Alan B
|
| posted on 29/1/04 at 09:41 PM |
|
|
A friend of mine who is professor of mathematics says no point in changing it IS 50:50......
They are two events which are NOT connected
Take it further...say there were 10 doors...and it was narrowed down one at a time...whether you change at any time or not it will still come down to
a final 50:50 chance...
|
|
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 09:48 PM |
|
|
alan
the book i will scan from shows quotes from several phd's that said it was wrong too - but the chart says otherwise.
i think its kinda compounded odds, rather than odds at the start...
atb
steve
|
|
|
Alan B
|
| posted on 29/1/04 at 09:52 PM |
|
|
Steve, I think we have VERY polarized opinion on this one...
Now about the goat?....
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 09:55 PM |
|
|
well, im crap at maths - so pick the bones out of this
atb
steve
 
Rescued attachment big1.jpg
|
|
|
I love speed :-P
|
| posted on 29/1/04 at 10:01 PM |
|
|
is this me being thick, or something, but anrt they different Scenarios the origanal said that He then opens a door, or simlar but in the images u
have scanned it doesnt sy anything about getting rid of one option therefore, if u do get rid of one option, the odd go 2 50:50, now i am probly wrong
but i still tried
p.s. sry if that doesnt make sence
phil
Don't Steal
The Government doesn’t like the competition
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 10:10 PM |
|
|
the question was:
3 doors, make a choice (3 to 1)
its not that door
there are two doors left
confirm original choice or change (2 to 1 regardless of change or stick - supposedly)
thats it.
[Edited on 29/1/04 by stephen_gusterson]
|
|
|
blueshift
|
| posted on 29/1/04 at 10:12 PM |
|
|
I can't believe everyone is so confused about this. Quite clearly there is no reason to change you choice afterwards, to suggest otherwise is
quite absurd.
I have AB maths and further maths A-level and a degree in computing, for what it's worth. Probability was always one of my strengths.
Imagine this situation: you have two contestants instead of one. Contestant A picks door (box, whatever) number 1, contestant B picks door 2. The host
opens door 3 and reveals a goat. Now should both contestants change their choice? will they both be more likely to win a car if they do? of course
not. When the host reveals the goat the probability that you chose the correct door changes to 1/2. there is one correct option out of two possibiles,
where before it was one correct option from three possibles. one of the possibilties has been eliminated.
[Edited on 29/1/04 by blueshift]
|
|
|
Alan B
|
| posted on 29/1/04 at 10:13 PM |
|
|
Hmmm..impressive, but bollox IMO...
By the time you reach the final choice one of the three branches with a bad choice has been eliminated so it's still 50:50....
|
|
|
ijohnston99
|
| posted on 29/1/04 at 10:15 PM |
|
|
To quote Prof Gusto,
Thats bollox!
All the chart does is show that he's wrong.
There are 2 other branches missing from there and those are the ones that say.
You stick you've won a car!
The odds are still 50:50. Showing a chart and quoting some crap formula doesn't make it any different.
Also putting some none existant quotes from professors doesn't make it right either.
All it proves is that if you put bollox together like this. Some mug'll believe it.
It's like saying you can build a car for £250, just because it's in print doesn't make it true.
Dr Johnston
[Edited on 29/1/04 by ijohnston99]
|
|
|
I love speed :-P
|
| posted on 29/1/04 at 10:16 PM |
|
|
quote:
Originally posted by stephen_gusterson
You are in a game show. you can win a car as a prize. The game show host tells you there is a car behind one of three doors. The other two doors have
goats behind. You choose a door. He then opens a door which isnt the one you chose. There is a goat there. The host asks if you want to change your
mind.
This infers that there r 3 doors, 1 open and 2 closed, the open door has a goat behind it, so this leaves 2 doors, one with a goat and one with a car,
no matter what u do, there on r only 2 doors and one option, which has not been influanced by the 1st door opening, therefore it can only be 50:50
p.s. sry agin if it doesnt make sence
Don't Steal
The Government doesn’t like the competition
|
|
|
CairB
|
| posted on 29/1/04 at 10:18 PM |
|
|
Shouldn't you lot be building cars?
Steve,
From your avatar it looks like the guy next door has finished his.
No offence mean't.
PS - I'm with Blueshift etal on this one.
You can create any incorrect result with a flawed model.
Cheers,
Colin
[Edited on 29/1/04 by CairB]
|
|
|
blueshift
|
| posted on 29/1/04 at 10:18 PM |
|
|
indeed (to my brothers in the 50:50 gang).
if anyone is still confused about this tomorrow I will draw a real probability diagram.
|
|
|
I love speed :-P
|
| posted on 29/1/04 at 10:24 PM |
|
|
the more i look at it the more i think u r wrong Prof Gusto, i think u have been had, sry mate
Don't Steal
The Government doesn’t like the competition
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 10:38 PM |
|
|
I didnt imply I thought it was right or wrong. I said my maths was crap.
I have just put it up to see what others thought.
I think its likely to be 50 50 too - the only way it can be made otherwise is due to it being a double choice, compounded with the original odds. but
at the time you make the final choice, its clearly 50 50.
now, CairB...
my next door neighbour bought his VW ready made
I am however the owner of an 80% complete locost derived car.
In 3 years on lists like this I can take a daft comment or two 
my fave word applies here methinks
quote:
Originally posted by CairB
Shouldn't you lot be building cars?
Steve,
From your avatar it looks like the guy next door has finished his.
No offence mean't.
Colin
[Edited on 29/1/04 by stephen_gusterson]
[Edited on 29/1/04 by stephen_gusterson]
|
|
|
Alan B
|
| posted on 29/1/04 at 10:48 PM |
|
|
quote:
Originally posted by stephen_gusterson 80% complete.....
So, as someone else on here said.....only 80% to go then...
|
|
|
JoelP
|
| posted on 29/1/04 at 10:52 PM |
|
|
ok, i'll reserve judgenment till the beer wears off, but i must suggest this:
imagine he said in the beginning, 'make a choice out of three, but whatever you pick im gonna get rid of a goat in a minute anyway.'
effectively there is only one goat from the beginning. if you forget about him revealing a goat, it is 1 in 3, but since he is gonna get rid of one
goat, even if you pick the other goat, it becomes 1 in 2.
however, the diagram in the book does merit a few minutes thought, cos if there are no flaw or omissions then i am wrong.
BTW im a clever twat myself, or so i thought till earlier!
|
|
|
JoelP
|
| posted on 29/1/04 at 10:58 PM |
|
|
the basis of the flaw...
ok. after a few minutes thought...
essentially, what ever you pick there will be a goat and a car at the end. on his diagram, there are two lines for goats, but at the end there is only
one goat. hence the opposite (changing ones mind) is a car twice over.
i cant reason it thru to the end (the aforementioned beer...) but thats something to do with it.
my main argument against this is this. the act of changing ones mind alledgedly changes the odds from 1/3 to 1/2. So what if you changed it back
again? it must remain 1/2 because there is really only 2 doors to choose from. And the same choice cant have two different odds.
again, reason leaves me, i suspect this is why the problem remains debated, waiting for some clever twat to prove it either way.
i'll edit this in the morning anyway...
infact, now will do for the edit!
the chart is crap! there is no mention of a goat being revealed!!! so obviously a 2/3 against becomes a 2/3 for when you change your mind! TBH, the
lunatic that first suggested this 'problem' would've been laughed out of the university that gave him his phd! its nonsense!!!!
[Edited on 29/1/04 by JoelP]
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 11:01 PM |
|
|
I wouldnt even think of disagreeing with you Alan 
quote:
Originally posted by Alan B
quote:
Originally posted by stephen_gusterson 80% complete.....
So, as someone else on here said.....only 80% to go then...
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 11:03 PM |
|
|
quote:
Originally posted by JoelP
ok. after a few minutes thought...
essentially, what ever you pick there will be a goat and a car at the end.
[Edited on 29/1/04 by JoelP]
No
there is a 1 in 3 chance at the start that you pick the car and you nver have to be shown a goat.......
atb
steve
|
|
|
CairB
|
| posted on 29/1/04 at 11:03 PM |
|
|
No problem Steve, just sayin what I see,
But... can anyone help me out?
I want to build a shelving system with 7 mechanisms that can transport items from one shelf to another in my garage.
Each mechanism can only stop at 6 shelves max.
How many shelves high can I have with the restraint that I must be able to get an item from any shelf to another by only using one of mechanisms in
the move?
|
|
|
JoelP
|
| posted on 29/1/04 at 11:19 PM |
|
|
again i say, whatever you pick, your choice will be present at the end, along with one other. 50 50. simple as.
if he just opened your door, and said right or wrong, then it would remain 1 in 3. but that isnt going to happen cos whatever you pick he is gonna
remove one goat, so at the end it becomes 50 50 again.
steve, i am sorry for arguing so fanatically about this, but the issue that decided it for me is the fact that changing your mind does not change the
fact there are two doors and one car at the end.
there are times when you sit here thinking am i right or wrong, someones got the wrong end of the stick.
the only reason i argue rather than shutting up is that i am rather good at maths, i got A grade GCSE when i was 11. dont mean to sound clever,
y'all know im pretty crap with cars, but maths is my thing.
can someone please explain how one car and two doors can possible not be one in two? thats the crux of the matter, the crap about revealing a wrong
answer is simply included to muddy the water. thats the essence of this problem.
good night anyway, it'll be hours till i get this out of my mind....  
|
|
|
stephen_gusterson
|
| posted on 29/1/04 at 11:27 PM |
|
|
Surely, thats not the first passat you have seen to have to comment on it?
quote:
Originally posted by CairB
No problem Steve, just sayin what I see,
But... can anyone help me out?
I want to build a shelving system with 7 mechanisms that can transport items from one shelf to another in my garage.
Each mechanism can only stop at 6 shelves max.
How many shelves high can I have with the restraint that I must be able to get an item from any shelf to another by only using one of mechanisms in
the move?
[Edited on 29/1/04 by stephen_gusterson]
|
|
|
Alan B
|
| posted on 30/1/04 at 12:01 AM |
|
|
This is from another forum....(Preston north end mailing list actually......posted by a maths professor.....)
The easiest way to see it is to list the possibilities. Let's assume that the three items are A, B and C. You choose one, get shown another and
decide to stick with your original choice or choose the third. SInce the A, B and C are arbitary let's make A the desirable item.
Options:
1) Pick A, shown B, stick with A - CORRECT
2) Pick A, shown B, choose C - INCORRECT
3) Pick A, shown C, stick with A - CORRECT
4) Pick A, shown C, choose B - INCORRECT
5) Pick B, shown C, stick with B - INCORRECT
6) Pick B, shown C, choose A - CORRECT
7) Pick C, shown B, stick with C - INCORRECT
8) Pick C, shown B, choose A - CORRECT
8 options, 4 giving you the desirable result. Of those 4, 2 came from sticking and 2 from choosing.
|
|
|
blueshift
|
| posted on 30/1/04 at 12:13 AM |
|
|
That sounds like it was written by someone with their head screwed on right.
|
|
|
GO
|
| posted on 30/1/04 at 12:20 AM |
|
|
Nice one big Al.
We had this argument at work a few months back, it drove me crazy! Which is why I've sat back and watched the discussion flow.
I've always believed it must be 50:50.
What got me though was there's loads of simple computer programs people have written to illustrate the problem. Sad git (and excellent
procrastinator) that I am, I sat at my desk and clicked through one of these progs several (hundred) times, always sticking with the same choice, in
fact always sticking with the same door. When I finally conceded the point, shortly after my finger went numb, the results showed that I'd lost
almost exactly 2/3 of the times. I put it down to dodgy coding and that I'd write my own version one day to prove it was all hairy gonads. Never
got around to it though.
Anyone wanna write their own version, I know there's enough IT boys on here.
[Edited on 30/1/2004 by GO]
|
|
|
