stephen_gusterson
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| posted on 30/1/04 at 06:51 PM |
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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]
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200mph
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| posted on 30/1/04 at 07:03 PM |
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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
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I love speed :-P
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| posted on 30/1/04 at 07:06 PM |
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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
yes i think i do 
Don't Steal
The Government doesn’t like the competition
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JoelP
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| posted on 30/1/04 at 07:35 PM |
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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]
ignore me, wrong again... i even contradicted myself this time! if its become one in two, how can it not have affected his chances of survival?!?!
BTW, that link of chris' is both good, annoying and nonsensical. try it, you really have to...
especially question 4 on the impossible one at the top...
[Edited on 30/1/04 by JoelP]
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stephen_gusterson
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| posted on 30/1/04 at 07:38 PM |
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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.....
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DavidM
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| posted on 30/1/04 at 10:19 PM |
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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.
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David Jenkins
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| posted on 30/1/04 at 10:43 PM |
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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]
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MikeP
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| posted on 31/1/04 at 02:03 AM |
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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!
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TPG
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| posted on 31/1/04 at 03:07 AM |
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.........is it possible to restart the nutsack thread that was recently running?.........
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steve m
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| posted on 31/1/04 at 03:20 AM |
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what type of car was it anyway ??
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JoelP
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| posted on 31/1/04 at 09:23 AM |
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bollox...
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Liam
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| posted on 30/3/04 at 09:57 PM |
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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
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Liam
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| posted on 30/3/04 at 10:14 PM |
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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
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heinlein
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| posted on 6/4/04 at 10:24 PM |
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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.
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Viper
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| posted on 6/4/04 at 11:34 PM |
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My better half has just finished that book.
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