a Get a Get a Get a Get a
Get a
Goat Car Goat Car Car
Goat
Ok, so if you look at the branches of the tree then, on the three occasions
that you stick, you get a goat 2/3 times, and a car 1/3 times.
On the three occasions that you change, you get a car 2/3 times, and a goat
1/3 times.
I sympathise, cos I was up half the night thinking "No, that can't be
right".
The example is used to illustrate how some maths problems are completely
unintuitive
wotsayyou
Shit. Ok, the formatting has completely messed up and the above post is
about as useful as a condom machine in the Vatican. Hopefully it helps at
least a little
George Smith
>and the host removed 998 bad doors, leaving doors A and B. The choice to you
>is the same. It doesn't matter to you which one I chose; you have a choice
>of two doors, one of which has the car.
I was going over this thinking the same thing too, then I realized this...If
you pick one door out of a tousand, you obviously have a 1 in 1000 chance of
picking the door with the car. Now for the host, even if he *didn't* know
which door held the car, would still have a 999 in 1000 chance of finding the
car behind one of the doors you didn't pick. That still applies even after he
opens 998 doors that he knows hide goats, you see. If you just come up at that
time without knowing anything that has gone before, yeah, you'd have a 50-50
shot at picking the right door, but knowing that your door was going to be one
of the remaining two out of 1000 *no matter what*, your odds are far, far
greater by switching to the other door, which still has that 999 in 1000 chance
of being the car.
Just took a while to get my mind around it...
George
"I would rather be exposed to the inconveniences attending too much liberty
than to those attending too small a degree of it." - Thomas Jefferson
jean.paulo
--
Jean.paulo
"Dr. Randall Flagg" <cujogro...@yahooo.com> a écrit dans le message de
news:2l5crrF...@uni-berlin.de...
To the absurd demonstration :
Suppose the host open the door WITH the car.
Do you still have 2/3 chances to get it when switching your choice, then ?
Ami Fairchild
>
> Shit. Ok, the formatting has completely messed up and the above post
is
> about as useful as a condom machine in the Vatican. Hopefully it helps
at
> least a little
I've seen this thing often enough to know that switching is always the
right choice ;-).
Have a great day!
Ami -- I don't *understand* it, though . . . .
_____________________
"I try to take one day at a time, but sometimes, several days attack me
at once."
Algomeysa2
> Both are the same. You can't apply the logic to just one door and ignore
the
> other.
No, you're still not getting it. What's throwing you is that there's no
chance involved in the host revealing a goat; he knows where all the goats
are, and always reveals a goat.
This has been debated at great length since Marilyn Vos Savant posted it in
her column years ago.
It really does pay to switch doors.
Bev Vincent
"jean.paulo" <Jean.pau...@free.fr> wrote in message
news:ccl6p4$ini$1...@news-reader2.wanadoo.fr...
> Do you still have 2/3 chances to get it when switching your choice, then ?
No, now you have a 100% chance -- because you change to the one wthe host
just opened!
Dr. Randall Flagg
When you show only three cases, yes, indeed. But there are more than three
cases.
When you pick a CAR, you show only one branch, but there are really two.
When you pick the CAR, the host could remove EITHER goat . . .
CASE 1:
When the host removes the first goat:
CAR GOAT1 GOAT2 You picked the CAR, host removes GOAT1
Stick = CAR
Change = GOAT2
CASE 2:
When the host removes the second goat:
CAR GOAT1 GOAT2 You picked the CAR, host removes GOAT2
Stick = CAR
Change = GOAT1
You could have picked a goat, however. Had you picked the first goat, the
host could remove only the second goat, thus there's only one case here:
CASE 3:
CAR GOAT1 GOAT2 You picked GOAT1, host must remove GOAT2
Stick = GOAT1
Change = CAR
Had you picked the second goat, the host could remove onlhy the first goat,
so there's only one case here, too:
CASE 4:
CAR GOAT1 GOAT2 You picked GOAT2, host must remove GOAT1
Stick = GOAT2
Change = CAR
Stick: twice get the car; twice get a goat
Change: twice get the car; twice get a goat
Algomeysa2
> Stick: twice get the car; twice get a goat
> Change: twice get the car; twice get a goat
No. No, no, and no.
try this and see:
http://planettom.home.mindspring.com/vossavant.htm
Also see this:
http://www.randomhouse.ca/readmag/volume4issue1/excerpts/curious.htm
And if you need more, do a goggle search on "marilyn vos savant goats doors
monty hall" and read through the dozens of webpages on this subject
Dr. Randall Flagg
> "Dr. Randall Flagg" <cujogro...@yahooo.com> wrote in message
> news:2l7mdnF...@uni-berlin.de...
>
>> Stick: twice get the car; twice get a goat
>> Change: twice get the car; twice get a goat
>
> No. No, no, and no.
>
snipped. So, tell me which case you are disputing, and why that case is
wrong.
stePH
shitty seaside vacation for what's behind door number two.
Bev Vincent
Three doors: Goat 1, Goat 2, Car. Let's say they're behind Doors 1, 2 and 3
respectively.
From the onset, you have a 1/3 chance of picking the car.
Case 1: You choose Door 1. The only thing the host can do is show you Goat
2; he's not going to show you the car. You either stay (and lose) or change
(and win).
Case 2: You choose Door 2. Exactly the same as case 1. He shows you Goat 1.
You stay (and lose) or change (and win).
Case 3: You choose Door 3. He shows you a goat--doesn't matter which one.
This is probably the crucial factor; there aren't two scenarios, one where
he shows you Goat 1 and one where he shows you Goat 2. A goat is a goat in
this case. This is the 1 out of 3 where you were right from the beginning.
You stay (and win) or change (and lose).
Two times out of three you change and win.
Dr. Randall Flagg
Yes, Yes that's it. Still 4 cases, BUT . . .
Case 1: 1/3
Case 2: 1/3
Case 3: 1/3 and here's the but . . .
The host still has two choices about which goat to eliminate, but they
aren't really separate cases, there're more like subcases . . .
Case 3a: 1/3 * 1/2 = 1/6
Case 3b: 1/3 * 1/2 = 1/6
Summing up . . .
1/3 Change and Win
1/3 Change and Win
1/6 Stick and Win
1/6 Stick and Win
For a grand total of . . .
2/3 Switch and Win
1/3 Stick and Win
I think I beated that goat to death (beating the car to death is another
story . . .). Sooooo, like in the old cigarette commercial . . .
I'd rather switch than fight (OK, backwards from the old cigarette
commercial).
Hmmmph. Y'all had me thinkin' so much about this that I've had several posts
without knitting quotes.
jean.paulo
The decisive factor is 'Always invert your previous decision'
If you do ALWAYS decide to change BEFORE the door is opened, then the
result is really 666 car won over 333 loss.
But if you invert at random (Do I invert my choice or not ?) after the door
is opened
then the final result if 500/500.
So in fact, the question is biaised. You should ask :
Chose a door, and then when the wrong door is opened you MUST choose
now the other door. Then the chances are 3/2 to win.
But the question is really 'now that the third door is opened, do you want
to invert
or not ? ' Then the chances are 50/50§
As in the game you cannot predict, the true answer is still 50/50.
--
Jean.paulo
Algomeysa2
nope.
http://www.randomhouse.ca/readmag/volume4issue1/excerpts/curious.htm
Otter
<Algomeys...@mindspring.comNOPESPAM> wrote:
>nope.
>
>http://www.randomhouse.ca/readmag/volume4issue1/excerpts/curious.htm
Interesting, but isn't that chart flawed? As the riddle goes, one of
the doors with the goat is revealed, so there are two doors remaining,
THEN you are asked if you want to change.
The chart as shown shows all THREE doors. Surely to come up with a
solution you must remove one of the doors with a goat behind it from
the chart to accurately represent what you are faced with at the time
you are given the option to chose, that is TWO doors.
The door you picked originally either has a car or a goat, and the
other door either has a car or a goat. So it's still 50/50.
Now if I chose a door, then Monty asked me if I wanted to change,
BEFORE revealing what is behind a door, then I'd agree you have a 2/3
chance on winning the car.
Two doors left, One prize, one booby prize = 50/50 chance of getting
the prize. Look at it this way, I put a quarter in my hand while it's
behind my back, I ask you to choose which one has the quarter in it.
You have a 50/50 chance of getting it right. Does my asking if you
want to change your choice better your odds? No. It's still 50/50.
-Otter
Dr. Randall Flagg
Yes, but there were THREE doors. You need to revise your illustration to
include a third hand.
Choose 1 of three hands. The odds of getting the quarter are 1/3, correct?
_____________ ____________________
| 1 hand = 1/3 | | 2 hands = 2/3 chance |
| chance | |
|
------------------ -----------------------------
If you picked the hand with the quarter, then you lose by switching.
QUARTER EMPTY EMPTY
If you picked an empty hand, then the quarter is in one of the other 2
hands.
The odds of the quarter being in one of the other 2 hands is 2/3, correct?
EMPTY QUARTER EMPTY
EMPTY EMPTY QUARTER
|_____________________|
|
Do you still agree there's a
2/3 chance that the quarter is in
one of these two
hands?
If were're in the first case (EMPTY, QUARTER, EMPTY), I open the third hand
because I know it's empty. In a sense, the remaining closed hand represents
both of these hands.Or a bigger hand now holds both hands:
EMPTY QUARTER
EMPTY QUARTER
Considering this, go back to the beginning. With three hands the quarter
must be in one of them. This gives us three cases:
Hand you choose Other hands
============= ===================
1. QUARTER EMPTY EMPTY
2. EMPTY QUARTER EMPTY
3. EMPTY EMPTY QUARTER
Case 1: Stick: WIN Switch: LOSE
Case 2: Stick: LOSE Switch: WIN
Case 3: Stick: LOSE Switch: WIN
If the chart isn't helping, then wwo final comments:
1. This is not the same as simply presenting two hands. The number of hands
in the beginning is important. The hand you chose has a 1/3 chance (which I
think you'll agree). The hand that remains, however, doesn't have a 1/3
chance. I'm not opening a hand at random; I'm purposely opening an empty
hand. I'm not expressing this eloquently, but this is where the confusion
is. That one closed hand represents the two hands you didn't choose.
OK, in a sense, here is what we're doing:
I have three hands. In each is a piece of paper. On one piece of paper is
written CAR; GOAT is written on each of the other two pieces of paper. Each
piece of paper is a chance of winning.
You pick a hand, say the first hand; this hand has one chance to win. I
(figuratively, not literally) put whatever is in the third hand into the
second hand, and drop the third hand. Thus, the second hand now has TWO
chances in it, while the hand you chose has only 1.
Are you with me so far? one chance in one hand; two chances in the
other hand
Now, you can keep the hand with one chance, or switch to the hand with two
chances.
If you stick, you have one chance of three of winning the car.
If you switch to the other hand, you of 2 chances of three of winning the
car. You get BOTH chances in the second hand; if one is bad, we throw it
away and you get the other chance. All Monty did was to throw the bad chance
away for you.
If that didn't help, my second final comment is:
Read Algomeysa2's post where this was taken to an extreme for illustration
of the concept. Instead of 3 doors, the illustration used 1000 doors. You
might need to read that post several times. (Same thing, 999 chances are put
behind a single door; you get them all, throwing away the bad ones.)
George Smith
>the doors with the goat is revealed, so there are two doors remaining,
>THEN you are asked if you want to change.
Yes, but when you chose a door, you had a one in three chance of getting the
right door, meaning that the host had a two-thirds chance of having the car
behind oner of the other doors. Now, he's *always* going to open one door with
a goat behind it, so the other door still has a two thirds chance of having the
car.
Look at it like this. The fact that you know what the original odds were and
what your chances were make the difference in the odds for you. If you were
walking into the choice cold after one door is opened with no knowledge of what
door was chosen in the first place, yes, you'd have a 50-50 shot at picking the
right one. But *knowing* which door was taken originally, thereby locking in
the one-third odds on that door, gives you the edge.
jean.paulo
> nope.
>
> http://www.randomhouse.ca/readmag/volume4issue1/excerpts/curious.htm
>
>
How positive !
We were speaking about :
Curious incident of the dog in the night time.
Ridle of the TV game show.
The book is definitively not honest with that riddle. It is also not globaly
honest when the author pretend to be an autist writing a book.
I would be very surprised if he IS one.
The ridle is based on TWO situations not explicitly defined:
Either you KNOW that you'll have the choice to switch, or you don't.
If you KNOW, then you are presented with two unegal choices :
- You choose never to switch. In this case, you select one door out of
three, and there is
1/3 chances to win.
- You choose to SWITCH from the beginning (But this imply that you KNOW THE
RULE beforehand)
Then you are in fact picking TWO doors in the same time, and one of them
will always be
eliminated (and always a bad one)
So this clearly gives you a 2/3 chances to win. And that is the false answer
given by the book.
But all the beginning of the ridle implies that YOU DON'T KNOW that you will
be able to switch, and that it it happens only ONCE.
The book says :
'Then (after the door is opened) he ask you if you want to change your mind'
At this point, you're still in a context when you obviously DID NOT KNOW
before that you'll be able to change.
Then, abruptly, the question changes to 'when you always switch'. But this
could only happens if you knew the rule.
In other words, the presentation implies that if 1000 people are playing the
game, none of them have been told the option.
Then the conclusion implies that they ALL know that they WILL have the
option to change.
I have written a small VB program to test random cases. And to my surprise,
the program
DID show 650 win and 350 loss IF YOU SELECT ALWAYS Switch. And only 350
win if you never switch.
Now, I understand why.
But the program also shows that if you don't A PRIORI select 'always
switch', but that the
unknowing gamer choose WHEN the option is offered, then there are 500 win
and
500 loss.
All the diagrams in the world will never show you that the answer depends on
WHEN you are made aware of the rule.
This is NOT a mathematical curiosity, it is just a cheat !
Dark Tower IV used 'dirty' ridles to get rid of the evil car. This is one
such dirty ridle.
Or 'What do I have in my pocket' ridle of Bilbo
A more cleaner one is that of the Sphinx :
You are before two doors. One will lead you to your safe destination, the
other to death.
The Sphinx tell you that there are two people in front of the door (Pierre
and Paul).
They will answer ONLY ONE question. One of them is always telling the truth,
the other one always lying. Of course, you don't know which one !
How can you win your live ?
Or a more mathematical ridle :
1 + 1 + 1 = 10
When ?
--
Jean.paulo
Harry Teague
> >
>
> The book is definitively not honest with that riddle. It is also not
globaly
> honest when the author pretend to be an autist writing a book.
> I would be very surprised if he IS one.
The author of the book (Mark Haddon) never claims to be autistic himself -
read the "postscript" to the book. He has, however, extensively worked with
autistic people.
As the uncle of an autistic nephew, I'd say that his portrayal of the
behaviour and thought processes of autistic children is extremely realistic.
Regards,
Harry.
jean.paulo
"Harry Teague" <ha...@nowhere.com> a écrit dans le message de
news:qIyIc.3842$tY2....@fe46.usenetserver.com...
The copy I got (Vintage UK) does not have any postscript, just an appendix
with various mathematics.
--
Jean.paulo
Harry Teague
>
> The copy I got (Vintage UK) does not have any postscript, just an appendix
> with various mathematics.
>
I tell a lie; it's not a postscript to the book; it's the "about the author"
information on the inside cover.
Regards,
Harry.
wotsayyou
> Or a more mathematical ridle :
>
> 1 + 1 + 1 = 10
> When ?
>
When you are adding numbers that are base 3 and not base 10.
--
> Jean.paulo
>
>
>
wotsayyou
Say the left hand door is the safe one.
Ask Pierre this question
"If I ask Paul if the left hand door is safe, will he say YES ?"
If Pierre is a liar, he will have to say that Paul would say "No" (since
Paul would actually say "Yes", as he always tells the truth).
If Pierre is not a liar, he will have to say that Paul would say "No" (and
indeed Paul would actually say "No", as he always lies)
jean.paulo
0 +1 = 1
1+ 1 = 2
2 + 1 = 10
--
Jean.paulo
"wotsayyou" <jonho...@hotmail.com> a écrit dans le message de
news:LYUIc.824$X0...@newsfe1-gui.ntli.net...
jean.paulo
--
Jean.paulo
"wotsayyou" <jonho...@hotmail.com> a écrit dans le message de
news:RYUIc.825$X06...@newsfe1-gui.ntli.net...
>
> "jean.paulo" <Jean.pau...@free.fr> wrote in message
> news:cctoqf$oui$1...@news-reader1.wanadoo.fr...
> > >
> > > nope.
> > >
> >
> >