TheGecko
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| posted on 30/1/04 at 01:29 AM |
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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
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JoelP
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| posted on 30/1/04 at 08:30 AM |
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Gentlemen!!!
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....
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TheGecko
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| posted on 30/1/04 at 08:31 AM |
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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
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JoelP
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| posted on 30/1/04 at 09:05 AM |
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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]
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Cussed
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| posted on 30/1/04 at 09:31 AM |
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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
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DaveFJ
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| posted on 30/1/04 at 09:33 AM |
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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
Dave
"In Support of Help the Heroes" - Always
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JoelP
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| posted on 30/1/04 at 09:42 AM |
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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?
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DaveFJ
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| posted on 30/1/04 at 09:54 AM |
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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
Dave
"In Support of Help the Heroes" - Always
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stephen_gusterson
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| posted on 30/1/04 at 10:09 AM |
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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
The question IS posed that a door is opened with a goat behind it. So, the odds count taking a goat into account as the first door.
atb
steve
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stephen_gusterson
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| posted on 30/1/04 at 10:11 AM |
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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]
Im sitting on a fence - I just posted what was in a book.
Its maryln vos savant that says its 2:3. This person is in the guniess book of records as having the highest IQ ever recorded. If you google her,
stuff comes up
atb
steve
[Edited on 30/1/04 by stephen_gusterson]
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JoelP
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| posted on 30/1/04 at 10:16 AM |
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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!
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DaveFJ
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| posted on 30/1/04 at 10:19 AM |
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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
Dave
"In Support of Help the Heroes" - Always
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JoelP
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| posted on 30/1/04 at 10:20 AM |
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quote:
Originally posted by protofj
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
i can live with that like the new sig though! oops its not a sig, oh well...
[Edited on 30/1/04 by JoelP]
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DaveFJ
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| posted on 30/1/04 at 10:25 AM |
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It is now
Dave
Dave
"In Support of Help the Heroes" - Always
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Cussed
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| posted on 30/1/04 at 10:30 AM |
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wager ?
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
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JoelP
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| posted on 30/1/04 at 12:27 PM |
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funny thing is, last night i was thinking of betting my car!!! luckily i didnt...
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splitrivet
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| posted on 30/1/04 at 12:34 PM |
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quote:
Originally posted by Cussed
we'll need one other item of livestock and we can conduct an experiment.
Eddie
Would some fly's do.
Cheers,
Bob
I used to be a Werewolf but I'm alright nowwoooooooooooooo
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MikeP
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| posted on 30/1/04 at 04:31 PM |
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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?
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JoelP
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| posted on 30/1/04 at 06:03 PM |
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i know that one! just deleted it though cos i dont want to spoil another 8 pages of fun!!!
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splitrivet
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| posted on 30/1/04 at 06:11 PM |
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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 used to be a Werewolf but I'm alright nowwoooooooooooooo
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I love speed :-P
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| posted on 30/1/04 at 06:19 PM |
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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
Don't Steal
The Government doesn’t like the competition
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Chris Leonard
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| posted on 30/1/04 at 06:32 PM |
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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
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Chris Leonard
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| posted on 30/1/04 at 06:42 PM |
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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?
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Chris Leonard
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| posted on 30/1/04 at 06:49 PM |
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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
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JoelP
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| posted on 30/1/04 at 06:51 PM |
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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]
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