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One Girl - One Boy

Best Answer Riddari , 13 July 2007 - 07:38 PM

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348 replies to this topic

#61 Martini

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Posted 04 August 2007 - 03:05 AM

this is why i have a problem with this aspect of statistical analysis. let's see what we have here.

1] if you know the oldest child is a girl, the probability the other is also a girl is 1/2.
2] if you know the youngest child is a girl, the the probability the other is also a girl is 1/2.
3] if you do not know which child is a girl, the probability changes to 1/3?

It's not that the probability changes to 1/3, it's that the problem is altogether different.

It becomes easier to understand if you take the time to do the experiment I mentioned earlier using 100 pairs of pennies.

If you randomly lay out 100 pairs of pennies, about 50 of the pairs will have the first penny being heads up. Of those 50 pairs, about half of those will also have the second penny heads up. In other words, following your correct logic in scenario 1, "if you know the oldest child (first penny) is a girl (heads up), the probability the other is also a girl (heads up) is 1/2".

However, about how many of the pairs will have either penny being heads up? About 75 of the pairs will have at least one heads up penny. So, now we have 75 pairs of pennies with at least one heads up penny. Of those 75 pairs, 1/3 of the pairs will be HH, 1/3 of the pairs will be HT, and one third will be TH. So, out of 75 pairs where at least one of the pair is heads up, about 25 0f the pairs will be both heads up. 25 is one third of 75.
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#62 dsu

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Posted 04 August 2007 - 05:00 AM

Martini, Unreality - you guys are good logicians. But when good logicians develop blind spots, their ego prevents them from listening to anyone else!

One point I and others have repeatedly tried to point out:

In your example of 100 pair coins, 75 pairs have either one head. Ok fine. BUT WHY? Why 75??? Because the sum total is 100, you are getting 75 out of 100. If it were 200, you would have got 150 out of 200. This 75/150 is dependant on the fact the total is 100/200! But then you remove 25, and out of the 75 you start divisions. You cannot do that! As soon as the total becomes 75, all other numbers change! You cannot first use probability, and then select a portion and start doing statistics on that! As soon as you select, the probability changes! The numbers in a probability is always related to the total.

I tried to state this in various ways - do not mix permutation and combination, etc. Maybe my command of the language is not good; but the point is you cannot declare a probability, and remove some of them, and take the same numbers in another reckoning! As soon as you eliminate some, the probability changes! This is one of the paradoxes you have been stating again and again, so much so, that you firmly believe in it like the holy grail - even if your common sense tells you otherwise!
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#63 Martini

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Posted 04 August 2007 - 05:28 AM

Martini, Unreality - you guys are good logicians. But when good logicians develop blind spots, their ego prevents them from listening to anyone else!

Uh huh.

One point I and others have repeatedly tried to point out:

In your example of 100 pair coins, 75 pairs have either one head. Ok fine. BUT WHY? Why 75??? Because the sum total is 100, you are getting 75 out of 100.

Are you saying that the original 100 should still be relevant? It's not. The TT pairs don't matter because the riddle states there is at least one girl. It is then impossible to have a boy/boy (TT) pair. Out of all of the possible penny pairs (after all the TT pairs are removed), 1/3 will be HH, 1/3 will be TH, and 1/3 will be HT. PLEASE try the experiment yourself and then get back to us!

Speaking of blind spots, you have said in this thread that "We are confusing sequential events with simultaneous events" and then went on to state nonsense about how that makes a difference.

You then went on to state "If we toss a coin, the probability of getting a head is 1/2. If we toss two coins together, the probability of getting 2 heads is 1/3 (3 possibilities, head-head, tail-tail and head-tail)" with the illogical premise that HT = TH/HT. If you don't believe that the probability of getting two heads in a row are 1/4, again, PLEASE do the experiment yourself!

If you try the experiment and you are still convinced the pennies are lying, then take a look at the following links:

<!-- m --><a href="http://www.learninghaven.com/articles/r" target="_blank">http://www.learninghaven.com/articles/r ... asers.html</a><!-- m -->

<!-- m --><a href="http://en.wikipedia.org/wiki/Brain_teaser" target="_blank">http://en.wikipedia.org/wiki/Brain_teaser</a><!-- m -->

<!-- m --><a href="http://mathforum.org/dr.math/faq/faq.boy.girl.html" target="_blank">http://mathforum.org/dr.math/faq/faq.boy.girl.html</a><!-- m -->

If the above reputable sites aren't enough to convince you that you're wrong, don't tell us that we are the ones having our egos preventing us from listening to someone else.
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#64 dsu

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Posted 04 August 2007 - 05:50 AM

Ok, I finally admit I was wrong. I tried to search for those relevant sites for arguments for or against,, without results.

My sincerest apologies to everyone
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#65 Martini

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Posted 04 August 2007 - 05:54 AM

No apologies necessary. It takes a big man to admit he's wrong. Just glad this one's finally over. Or am I speaking too soon?
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#66 willy

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Posted 04 August 2007 - 04:42 PM

as i stated, i know statistical analysis pretty well and i understand all the examples that you guys have given. for those of you who have changed sides, you are following the standard view. after you read the argument proposed at this website, tell me again how it's not ambiguous. here's a quote to get you interested. note the source. also note the date. this debate has been going on a long time.

Scientific American Declared this Problem Ambiguous in the 1950's
Nick Ratti Jr. of Bristol, RI, sent me a photocopy of a clipping from "Scientific American, circa 1950's." (There is no page number, date, or other information on the clipping itself.)

Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.

Eldon Moritz reports that he also received a copy of this column, and was told that appeared in the October 1959 issue, and was written by Martin Gardner.

here is the site. it has to do with an 'ask marilyn' question. i'm sure you can guess what the question was but check the link.

http://www.wiskit.com/marilyn/boys.html

this is why you guys have spent so long discussing it. it is ambiguous.
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#67 Martini

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Posted 04 August 2007 - 04:58 PM

this is why you guys have spent so long discussing it. it is ambiguous.

No, it's not why we've been spending al ot of time discussing this. We've been spending a lot of time discussing it because of others misunderstanding probability.

Look again at what your article says:

"Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3."

Then look at what it specifies about how the answer of 1/2 is obtained. It says the same thing we've been saying all along:

"And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2."

We have already pointed out that if the child is named or if the riddle stated "the first child" then the answer would have been 1/2. The riddle as given in the OP did not name the child, hence, the riddle and the answer are not ambiguous and has the answer that your cite says it would under the given circumstances: 1/3.

The very article you quoted confirms that the logic you don't understand is the correct logic:

this is why i have a problem with this aspect of statistical analysis. let's see what we have here.

1] if you know the oldest child is a girl, the probability the other is also a girl is 1/2.
2] if you know the youngest child is a girl, the the probability the other is also a girl is 1/2.
3] if you do not know which child is a girl, the probability changes to 1/3?

sorry, but this defies any logic i understand.

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#68 unreality

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Posted 04 August 2007 - 09:44 PM

mmkay, another proof of 1/3:

you have a bag. it is not see through. You have two marbles in the bag. Marbles can be red or yellow.

The marbles are already there. They are inside the bag. There are two of them. There is nothing else in the bag.

One of them is red. What is the probability that both marbles are red?

these are the options of what is inside the bag:

Red and Yellow
Yellow and Red
Red and Red
Yellow and Yellow

but at least one has to be red... so we can say goodbye to "Yellow and Yellow". Now lets look at it again:

Red and Yellow
Yellow and Red
Red and Red

tada! there is a 1/3 chance they are both red.

I FOUND THE SOURCE OF THE ARGUEMENT!

everyone on both sides agree to remove (Yellow - Yellow)... leaving however many things are left. Thus, people who believe 1/3 believe in this set:

Red and Yellow
Yellow and Red
Red and Red
Yellow and Yellow

When you take one away, you have three to pull one out of. 1/3.

The people who believe in one-half answer believe this is the set:

Yellow and Red
Red and Red
Yellow and Yellow

when you remove 1, there are two left to pull one out of.

Basically, the people who think it is 1/2, they dont think (Red/Yellow) and (Yellow/Red) are both equal chances. The fact is, having one of each is twice the probability of having both. It all comes down to this:

People who think it is 1/3:
Mixed (yellow and red, or red and yellow): 1/2
Both yellow: 1/4
Both red: 1/4

While people who think it is one half:
Mixed: equal (1/3)
both yellow: equal (1/3)
both red: equal (1/3)

thus when you take away yellow, they think the answer is 1/2.

but it isnt

The probability for Mixed is twice as much... the sample space is this:

Red and Yellow
Yellow and Red
Yellow and Yellow
Red and Red

NOT THIS:

Mixed
Yellow and Yellow
Red and Red

therein lies the rift between the two sides. Hopefully the people who think it is 1/2 will finally understand now... that the answer is most definitely 1/3
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#69 PDR

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Posted 05 August 2007 - 12:10 AM

so I've read all the responses, and still believe that 1/2 is the correct answer. Here's one way of looking at it that I don't think anyone has brought up:
We know there are 4 possibilities:
GG
BG
GB
BB
we can throw out the last one - BB, leaves 3 combinations, so the thinking is 1/3 is what's left. The fallacy as I see it, is we don't know which one is the girl. Since we know one of them is a girl, we have 2 possibilities - one where the first child is the girl we're told about, and one where the second child is the girl.

So if the girl is the first child, the possibilities are:
GG
GB
or 1/2 the time the second child is also a girl

If the girl is the second child, the possibilities are:
GG
BG
or 1/2 the time the first child is also a girl

since the first scenario results in 1/2 and the second scenario results in 1/2, we have:
(50% * 1/2) + (50% * 1/2) = 1/4 + 1/4 = 1/2
Thus, like many have said - the odds are clearly 1/2 that the other child is also a girl.

another way to look at it - if we were told that the first child was a girl - what are the odds the second is a girl - again 1/2 (we throw out BB & BG, and are left with GG or GB). If instead we were told the second child was a girl, what are the odds the first one is also a girl - again 1/2 (we throw out BB & GB, and are left with GG or BG). Simply knowing or not knowing which one is the girl does not change the odds for the other one - which is a random event not dependent on the sex of the other child.
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#70 unreality

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Posted 05 August 2007 - 12:15 AM

You are assuming we are told which girl is first and second...

WE ARE NOT TOLD IF THE GIRL IS FIRST OR SECOND!

We are just told that ONE OF THEM is A GIRL...

we agree on this:

GG
GB
BG

we dont know if the girl is first, or the girl is second. I agree. If we KNEW the girl was first:

GG
GB

it would be 1/2

if we KNEW the girl was second:
GG
BG

it would be 1/2

BUT WE DO NOT KNOW!

GG
BG
GB

It is 1/3
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