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Author: Subject: Girls!
PHILOU Zrealone
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smile.gif posted on 18-5-2004 at 16:10


Anyway, girls, continue to show to the face of the world you aren't affraid of change, fire and that all this is a myth nourished by other myths.

"DestroyThe
BiG
MYTH!!!"


:cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool:
:cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool::cool:




PH Z (PHILOU Zrealone)

"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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PHILOU Zrealone
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[*] posted on 18-5-2004 at 16:16


Destroy The BiG MYTH

For unknown reason it was screwed up in previous post :(

Even the text program is sexist ;););)




PH Z (PHILOU Zrealone)

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Organikum
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[*] posted on 18-5-2004 at 17:42


Destroy the BigMac ?

Ah! French! Fight cultural imperialism I understand, welldone!

Mr. Freud had a good laugh btw..........


;)
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thalium
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[*] posted on 12-11-2004 at 10:32


I'm a girl too. The only problem is that my mother doesn't accept that I'm interested in chem and physics (more in chem.. MUCH MORE in chem). She'd like me to play the piano:mad: and study literature and arts and classic music:mad::mad::mad:. Damn...so this may be what happens to many girls: the influnece of their mothers (maybe torture too...who knows...). I didn't let myself be influenced by my mom:D



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Saerynide
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[*] posted on 12-11-2004 at 19:04


Omigod, my mom used to make me play piano too. I have lonnnnngggg quit though. *shudder* I have no musical talent :D What is it with moms and having their daughters learn music?



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thalium
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[*] posted on 13-11-2004 at 00:30


I didn't even TRY to learn...I'd learn to play the giutar tough...rocker thing (when I'll have some time for it). I don't know why...maybe just to amuse the guests when they invite people:P



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TomThumb
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[*] posted on 13-11-2004 at 08:38


off subject, spare me to torment and explain what "IIRC" means :/


--> If I Remember Correctly.
--> If I Recall Correctly. ^^

[Edited on 13-11-2004 by chemoleo]

[Edited on 15-11-2004 by Ramiel]




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MadHatter
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[*] posted on 14-11-2004 at 04:31
Sociology


I remember that course well. There was only 3 males in the entire class ! We were the
engineering types taking that course as part of a well-rounded degree program. Actually,
there was quite a bit of statistics involved and I managed to get a B for that class. The other 2
males were bitching about this requirement while I felt I had hit the mother lode for women !
They were mostly nursing students who also bitched about other required courses -
namely Calculus I. I was more than happy to offer my assistance to these lovely young
ladies !

P.S. That's 1 of the few textbooks I kept. Later on, my niece found it useful in her
report on Margaret Meade - a well known sociologist.

[Edited on 14-11-2004 by MadHatter]




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chloric1
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[*] posted on 14-11-2004 at 06:34
Social change


Its the beginning of the 21st century and certain generations tend to want to cling to 19th century ideals. My daughter will be born this February and my wife has no problem with me teaching her math and science. It is the time for change otherwise we will no be able to reach the status of a "civilized" society.



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Jome
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mad.gif posted on 17-11-2004 at 16:25


I hope so. In my opinion the "I have to be like everyone else" thing in most girls is very sad, most interesting personalities certainly result where it does not.











:mad:




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Twospoons
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[*] posted on 17-11-2004 at 17:06


I love it when my daughter comes to me and asks if we can do a 'science experiment'. At which point I wrack my brains for something easy, immediate, and impressive (making pH indicators out of flowers went down extremely well).

At her 7th birthday I got to be 'Professor Snape' :P and amuse all the party goers with a hands on 'potions' class.

Professionally I'm an electronic engineer, and it saddens me that so few women take up a career in engineering.
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[*] posted on 17-11-2004 at 18:10


I was talking to my sister the other day (trying to discuss a probability problem), and she rather rapidly declared that she found the whole concept of averages to be almost useless. Draw what conclusions you may.

Not that it's easy to say exactly what a numerical probability does mean. Not easy at all.

BTW, here's the problem: You are given two boxes, one containing twice the amount of money (or whatever) as the other. That's all you know. You open one box, and, after examining it's contents, you are given the choice of either keeping it, or switching to the other box. Which do you choose? Your object is to maximize your earnings. (My sister, interestingly, was interested in minimizing dissapointment.)

You reason as follows. The opened box contains X dollars. Since you had an equal chance of picking either box, half the time the other box will contain 2X, and half the time it will contain 0.5X. On average, then, switching to the other box will get you 1.25X.

Does this make sense?
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chemoleo
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[*] posted on 17-11-2004 at 18:30


No it doesn't. :P
As it's random, one could say, oh, let's always take the first one. Therefore the average is (2x + 1x)/2 = 1.5 x, as the choice on 1x or 2x is random.
This is your average earning. Regardless which boxes you pick (i.e. random, or selectively the first or second).
Nice one :)
anyway...there are some related threads in Whimsy, so your little confusing problems may be better suited there.
I can see why your sister doesnt like averages, if you treat the poor lass with that!




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HNO3
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[*] posted on 30-11-2004 at 18:49


Can you pick up hte other one? If you can, feel which one is heavier, and take it:P:P:P:P;):D:P:P:P:P:cool::cool::cool:



\"In the beginning, God...\" Wait a minute, God doesn\'t exist!!!!!!!!!! \"OK, in the beginning, ummm, hydrogen...\" Wait a minute, what about the laws of thermodynamics? \"OK, in the beginning, ummm.....UMMMMM, what\'s left to choose from?
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I am a fish
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[*] posted on 1-12-2004 at 06:31


A similar problem is as follows:

There are three identical boxes, one of which contains a prize. You are asked to select a box. Before opening it however, one of the other boxes is opened and revealed to be empty. You are then given the option of changing your choice of box. What should you do to maximise your chance of winning?

The "obvious" answer is that switching boxes won't affect your chances of winning, as the game is entirely random. However this is wrong. The probability that the prize is in one of the two boxes that you didn't initially choose is 2/3. Once one of the boxes is opened, this probability remains the same and so there is still a 2/3 chance that the prize is in the remaining unselected box. Therefore, switching boxes doubles your chance of winning.

[Edited on 1-12-2004 by I am a fish]

[Edited on 1-12-2004 by I am a fish]




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[*] posted on 1-12-2004 at 11:53


Quote:
Originally posted by I am a fish
A similar problem is as follows:

There are three identical boxes, one of which contains a prize. You are asked to select a box. Before opening it however, one of the other boxes is opened and revealed to empty. You are then given the option of changing your choice of box. What should you do to maximise your chance of winning?

The "obvious" answer is that switching boxes won't affect your chances of winning, as the game is entirely random. However this is wrong. The probability that the prize is in one of the two boxes that you didn't initially choose is 2/3. Once one of the boxes is opened, this probability remains the same and so there is still a 2/3 chance that the prize is in the remaining unselected box. Therefore, switching boxes doubles your chance of winning.


It sounds like the obvious answer is still correct. After one box is revealed empty, there is an equal chance of finding the prize in either remaining box. The misleading part is "the prize is in one of the two boxes that you didn't initially choose". Initial choice is irrelevant; only final choice determines whether or not you have a box with a prize. Once one box is revealed empty, there is a probability of 1 of the prize being in the unopened boxes, 1/2 for each.

[Edited on 12-1-2004 by Polverone]




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I am a fish
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[*] posted on 1-12-2004 at 14:29


Quote:
Originally posted by Polverone
Initial choice is irrelevant; only final choice determines whether or not you have a box with a prize.


You're falsely assuming that the process of opening one of the empty boxes is entirely random. If the player initially chooses incorrectly (as she has a 2/3 chance of doing), then she effectively dictates which of the boxes must be opened (as neither the winning box nor the chosen box can be opened). Therefore, information is passed onto the player, from which she can can make a better than random guess (by switching her choice).

[Edited on 2-12-2004 by I am a fish]




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[*] posted on 1-12-2004 at 17:10


I wasn't assuming that the initial box-opening is random, but that it was rigged to always be empty by re-distribution of the box contents rather than selection of an existing empty box. I didn't pay enough attention to the physical reality of the situation; in an electronic gambling game (for example), it would be common to implement this so that you just show an empty box and then let each remaing box have a 1/3 probability of containing the prize. Of course in electronic gambling games it's also standard fare to rig things so that there is no real probability involved, just a guaranteed prize after X thousand plays, so I really should keep physical and electronic worlds better separated in my thinking :(

[Edited on 12-2-2004 by Polverone]




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Twospoons
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[*] posted on 2-12-2004 at 13:45


The initial choice is totally irrelevant, as one empty box will always be removed. This leaves two boxes, one prize, and a choice out of two boxes. "Option to change your initial choice" is just messing with words - what you really have is "Pick one of these two remaining boxes, one of which contains a prize". So your chance of winning is exactly 1/2, whether or not your second pick is different to your first.

If you don't believe it get three boxes, one prize, and try it 100 times.
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[*] posted on 5-12-2004 at 05:33


You're confusing the mean probability with the individual probabilities. If you make a random choice between the original box and the other box, the probability of winning will be 1/2. However, the individual probabilities are 1/3 for the originally chosen box, and 2/3 for the other box.



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HRH_Prince_Charles
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[*] posted on 5-12-2004 at 07:13


The initially chosen box has a 1/3 probability of containing the ball. The probability that the ball is one one of the 2 remaining boxes is 2/3.

When one of the remaining boxes is shown to be empty and removed, knowledge of the system has changed: the ball is equally likely to be in either of the 2 remaining boxes; the probability of it being in either is 1/2.

Probability is based on knowledge of the possible states of the system: as the knowledge changes, so does the proabability.




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[*] posted on 5-12-2004 at 14:28


Chemoleo's shown that the reasoning in the problem I mentioned is wrong (which is intuitively clear). He hasn't shown WHY it's wrong, though. Any takers?

Polverone: Your point regarding electronic gaming is well taken. The discussion below assumes that the prize begins in one box and stays there.

HRH, Twospoons: you're not alone in falling for the 50:50 story. The issue was raised in Parade magazine, and several mathematicians wrote in claiming something similar. It's still wrong.

Intuitively, since it is always possible to remove a nonchosen empty box, the probability of you having chosen the nonempty box cannot be changed by the new information given (we ALREADY KNEW one of the nonchosen boxes was empty, so no new information about the chosen box is available).

Indeed, your 50:50 arguement should work regardless of the numer of boxes. Imagine a large number, say N, of boxes, one prize, and one initial choice. The person running the game then removes N-2 of the nonchosen empty boxes. Obviously, if N is, say, Avagadro's number, the chances of your initial choice being right is absurdly small, so you can safely assume that the prize is in one of the boxes you did not choose. But that means that the prize is in the one nonchosen box left after the others are removed, so switching is almost always going to get you the prize.

Formally, probability problems arising from a finite and uniform distribution can always be solved by listing all the possibilities and counting. There are three possibilities for the initial layout of the boxes/prize:
p n n
n p n
n n p
(p being the prize, n being an empty box)
These are all equally likely. Assume that you initially choose the first box (the other cases are similar). The three cases go to:
Pnn
Npn
Nnp
(the chosen box is capitalized)
Since the initial cases are equally likely, so are these. One non-empty, nochosen box is removed, and the three cases lead to these descendents:
Pn
Np
Np
Once again, since the previous states are equally likely, so are these. But then it is clear that in 2 out of three possible and equally likely cases, switching to the other box will win the prize.
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[*] posted on 6-12-2004 at 05:41


Geomancer: I am very sorry to say that you are correct.



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Twospoons
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[*] posted on 6-12-2004 at 16:31


Your mistake is in assuming the order of the boxes is important.
Pnn
Npn
Nnp

the last two are indentical states, as it doesn't matter which of the other two boxes actually contains the prize since the other is going to be deleted! Leaving just
Pn
Np

50:50 chance


Look at it another way : why would it matter if you make a choice before one of the empty boxes is removed? Your final choice is still one of two!

[Edited on 7-12-2004 by Twospoons]
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[*] posted on 6-12-2004 at 16:38


Twospoons:

Geomancer's solution is correct, there are 3 possible outcomes; it just so happens that 2 of those outcomes are identical, giving 2 chances of the same outcome: that is why the probability is twice as high.

Geomancer's head is probably swelling as we speak.




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