I read a book yesterday - even tho there was hardly any pictures in it.
It won the whitbread prize and is an excellent read - The Curious Incident of the dog in the night time - by Mark Haddon.
In it is a rather stange maths related question on probabilities.
Its called the monty hall problem.
it goes like this.
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.
do you stick, or change to get the best possibility of winning? Or is it the same?
atb
steve
i,d keep the goat
i'd whack him with a kipper and tell him to open the bloody door i chose.
Ned.
quote:
Originally posted by theconrodkid
i,d keep the goat
Stevo did you honestly expect a serious reply!
if the host doesnt change what is behind the doors aftershowing you his goat 
you will have 1 in 2 chance of guessing the right door if you opt for
the change your mind option, as opposed to a 1 in 3 chance of getting it right when you innitially choose which door to open.
hope that makes sense!
No point in changing...all it tells you is that you now have a 1 in 2 chance instead of a 1 in 3 when you started...
Seriously though,...the goat..naked or kinky undies on?
Damn Col....beat me to it...
hi.... this is in fact a serious reply to your question.
My girlfriend was studying Economic last year, and she posed me this question, although subtely different.
If I remember correctly, you should always change. this seemed really daft to me at the time, and although I like to think of myself as a smartish
guy, it stumped me for a bit.
Think she tried to explain it like she would to another girl!!
Anyways, will ask her again tonight, and see if i can confirm it.
theres another piece of useless information for anyone who ever enters a chatshow!!
Mark
quote:
Originally posted by Alan B
Seriously though,...the goat..naked or kinky undies on?
Col, you're right,supposedly mathematically you've got a better chance if you change your mind but I dont beleive it! Loads of people have
written computer simulations to prove it but I still think its rubbish.
Now, if it was a sheep behind the door it would be awhole nother matter...
As Colmaccol has already said; the answer is you should always change. It's one of the probability questions that always messes with heads.
If you stick with your original guess then you have a 1 in 3 chance of being right, if you change then you have a 1 in 2 chance of winning.
If you think about it too much steam will start to come out of your ears.
The other thing to consider is, do you want a goat or the car?
Other than this, is it a good book? I've been thinking of getting rid of some booktokens on it How would you rate it on the scale from Robin
Hood to Dax?
Eddie
when he opens the door he reduces the odds of you having chosen correctly to 1 in 2. it is 50 50 either way.
all that changes is that one remaining door is a goat, and one is the car. its the same whichever you choose.
you are right - i did forget where I was posting this question ! 
I will post a piccie later on of the correct answer (decision tree) - some of you are close, but not exactly right........ you should change - but
what are the overall odds improved to?
atb
steve
[Edited on 29/1/04 by stephen_gusterson]
surely the only way the odds can change is if he opens the door you selected and says goat! do you wanna change?
im gonna have a think since i was sure just now...
edit: heres the crack!
you've picked one, seen a goat in another door, and then you have a choice. there are two doors to choose from, so from here it is 50 50 which is
the car. so if you stay with your original choice, there is a 50% chance you were correct, and if you choose the other, again there is a 50% chance
you will now be right. you cant choose the door he opened, so thats part of the chance gone.
the key is that, if you didnt tell the dude what door you had chosen, he might open the door you chose immediately. then you would have to change. as
such its not random which door he opens. not sure how that affects it but it seems relevant in some way.
it seems pretty insurmountable, almost like a paradox, and paradox's dont exist in my books! at first it is definately 1 in 3, and at the end it
is definately 1 in 2. but changing will make no difference cos its already 1 in 2 whatever you choose.
its like the lottery, when the first 5 balls have been matched its down to 1 in 37 ish, not 1 in 14m, but changing the last ball makes no difference
then...
this reminds me of that ridiculous question about a dinner lady trying to return change to 3 people and the mad sum that results, if you remmeber that
one...
anyways steve, do you have a tree of this? its quite complex due to man opening the door not being random.
[Edited on 29/1/04 by JoelP]
Steve and all,
I will take a LOT of convincing that by changing your choice you have improved your chances.....as I see it is still 50:50 either door.
damn right alan!! i'd love someone to prove me wrong though cos it is quite baffling.
how come we always seem to agree on stuff though?!?
Easy mate...cos' we're f*****g right...
Ok,
I had this figured out a minute ago, but am all confused now!!
At the start the probability of you being correct is 0.33 (ish), i.e. a 1 in three chance of success (lets say you pick door number 1). That
consequently means that there is a 2/3 chance of being wrong.
So, there is 2/3 chance that the other two doors contain the car(either in door number 2 or number 3).
The guy tells you one of the other doors (number 3 for example) is a goat, which takes it away, meaning there is a 2/3 chance that the remaining door
(number 2) has the car.
Also follows, if in doubt, go for the middle one!!
Mark
[Edited on 29/1/04 by 9904169]
quote:
Originally posted by 9904169
The guy tells you one of the other doors (number 3 for example) is a goat, which takes it away, meaning there is a 2/3 chance that the remaining door (number 2) has the car.
Mark
[Edited on 29/1/04 by 9904169]
not really.
ok, between doors two and three there is a 66% chance, ok, as its two out of three. All that effectively happens is that as door three is removed,
all of that 66% chance transfers to door two.
Crazy I know, but true.
should put this in as an edit I know, but tuff.
being wrong from the beginning doesnt come into it, as at the start, with no other information, you have a one in three chance. So of course most of
the time (67%) you will be wrong.
it IS a 2 in 3 chance.
I will let you stew a bit and post the answer later.
apparently when this concept first came out it caused an uproar in the mathematicw world.
but the drawing I will post will prove its 2 in 3
atb
steve
ouch, have just hurt my arm trying to pat myself on the back
I think it's all about seperating the right odds from the wrong ones
[Edited on 29/1/04 by 9904169]
The answer is 42.
Simple!
Ian
Are these big doors to hide a car behind or small doors to hide a goat behind or is it a small car
. Personally I would use Bayes' theorem
depending on which door was opened to lower the odds on you getting the right choice.
P(C|host opens B) = p(C)*p(host opens B|C)/p(host opens B)
= (1/3)/(1/2)
= 2/3
This of course presumes you chose door A and were shown behind door B In which case you should change choice to door C.
yours, Pete.
A=1 B=2 C=3 and P is probability constant
[Edited on 29/1/04 by Peteff]
[Edited on 29/1/04 by Peteff]
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...
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
Steve, I think we have VERY polarized opinion on this one...
Now about the goat?....
well, im crap at maths - so pick the bones out of this
atb
steve
Rescued attachment big1.jpg
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
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]
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]
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....
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]
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.
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]
indeed (to my brothers in the 50:50 gang).
if anyone is still confused about this tomorrow I will draw a real probability diagram.
the more i look at it the more i think u r wrong Prof Gusto, i think u have been had, sry mate
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
quote:
Originally posted by stephen_gusterson 80% complete.....
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!
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]
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...
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 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?
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....
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?
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.
That sounds like it was written by someone with their head screwed on right.
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]
Two things:
1 - the book is a good read
2 - have a look at this URL for an illustrated explanation of the solution. My simple summary goes
like this:
At the beginning whatever door you choose has 1/3 chance of being right and 2/3 of being wrong. If you choose A, then B and C together have a 2/3
chance of being right since A is 1/3 and the probabilities must add up to 1, because we know there is a prize. It's important to consider B and
C together as if they are one door with a 2/3 probability of being right. Once Monty opens one of those two doors to show you a goat, that
doesn't change the 2/3 probability that the pair of doors hides the prize. It does, however, reduce the number of doors from two to one. Your
original choice still has 1/3 chance, and the other door is 2/3. The important point is that one losing door has been eliminated AFTER you made your
initial choice. That supplies you with extra information which is why the choice isn't a simple 50:50.
Hope this helps. Read the page referenced above or do a google search for lots (and lots) more pages on this problem. It's a great example of
how bad the human brain is at probabilties and how wrong 'intuitive' answers sometimes are.
Dominic
i must, unfortunately, abandon ship!!!
because it dawned on me last night in bed that i am wrong. so i apologuise for so ashamedly arguing the wrong point!
the reason i am wrong is because if you pretend there are 100 doors, you are almost certainly wrong whatever you choose. so when they reduce it to two
doors, you are still almost certainly wrong, so if you change to the other option you will almost certainly right!
damn i feel daft now...
its because 1/2 and 1/3 are so similar, but with 1/100 it seems more clear....
Later,
Having had a series of arguments with some co-wokers about this I did two things:
- I analyzed the problem again and came up with my own version of the decision tree diagram that Stephen G. posted. The one from the book is
WRONG!
- I wrote a quick "simulation" of 10,000 turns and came up with the following average figures; stick with your choice -> win car 50% of
time; change choice -> win car 50% of time.
I hereby withdraw my support for the argument If anybody wants a copy of my corrected decision tree diagram or the Excel spreadsheet I did the
simulation in, let me know.
It does depend exactly how the question is asked. If the door the host opens is ONLY EVER one with a goat behind it, then it's 50:50. If he
just opens one of the other two doors and there HAPPENS to be a goat then changing IS in your favour. In that circumstance he could also open a door
and reveal the car. Ading that possibility of losing changes things a bit.
Dominic
oh dear...
ive done a tree diagram, and it agrees with steve. i dont have a scanner so im gonna do it in paint now... back in 10.
[Edited on 30/1/04 by JoelP]
Gecko - are you sure that your excel sheet is correct? I'm sure that the answer is change for exactly the reasons that you stated in your
previous response i.e. 2/3 chance of success if you change (and not 1/2 as I wrote in a stupid moment on page 2 of this thread).
After being shown a wrong door you are really making a choice between
your original choice of 1 door (probability 1/2)
OR the other 2 doors (probability 2/3).
The other interesting things is people seem very quick to believe what they read on the internet (are any of these people who say they are, really
professors) and that plenty of professors know nothing outside their area. Why should a Prof of English Lit. know about probability?
I'f have thought that people on here would have finely-tuned bullshit detectors otherwise you could take some dubious/dangerous advice on putting
your car together.
There really is no substitute for using your own head.
Eddie
Sorry to have to say this but - your all talking bollox
the seccond choice between 2 doors is a seperate probability question and cannot be associated with the first. therefore the choice is 50/50.
(and I have a degree in modelling with mathematics - so there )
Dave
ho hum, dont we love discussions!
protofj, do you believe that the initial choice is 1/3, becoming 1/2 when he removes a goat?
Not beleive - I KNOW!
yes the initial choice is 1/3
but it does not BECOME 1/2
the second choice is a completely seperate question and cannot be associated with the first.
the second choice is 1/2
Dave
quote:
Originally posted by TheGecko
Later,
It does depend exactly how the question is asked. If the door the host opens is ONLY EVER one with a goat behind it, then it's 50:50.
Dominic
quote:
Originally posted by JoelP
oh dear...
ive done a tree diagram, and it agrees with steve. i dont have a scanner so im gonna do it in paint now... back in 10.
[Edited on 30/1/04 by JoelP]
you are right to a certain extent - and believe me, after my recent folly i dont say that lightly!
the second choice isnt between random events.
if he jumbled up the doors at the second choice it would be pure 50 50 to get the car. but the two doors arent random. one you chose initially, and
that door had a 1/3 chance of having a car. the other door, he effectively choose for you. remember that at the end there must be one car and one
goat. so if you had correctly guessed the cars door, then the door he presents will be a goat. if you choose a goat, he will offer the door with the
car. as there was a 1/3 chance of you picking the car, there is a 1/3 chance he will have to offer a goat, and if there was a 2/3 chance you picked a
goat, there is a 2/3 chance he will have to offer the car, as at the end there must be one car and one goat.
when i say offer, i mean he takes one and leaves one. so the door he doesnt remove has a 2/3 chance of being a car.
AND FINALLY!
imagine there are 100 doors. you pick one, you're pretty gutted cos theres only a 1 in 100 chance you got it right. he then says, its either the
one you chose, or this one here, pointing at another. this is the same as removing 98 other doors. so you think, 'bloody hell! it was a very long
shot that i choose correctly first time, and if i was wrong he would have to offer the door with the car behind it, otherwise i couldnt win.' so
there is a 99/100 chance that the door he offered is the winning door.
and if no one believes that then fair doos to you, cos yesterday i didnt believe it either!
there is a 1/3 chance that everyone is talking bollox - or is that a 2/3 chance that someone is talking bollox?
Dave
remember the greatest minds in the world once thought the earth was flat
quote:
Originally posted by protofj
there is a 1/3 chance that everyone is talking bollox
Dave
remember the greatest minds in the world once thought the earth was flat
like the new sig though! oops its not a sig, oh well... It is now
Dave
How about at the next kitcar (remember those?) show three people bring along a door each, I'll volunteer my cat (as I don't have a goat)
we'll need one other item of livestock and we can conduct an experiment.
It should only take a couple of dozen goes to see which way the chances take us. Personally I'd place a good amount of money on the 2/3 win if
you change choice.
Any takers. I promise that I my cat isn't in on the trick.
Eddie
funny thing is, last night i was thinking of betting my car!!! luckily i didnt...
quote:
Originally posted by Cussed
we'll need one other item of livestock and we can conduct an experiment.
Eddie
LOL, fun thread. My thinking is similar to one that was posted earlier, but maybe rephrased will help:
The host will always show you a door after you've picked: it will not be the door you've picked, and it will always have a goat behind it.
This is possible since regardless of which door you pick, there will always be a door left for him with a goat behind it. Conclusion: you're
really picking from 2 doors, not three. Two doors, so it's 50/50, no matter which door you pick - changing choices or not.
Best odds are this puzzle was intended to get your goat .
My son brought this home yesterday:
The waiter brings a bill of $25 to a party of 3, who each pay with $10 bills totalling $30. He comes back with $5 change, gives each $1 and keeps a
$2 tip. So each patron has paid $10-$1 = $9, or $27, plus the $2 tip making $29. Where's the other dollar of the original $30?
i know that one! just deleted it though cos i dont want to spoil another 8 pages of fun!!!
Steve you still havent said whether this is an attractive goat or not.
Cos if its a tug Im betting nowt
Bob
[Edited on 30/1/04 by splitrivet]
just 2 add a bit more, but
i think that, there r 2 ways of reading the question
if u take the odds, of getting the car, at the beging, and no what is going 2 happen, if u change i think it could be 2/3, but if u take the odd,
after he has opened the door, the odds r 1/2,
like i say i am probly wrong, but it is 2/3 at the beging, and 1/2 after the door opens
phil
This logic problem is a variant that has been going round quite a while.
Another version is:
Tom, Dick and Harry are in prison. One of them is scheduled to die in the
morning, and the other two will be set free. Their guard knows which one
will die, but none of the prisoners does. The guard is under strict
instructions not to divulge the identity of the doomed man. Tom is desperate
for any information beyond the fact that his odds of death are one in three.
He begs the guard to throw him an informational bone. Finally, to shut him
up, the guard agrees to reveal only the following: the name of one of Tom's
fellow prisoners who will be set free rather than killed. The guard then
says that Dick will be set free. After the guard's revelation, what is the
probability that Tom will be killed?
None of the other options is correct.
One out of two.
One out of three.
Two out of three.
answer in a mo
First reaction is to say that the odds have changed to 2/1
However, the original odds must remain unchanged. To explain this further
its better to expand it: Imagine that you have in front of you 100 empty
boxes. I put a prize in one of the boxes (without you knowing which one).
You then get to choose a box.
What are the odds that you have the prize (A:100/1)
I now open 98 of the boxes in front of you and show you these are empty, so
there are just two boxes left: the one you originally chose and the one I
have.
So what are the odds that you have the prize? it is not 2/1 it remains
100/1 - the original odds must remain
Same with the prisoner and the same with the original posting
arguably?
Anyway if you want to fry your brains further try this link
http://www.funtrivia.com/dir/5449.html
some are rubbish others are pretty good
chris, in a nut shell you have summed it up.
with your prison question, it is 1 in 2 because the options are random. it also did not affect his chances of survival asking that question, because
the sentancee was decided in advance. afterall, he could've said that the prisioner asking the question was to be set free, which would be like
the presenter revealing the car to the contestant in the first problem.
[Edited on 30/1/04 by JoelP]
well, one goat is called kylie, and the other agatha.
which one do you prefer
atb
steve
quote:
Originally posted by splitrivet
Steve you still havent said whether this is an attractive goat or not.
Cos if its a tug Im betting nowt
Bob
[Edited on 30/1/04 by splitrivet]
I have to say, it depends on the way you look at it....a daft statement but true.
Saying its 50/50, refers to a different question/puzzle than saying its 1 in three or 2 in three. There are three different questions withing the
original, and I think peopel are answering different parts..
anyone agree??
Mark
quote:
Originally posted by 9904169
I have to say, it depends on the way you look at it....a daft statement but true.
Saying its 50/50, refers to a different question/puzzle than saying its 1 in three or 2 in three. There are three different questions withing the original, and I think peopel are answering different parts..
anyone agree??
Mark
quote:
Originally posted by JoelP
chris, in a nut shell you have summed it up.
with your prison question, it is 1 in 2 because the options are random. it also did not affect his chances of survival asking that question, because the sentancee was decided in advance. afterall, he could've said that the prisioner asking the question was to be set free, which would be like the presenter revealing the car to the contestant in the first problem.
[Edited on 30/1/04 by JoelP]
as I said, i kinda think this question has compounded odds, as choosing twice affects the result.
atb
steve
still am crap at maths tho.....
If the doors are hardwood, I'd quite like to win one of those.
I think therefore my odds of picking a door would be 1/1.
I was reading about this very problem in a book about statistics recently (yes, I know, sad git).
What isn't mentioned in the original problem is that this is based on a real TV programme. As such, this introduces a number of variables. The
author of the book listed 3 assumptions that have to be considered:
The first assumption is that Monty always gives guests a chance to switch, or at least that whether he gives them a chance does not depend on which
door they choose. For instance, if he offered a chance to switch only when the guest picks the door with the car, then not switching is obviously a
winning strategy.
The second assumption is that Monty always opens a door with a goat, never the door with a car.
The third assumption is that Monty makes all other relevant choices randomly, including the choice of behind which door to place the car and the
choice of which door to open when he has a choice between two doors with goats.
He reckoned that these assumptions mean that switching increases the contestant's chances of winning from 1/3 to 2/3.
This was illustrated with an example:
"Picture yourself in the role of Monty rather than in the role of the guest. In other words, imagine standing behind the doors rather than in
front of them. As the host, you know where the car is. Assume the car is behind door 3 and the guest has already made her first choice. Three
scenarios are possible. If the guest picked door 1, you are forced to open door 2. If she now switches, she will win the car. If the guest picked door
2, you are forced to open door 1. If the guest now switches, she will win the car. If the guest picked door 3, you will randomly open either door 1
or door 2. Out of the 3 scenarios, only in this last one will the switching result in the guest's not winning the car. In short, a guest who
switches will win in two out of the three cases, and a guest who stays will win in only one out of the three cases. Through Monty's eyes, it is
easier to see that switching increases the chance of winning from 1 in 3 to 2 in 3."
It all depends on how you look at it!
David
P.S. the book is "Reckoning with risk", by Gerd Gigerenzer.
[Edited on 30/1/04 by David Jenkins]
If anyone is keeping a count, I'm switching camps - I've been convinced by the "always change choices" crowd. It finally clicked
for me that Monty is doing a directed trim of the probability tree. Cool that it's so non-intuitive. And I also think Harry should be putting
his affairs in order...
Fun one Steve, thanks!
.........is it possible to restart the nutsack thread that was recently running?.........
what type of car was it anyway ??
bollox...
Ooooh - what with the trendy new forum-front-page already spoiling us, I just found this old thread. Cant believe I missed it, especially considering
I have the book (and rather enjoyed it).
I'm on the 2/3 camp (the correct one ). Intuition tried to fool me at first saying it doesn't matter whether you change your mind or
not, but with a bit of thought (not to mention the various proofs around) the answer is pretty obvious (imho).
Anyone saying the odds are 50:50 isn't completely answering the question. Yes, the final decision, treated completely seperately is a 50:50 one,
but that's an over-simplification of the problem. The problem is asking what is the overall probability of winning the car if you change
your original desicion. The answer, 2/3, is a product of the initial decision and the important information the host gives you (which is essentially
telling you exactly where the car is 2/3 of the time!).
Look at it this way: Imagine you play the game a million times. 2/3 of the time you initially choose a goat, right? So of course you have to change
your mind - dur!!! And in this case, the host then reveals the other goat (damn nice of him) so you can choose the car with 100% certainty.
So 2/3 of the time you have a probability of 1 of winning the car by changing your mind. The other 1/3 of the time, you initially chose the car so
you will loose it by changing your mind. That's the complete answer. Seems clearer to me every time I think about it.
This interesting thread then left me with a perplexing problem... what is the error in the analysis from Alan B's 'professor' which
seems to support the 50:50 argument? I couldn't see it, but a bit of thought and now i can - his analysis doesn't support the 50:50
argument! He has listed 8 'paths' and they're all correct, but his 8 paths dont all have equal probabilities. Your initial
choice is A, B or C, 1/3 probability each, but he has listed 4 paths for when A is chosen and 2 for each of the others. It's quite correct, but
the first 4 on his list have half the probability of the last 4. If you take into account the unequal probabilities of his 8 paths then you reach the
correct answer - considering all the paths in which you change your mind, the probability of getting a car is 2/3, and of getting a goat is 1/3. Any
statisticians care to check that?
The professor's 'tree' is essentially identical to the one in the book, just broken down more by considering each goat separately. The
paths in the book 'tree' all have equal probabilities.
Easy peasy
Liam
P.S. Any computer programmers getting 50:50, is your program doing what it should? To answer the question (should you change your mind) the program
must...
Have the 'contestant' choose a box at random.
Have the 'host' eliminate a goat from the remaining two boxes (in the case the contestant chose a goat, you have no choice here. In the
case the contestant chose the car you can remove either remaining box).
Have the contestant 'change their mind' and choose the remaining box instead of the first choice.
Do this and you winthe car 2/3 of the time as you should. If you simply run through the 'game' with the contestant making random desicions
you will win the car 50% of the time (as the probability tree in the book shows), but this does not answer the question - we want to know the odds if
you change your mind (or not).
Liam
I just joined your forum and considering I have always been interested in math and science I read this topic with interest. Glad to see you came up with the right answer; always switch. I believe Marilyn Vos Savant got more mail about this question than any other question with math professors among others telling her she was wrong, but it is fairly easy to prove that as long as the host doesn't pick the door with the car himself you will be twice as likely to win if you switch.
My better half has just finished that book.