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Testing random

Ian Kelly-2
On Jun 16, 2015 4:58 PM, "Ian Kelly" <ian.g.kelly at gmail.com> wrote:
>
> On Tue, Jun 16, 2015 at 4:30 PM,  <sohcahtoa82 at gmail.com> wrote:
> > On Tuesday, June 16, 2015 at 3:01:06 PM UTC-7, Thomas 'PointedEars'
Lahn wrote:
> >> This should give you pause: In real mathematics, events with zero
> >> probability can occur.
> >
> > Nobody will disagree with that.  The probability of me winning the
lottery is zero if I don't buy a ticket.
>
> I believe he's actually referring to this:
>
> https://en.wikipedia.org/wiki/Almost_surely
>
> Not that this has anything to do with the probabilities under
> discussion in this thread.

Actually, I take that back. The coin flip example in the Wikipedia article
is exactly what we've been discussing, taken to the ultimate extreme of an
infinitely long sequence.

Unfortunately for Thomas, the article does not agree with his position,
describing the event of never flipping a tail in such an infinite sequence
as "almost never". If, as Thomas maintains, the probability of this event
does not decrease as the length of the sequence increases, then the
probability for the infinite sequence would have to be non-zero since the
probability for any finite sequence is non-zero.
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Testing random

Ned Batchelder
In reply to this post by Thomas 'PointedEars' Lahn
On Tuesday, June 16, 2015 at 6:01:06 PM UTC-4, Thomas 'PointedEars' Lahn wrote:

> Ned Batchelder wrote:
>
> > [...]
> >     This is done empirically, by producing `nseq` sequences of
> >     `nrolls` rolls of the die.  Each sequence is examined to
> >     see if it has a zero.  The total number of no-zero
> >     sequences divided `nseq` is the probability.
>
> No, it is not.  It is the relative frequency for *this* number of trials and
> *this* run of the experiment.
>
> >     """
> >     no_zeros = 0
> >     for _ in xrange(nseq):
> >         seq = die_rolls(nrolls)
> >         if not any_zeros(seq):
> >             no_zeros += 1
> >     return float(no_zeros)/nseq
> >
> > for n in range(10, 101, 10):
> >     # Calculate the probability of getting no zeros by trying
> >     # it a million times.
> >     prob = probability_of_no_zero(n, 1000000)
> >     print "n = {:3d}, P(no zero) = {:.8f}".format(n, prob)
> >
> >
> >
> > Running this gives:
> >
> > $ pypy testrandom.py
> > n =  10, P(no zero) = 0.34867300
> > n =  20, P(no zero) = 0.12121900
> > n =  30, P(no zero) = 0.04267000
> > n =  40, P(no zero) = 0.01476600
> > n =  50, P(no zero) = 0.00519900
> > n =  60, P(no zero) = 0.00174100
> > n =  70, P(no zero) = 0.00061600
> > n =  80, P(no zero) = 0.00020600
> > n =  90, P(no zero) = 0.00006300
> > n = 100, P(no zero) = 0.00002400
> >
> >
> > As n increases, the probability of having no zeros goes down.
>
> Your programmatic "proof", as all the other intuitive-empirical "proofs",
> and all the other counter-arguments posted before in this thread, is flawed.  
> As others have pointed out at the beginning of this thread, you *cannot*
> measure or calculate probability or determine randomness programmatically
> (at least not with this program).  

You *can* estimate probability with a program, which is what is happening
here.

> I repeat: Probability is what relative
> frequency (which you can measure) *approaches* for *large* numbers.  100 is
> anything but large, to begin with.  

The number of trials in this program is not 100, it is 1 million.  You seem
uninterested in trying to understand.

> What is "large" depends on the
> experiment, not on the experimentator.  And with independent events, the
> probability for getting zero does not increase because you have been getting
> non-zeros before.  It simply does not work this way.

Again, if you look at the code, you'll see that we are not talking about
the probability of getting a zero on the next roll.  We are talking about the
probability of getting no zeros in an N-roll sequence.  I have no idea how you
have misunderstood this for so long.

I'll stop trying to explain it.

--Ned.

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Testing random

Thomas 'PointedEars' Lahn
Ned Batchelder wrote:

> On Tuesday, June 16, 2015 at 6:01:06 PM UTC-4, Thomas 'PointedEars' Lahn
> wrote:
>> Your programmatic "proof", as all the other intuitive-empirical "proofs",
>> and all the other counter-arguments posted before in this thread, is
>> flawed. As others have pointed out at the beginning of this thread, you
>> *cannot* measure or calculate probability or determine randomness
>> programmatically (at least not with this program).
>
> You *can* estimate probability with a program, which is what is happening
> here.

No.  Just no.

>> I repeat: Probability is what relative
>> frequency (which you can measure) *approaches* for *large* numbers.  100
>> is anything but large, to begin with.
>
> The number of trials in this program is not 100, it is 1 million.  You
> seem uninterested in trying to understand.

It still would _not_ a measure or a calculation of *probability*.  So much
for ?uninterested in trying to understand?.
 
>> What is "large" depends on the experiment, not on the experimentator.  
>> And with independent events, the probability for getting zero does not
>> increase because you have been getting non-zeros before.  It simply does
>> not work this way.
>
> Again, if you look at the code, you'll see that we are not talking about
> the probability of getting a zero on the next roll.  We are talking about
> the probability of getting no zeros in an N-roll sequence.  I have no idea
> how you have misunderstood this for so long.

You do not understand that it boils down to the same problem.  The
probability of only having sons is _not_ greater than that of having
sons and one daughter or vice-versa.  And for that it does _not_ matter
how many children you have *because* it does _not_ matter how many
children you had before.  The probability for a boy or a girl is *always*
the same.  You are _not_ due for a boy if you have many girls, and not for a
girls if you have many boys.  But that is precisely what your flawed logic
is implying.

Learn probability theory, and use a dictionary in Python when you want to
count random hits.

--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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Testing random

sohcahtoa82@gmail.com
On Tuesday, June 16, 2015 at 4:48:36 PM UTC-7, Thomas 'PointedEars' Lahn wrote:

> Ned Batchelder wrote:
>
> > On Tuesday, June 16, 2015 at 6:01:06 PM UTC-4, Thomas 'PointedEars' Lahn
> > wrote:
> >> Your programmatic "proof", as all the other intuitive-empirical "proofs",
> >> and all the other counter-arguments posted before in this thread, is
> >> flawed. As others have pointed out at the beginning of this thread, you
> >> *cannot* measure or calculate probability or determine randomness
> >> programmatically (at least not with this program).
> >
> > You *can* estimate probability with a program, which is what is happening
> > here.
>
> No.  Just no.
>
> >> I repeat: Probability is what relative
> >> frequency (which you can measure) *approaches* for *large* numbers.  100
> >> is anything but large, to begin with.
> >
> > The number of trials in this program is not 100, it is 1 million.  You
> > seem uninterested in trying to understand.
>
> It still would _not_ a measure or a calculation of *probability*.  So much
> for "uninterested in trying to understand".
>  
> >> What is "large" depends on the experiment, not on the experimentator.  
> >> And with independent events, the probability for getting zero does not
> >> increase because you have been getting non-zeros before.  It simply does
> >> not work this way.
> >
> > Again, if you look at the code, you'll see that we are not talking about
> > the probability of getting a zero on the next roll.  We are talking about
> > the probability of getting no zeros in an N-roll sequence.  I have no idea
> > how you have misunderstood this for so long.
>
> You do not understand that it boils down to the same problem.

Actually, no, they're not.  They're entirely different problems.  "What are the odds of getting 8 zeros in a row?" is a *COMPLETELY* different question from "What are the odds of getting a zero on the next roll?"


>The
> probability of only having sons is _not_ greater than that of having
> sons and one daughter or vice-versa.  And for that it does _not_ matter
> how many children you have *because* it does _not_ matter how many
> children you had before.  The probability for a boy or a girl is *always*
> the same.  You are _not_ due for a boy if you have many girls, and not for a
> girls if you have many boys.  But that is precisely what your flawed logic
> is implying.

Yes, we all know what the gambler's fallacy is, but that's not what anyone is arguing.

If you pick 8 random numbers between 0 and 9 (inclusive), then the odds of getting all zeros is (1/10)^8.  Do you agree with that?

The odds of getting NO zeros is (9/10)^8.  Do you agree with that?

Note that NEITHER of these scenarios say anything about a pre-condition.  The first question is *NOT* asking "If you picked 7 random numbers between 0 and 9 and got 0 for all 7, what are the odds of getting another 0?"  The answer to that is obviously 1/10, and anybody arguing something else would certainly be committing the Gambler's fallacy.

>
> Learn probability theory, and use a dictionary in Python when you want to
> count random hits.

I know enough probability theory to know that you're either wrong or you keep changing the problem to something nobody else has said in order to think you're right.

>
> --
> PointedEars
>
> Twitter: @PointedEars2
> Please do not cc me. / Bitte keine Kopien per E-Mail.


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Testing random

MRAB-2
In reply to this post by Thomas 'PointedEars' Lahn
On 2015-06-17 00:45, Thomas 'PointedEars' Lahn wrote:

> Ned Batchelder wrote:
>
>> On Tuesday, June 16, 2015 at 6:01:06 PM UTC-4, Thomas 'PointedEars' Lahn
>> wrote:
>>> Your programmatic "proof", as all the other intuitive-empirical "proofs",
>>> and all the other counter-arguments posted before in this thread, is
>>> flawed. As others have pointed out at the beginning of this thread, you
>>> *cannot* measure or calculate probability or determine randomness
>>> programmatically (at least not with this program).
>>
>> You *can* estimate probability with a program, which is what is happening
>> here.
>
> No.  Just no.
>
>>> I repeat: Probability is what relative
>>> frequency (which you can measure) *approaches* for *large* numbers.  100
>>> is anything but large, to begin with.
>>
>> The number of trials in this program is not 100, it is 1 million.  You
>> seem uninterested in trying to understand.
>
> It still would _not_ a measure or a calculation of *probability*.  So much
> for ?uninterested in trying to understand?.
>
>>> What is "large" depends on the experiment, not on the experimentator.
>>> And with independent events, the probability for getting zero does not
>>> increase because you have been getting non-zeros before.  It simply does
>>> not work this way.
>>
>> Again, if you look at the code, you'll see that we are not talking about
>> the probability of getting a zero on the next roll.  We are talking about
>> the probability of getting no zeros in an N-roll sequence.  I have no idea
>> how you have misunderstood this for so long.
>
> You do not understand that it boils down to the same problem.  The
> probability of only having sons is _not_ greater than that of having
> sons and one daughter or vice-versa.  And for that it does _not_ matter
> how many children you have *because* it does _not_ matter how many
> children you had before.  The probability for a boy or a girl is *always*
> the same.  You are _not_ due for a boy if you have many girls, and not for a
> girls if you have many boys.  But that is precisely what your flawed logic
> is implying.
>
> Learn probability theory, and use a dictionary in Python when you want to
> count random hits.
>
I think that different people are talking about different things in
this thread. You're talking about the probability of each event, while
everybody else is talking about the probability of certain combinations
of events.

If you have, say, two children, the possibilities are:

        boy, boy
        boy, girl
        girl, boy
        girl, girl

The probability of each boy or girl is 1/2.

The probability of only boys is 1/4 and of a son and a daughter is 1/4
+ 1/4 = 1/2.

Therefore, the probability of having only boys is less than the
probability of having a son and a daughter.


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Testing random

Chris Angelico
In reply to this post by sohcahtoa82@gmail.com
On Wed, Jun 17, 2015 at 10:36 AM,  <sohcahtoa82 at gmail.com> wrote:
> Yes, we all know what the gambler's fallacy is, but that's not what anyone is arguing.

The only instance of gambler's fallacy I'm seeing here is "PointedEars
didn't understand the last dozen emails, so he's due to understand the
next one". I've given up trying to explain.

ChrisA

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Testing random

Ethan Furman-2
On 06/16/2015 06:01 PM, Chris Angelico wrote:

> The only instance of gambler's fallacy I'm seeing here is "PointedEars
> didn't understand the last dozen emails, so he's due to understand the
> next one". I've given up trying to explain.

+1 QotW

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Testing random

Steven D'Aprano-11
In reply to this post by Ned Batchelder
On Tue, 16 Jun 2015 13:48:04 -0700, Ned Batchelder wrote:

> I apologize, I'm sure I've been using the mathematical terms
> imprecisely. We are all intelligent people, so I still believe we
> disagree because we are talking about different things.

Neil, I believe that your actual mistake is assuming that Thomas is
arguing in good faith. I see no evidence that he is, especially given the
content of his latest posts.

Multiple people have repeatedly explained the difference between his
argument and what everyone else is talking about. Others, including me,
have demonstrated empirically that he is mistaken, using both simulated
tests and direct calculation of the probabilities.

At this point, his insistence that we are making the gambler's fallacy is
clearly not a mere misunderstanding due to confusion. It is wilful and
deliberate misrepresentation of what we are saying.

Thomas is correct for a completely different question. Rather than
acknowledge that he has misunderstood the question, at every point he
doubles down harder and insists that he is right and we are wrong. We
have given *absolutely no reason* to think we have fallen for the
Gambler's Fallacy. Throughout this thread, Thomas has repeatedly picked
on trivial and unimportant errors in terminology as an excuse for
dismissing what others have said, while ignoring the substance of their
argument. When people have given mathematically indisputable and correct
arguments, he has ignored them, or misrepresented them.



--
Steven D'Aprano

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Testing random

Steven D'Aprano-11
In reply to this post by Thomas 'PointedEars' Lahn
On Wed, 17 Jun 2015 01:45:27 +0200, Thomas 'PointedEars' Lahn wrote:

> The
> probability of only having sons is _not_ greater than that of having
> sons and one daughter or vice-versa.

Take a family of four children. We can enumerate all the possibilities,
using S for son and D for daughter, there are exactly 2**4 = 16 of them:

SSSS
SSSD
SSDS
SSDD
SDSS
SDSD
SDDS
SDDD
DSSS
DSSD
DSDS
DSDD
DDSS
DDSD
DDDS
DDDD


There is exactly 1 outcome which is "all sons", 4 outcomes which is
"three sons and one daughter", 14 outcomes which is "at least one son and
one daughter", 15 outcomes which are "at least one son", and 1 outcome is
"no sons".

If you have *two* children, those probabilities are different:

SS, SD, DS, DD

All sons: 1/4, not 1/16.
Three sons and one daughter: 0, not 4/16.
At least one son and one daughter: 2/4, not 14/16.
At least one son: 3/4, not 15/16.
No sons: 1/4, not 1/16.


With four children, it is true that the probability of these are the same:

    SSSS vs SSDS

but that's not the question. The question is to compare the probability
of these:

    SSSS vs (SSDS or SSSD or SDSS or DSSS)




> And for that it does _not_ matter
> how many children you have

Of course it does. Assuming the births are independent and the
probability of a boy is 1/2, the probability of having "no boys" depends
on how many children you have:

Pr(no boys, given no children) = 1
Pr(no boys, given 1 child) = 1/2
Pr(no boys, given 2 children) = 1/4
Pr(no boys, given 3 children) = 1/8
Pr(no boys, given 4 children) = 1/16

and in general:

Pr(no boys, given n children) = 1/2**n



> *because* it does _not_ matter how many
> children you had before.  The probability for a boy or a girl is
> *always* the same.

That is completely irrelevant, as has been explained to you over and over
again. A little learning is a dangerous thing.


>  You are _not_ due for a boy if you have many girls, and not for a girls
> if you have many boys.  But that is precisely what your flawed logic is
> implying.

You are the only one talking about being "due" for a result. You are
ignoring what we are saying, ignoring the clear and detailed analysis we
have repeatedly given, ignored the simulations we have given, and keep
coming back to your arrogant *AND WRONG* accusation that we are making
the gambler's fallacy.

Time to man up and admit your mistake.



--
Steven D'Aprano

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Testing random

Jussi Piitulainen
In reply to this post by Ned Batchelder
Ned Batchelder writes:

> Thomas: let's say I generate streams of N digits drawn randomly from
> 0-9.  I then consider the probability of a zero *never appearing once*
> in my stream.  Let's call that P(N).  Do you agree that as N
> increases, P(N) decreases?

In probability theory, that could be phrased as the probability that N
unknown digits d_1, ..., d_N are all positive, assuming the digits are
independent (so learning one digit doesn't reveal anything about any
other digit), and for each d_k, the probability p_k of having a positive
digit is the same. Mathematicians often abbreviate these assumptions as
"i.i.d" for "independent" and "identically distributed".

Also assuming uniform distributions, p_1 = p_2 = ... = p_N = 9/10.

P(d_k > 0 for k = 1, ..., N) =
P(d_1 > 0 and d_2 > 0 and ... and d_N > 0) = (by independence)
P(d_1 > 0) * P(d_2 > 0) * ... * P(d_N > 0) =
p_1 * p_2 * ... * p_N = (by identical uniform distribution)
(9/10)^N

In mathematics, (9/10)^N decreases as N increases, so one should indeed
agree. Using more impressive notation and terminology correctly will not
change the analysis.

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Testing random

Thomas 'PointedEars' Lahn
In reply to this post by Thomas 'PointedEars' Lahn
MRAB wrote:

> On 2015-06-17 00:45, Thomas 'PointedEars' Lahn wrote:
>> Learn probability theory, and use a dictionary in Python when you want to
>> count random hits.
>>
> I think that different people are talking about different things in
> this thread.

I think that you, too, do not know what you are talking about.

> You're talking about the probability of each event, while
> everybody else is talking about the probability of certain combinations
> of events.

No.

> If you have, say, two children, the possibilities are:
>
> boy, boy
> boy, girl
> girl, boy
> girl, girl
>
> The probability of each boy or girl is 1/2.
>
> The probability of only boys is 1/4 and of a son and a daughter is 1/4
> + 1/4 = 1/2.
>
> Therefore, the probability of having only boys is less than the
> probability of having a son and a daughter.

And half of the population is homosexual as with 50 % males and 50 % females
there is an equal number of homosexual pairs (male?male, female?female) as
heterosexual ones (male?female, female?male).  NOT.

Final posting by me in this thread as the misconceptions about probability
are apparently too deeply embedded in the general public to be resolved
here.

In summary:

1. Combinations matter, not permutations.

2. <https://en.wikipedia.org/wiki/Law_of_averages>

3.
<http://www.teacherlink.org/content/math/interactive/probability/interactivequiz/home.html>
   (Whereas I predict that the ignorant will see the correct answer to
    question 3 as proof of the correctness of their misconception.)

EOD.

--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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Testing random

Christian Gollwitzer
Am 17.06.15 um 08:53 schrieb Thomas 'PointedEars' Lahn:
> 3.
> <http://www.teacherlink.org/content/math/interactive/probability/interactivequiz/home.html>
>     (Whereas I predict that the ignorant will see the correct answer to
>      question 3 as proof of the correctness of their misconception.)

I'v been to the UK recently for cycling holidays. A nightmare! So many
ignorant people all driving on the wrong side of the road!

> EOD.


        :)


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Testing random

Chris Angelico
On Wed, Jun 17, 2015 at 5:22 PM, Christian Gollwitzer <auriocus at gmx.de> wrote:
> I'v been to the UK recently for cycling holidays. A nightmare! So many
> ignorant people all driving on the wrong side of the road!

You might argue that you drive on the right side of the road, but if
you call what British people drive on the "wrong" side, then you don't
understand the language. It's called "English", remember, not
"American", and the word you want is "left". Get it *right*... oh
wait.

ChrisA

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Testing random

Tim Golden-4
In reply to this post by Christian Gollwitzer
On 17/06/2015 08:22, Christian Gollwitzer wrote:
> Am 17.06.15 um 08:53 schrieb Thomas 'PointedEars' Lahn:
>> 3.
>> <http://www.teacherlink.org/content/math/interactive/probability/interactivequiz/home.html>
>>
>>     (Whereas I predict that the ignorant will see the correct answer to
>>      question 3 as proof of the correctness of their misconception.)
>
> I'v been to the UK recently for cycling holidays. A nightmare! So many
> ignorant people all driving on the wrong side of the road!

Yes, we do get many foreign tourists at this time of the year :)

TJG


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Testing random

Mark Lawrence
In reply to this post by Chris Angelico
On 17/06/2015 02:01, Chris Angelico wrote:
> On Wed, Jun 17, 2015 at 10:36 AM,  <sohcahtoa82 at gmail.com> wrote:
>> Yes, we all know what the gambler's fallacy is, but that's not what anyone is arguing.
>
> The only instance of gambler's fallacy I'm seeing here is "PointedEars
> didn't understand the last dozen emails, so he's due to understand the
> next one". I've given up trying to explain.
>
> ChrisA
>

An alternative explanation is that he's just a plain, old fashioned
troll, as pointed out by Denis McMahon some weeks ago.  Now what is the
probability of that? :)

--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence


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Testing random

Laura Creighton-2
In reply to this post by Ian Kelly-2
In a message of Tue, 16 Jun 2015 16:58:26 -0600, Ian Kelly writes:

>On Tue, Jun 16, 2015 at 4:30 PM,  <sohcahtoa82 at gmail.com> wrote:
>> On Tuesday, June 16, 2015 at 3:01:06 PM UTC-7, Thomas 'PointedEars' Lahn wrote:
>>> This should give you pause: In real mathematics, events with zero
>>> probability can occur.
>>
>> Nobody will disagree with that.  The probability of me winning the lottery is zero if I don't buy a ticket.
>
>I believe he's actually referring to this:
>
>https://en.wikipedia.org/wiki/Almost_surely
>
>Not that this has anything to do with the probabilities under
>discussion in this thread.
>--
>https://mail.python.org/mailman/listinfo/python-list

I think he is talking about what mathematicians call 'discrepancy':
https://en.wikipedia.org/wiki/Low-discrepancy_sequence

Laura

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Testing random

Cecil Westerhof
In reply to this post by Christian Gollwitzer
On Wednesday 17 Jun 2015 09:30 CEST, Tim Golden wrote:

> On 17/06/2015 08:22, Christian Gollwitzer wrote:
>> Am 17.06.15 um 08:53 schrieb Thomas 'PointedEars' Lahn:
>>> 3.
>>> <http://www.teacherlink.org/content/math/interactive/probability/interactivequiz/home.html>
>>>
>>> (Whereas I predict that the ignorant will see the correct answer
>>> to question 3 as proof of the correctness of their misconception.)
>>
>> I'v been to the UK recently for cycling holidays. A nightmare! So
>> many ignorant people all driving on the wrong side of the road!
>
> Yes, we do get many foreign tourists at this time of the year :)

??_???

--
Cecil Westerhof
Senior Software Engineer
LinkedIn: http://www.linkedin.com/in/cecilwesterhof

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Testing random

Laura Creighton-2
In reply to this post by MRAB-2
Stick to dice.  Stay away from children.  One thing we know of, for sure,
is that certain breeding pairs are more likely to produce males, and
others are more likely to produce females.  We will ignore  those born who
are of indeterminate sex, for this discussion.

In human beings, as well as a whole lot of animal species, infant-and-
chidhood mortality is not sex neutral.  More baby boys die than baby
girls.  But when you look at breeding age men and women you will find
that the ratio is a lot closer to 50/50 than the birth age ratio.
Human beings, as a population currently produce more males than
females.

Now one way that could happen is if all human beings are predisposed to
have slightly more boy babies than girl babies when they breed.  But
this isn't what we think happens.  

Fisher's Principle says:

        Suppose sexually active males are less common than females in
        a population.

        A male then has better mating prospects than a female, and
  can expect to have more offspring.  (Because some lucky males
        can expect to mate with more than 1 female).

        Therefore parents genetically disposed to produce males tend to
        have more than average numbers of grandchildren born to them.

        Thus genes for producing males will spread.

        But this means that more males will be produced.  Which means
        that at some point, the first priciple no longer holds --
        sexually active males are no longer less common than females
        in the population.  You can reach a state where females are
        less common.

        But the same thing kicks in, once more. And this time parents
        who produce daughters tend to get more offspring.  If you produce
        a son, who doesn't get to breed at all, from an evolutionary
        standpoint you have wasted your time in a bad investment.

        Thus the sex ratio will tend to 50/50.  It is self-adjusting
        that way.

-----------

Now, human beings, despite being very interesting to us, are very bad to
study this way.  There are too many other complicating factors in human
reproduction, and human beings take too long to reach sexual maturity
anyway.  But you can study this in animals where you can artificially
bias the population any way you like, and who reach sexual maturity
quickly so you can watch and see if it plays out like Fisher says it
would.  And in lots of places that has been  studied, it does.

It is a little hard to see how you would get the proper correspondence
between infant mortality rates and sex ratio -- this species produces
precisely enough extra girl babies to offset the fact that more girls
die as infants -- unless something like this was going on.  And this
is measured to be so for some animals studied other than humans.

Bets are off in species where most of the males don't get to breed
anyway.  The logic doesn't work for them.

But what this shows is that the statement that a given birth is as likely
to be a son as a daughter (or 52% likely to be a son and 48% likely
to be a daughter) only works for large populations.  It's not going to
help you know if your next child is going to be a son or a daughter.
The parents of 4 boys who are trying, once again, thinking 'this time
it must be a girl!' may be up against something rather worse than
the gamblers fallacy. The odds that they have a girl may be substantially
lower than 50% -- it is quite possible that they are set up to only produce
boys, or to produce boys 80% of the time.  We have no way of knowing
this, because right now we don't know exactly how a bias in producing
one sex or another is expressed.  It is also quite possible that the
parents of 4 boys who want a girl have just been unlucky, because
they have no bias one way or another.  Or extremely unlucky because
their bias is for daughters, but they keep getting results against the
odds.

All we know is that notion that every human birth has a 50/50
chance of being male or female is wrong.

Laura


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Testing random

Steven D'Aprano-8
In reply to this post by MRAB-2
On Wed, 17 Jun 2015 08:33 pm, Laura Creighton wrote:

> Stick to dice.  Stay away from children.  One thing we know of, for sure,
> is that certain breeding pairs are more likely to produce males, and
> others are more likely to produce females.  We will ignore  those born who
> are of indeterminate sex, for this discussion.
[...]
> All we know is that notion that every human birth has a 50/50
> chance of being male or female is wrong.


This is certainly true. In human beings, certain families tend to run with
all girls or all boys, more often than you would expect from just chance.
This suggests that at least some people are predisposed to have boys, or
girls, rather than equal chance of both.

Oh, and those intersex children? Surprisingly common: Wikipedia suggests
that according to some definition, as many as 1.7% (about one in sixty)
newborns may have ambiguous genitalia, or something other than XY/XX
chromosomes. The frequency of non-XX, non-XY children alone is 3 in 5000.

While it is important to know these facts, to understand that human beings
(like other mammals) do not divide neatly into two distinct sexes, I think
that's a complication not really necessary for basic statistics classes. As
a first approximation, the idea that the proportion of boys and girls are
both roughly equal at 1/2 isn't too far off the truth.

After all, we make other simplifying assumptions about probability too.
There are magicians who are capable of forcing coins to land the required
way up, and somebody once built a machine capable of tossing a coin with
the precise equal force and velocity every single time. Dice are rarely
unbiased, and neither are roulette wheels. Nevertheless, we ignore those
factual biases for the sake of simplicity.


> In human beings, as well as a whole lot of animal species, infant-and-
> chidhood mortality is not sex neutral.  More baby boys die than baby
> girls.

Correct. And more Y sperm die than X sperm. Together, these two factors lead
to a small but real bias towards girl children, at least on countries that
don't practice wide-spread sex-specific abortion.


> But when you look at breeding age men and women you will find
> that the ratio is a lot closer to 50/50 than the birth age ratio.
> Human beings, as a population currently produce more males than
> females.

Surely that depends on where (and when) you are?

I know that, today, both India and China both abort far more female fetuses
than male ones, leading to a large excess of men. But in countries that
don't practice selective abortions, my understanding is that there is a
small excess of women at virtually all ages, especially among the elderly.
Men tend to die earlier than women in every age bracket.



--
Steven


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Testing random

Laura Creighton-2
In a message of Wed, 17 Jun 2015 22:47:37 +1000, "Steven D'Aprano" writes:
>There are magicians who are capable of forcing coins to land the required
>way up, and somebody once built a machine capable of tossing a coin with
>the precise equal force and velocity every single time. Dice are rarely
>unbiased, and neither are roulette wheels. Nevertheless, we ignore those
>factual biases for the sake of simplicity.

Simlplicity is fine.  But if you use this for real world decisions,
such as 'Let's try one more time, I really want a daughter!'
because you didn't know that the simplicity is there, or rather
that what is true in populations isn't true because it is also
true for each individual in a population .. well, that can be tragic.

>Correct. And more Y sperm die than X sperm. Together, these two factors lead
>to a small but real bias towards girl children, at least on countries that
>don't practice wide-spread sex-specific abortion.

It is not clear, last I checked, that 'more Y sperm die' matters, because
most of the time, for reasons we still don't know, the ratio of male
and female sperm produced is not 50/50 either.  So it may matter.  But
maybe not.

>> But when you look at breeding age men and women you will find
>> that the ratio is a lot closer to 50/50 than the birth age ratio.
>> Human beings, as a population currently produce more males than
>> females.
>
>Surely that depends on where (and when) you are?
>
>I know that, today, both India and China both abort far more female fetuses
>than male ones, leading to a large excess of men. But in countries that
>don't practice selective abortions, my understanding is that there is a
>small excess of women at virtually all ages, especially among the elderly.
>Men tend to die earlier than women in every age bracket.

No.  Before breeding age there is an excess of males.

I am excluding the notion of abortion here.  (Another reason why humans
make lousy experimental subjects for this sort of study.)  What I am
saying is that a whole lot more boy babies get born than female ones.
If infant mortality was sex-neutral, we would expect that, when we
took a census of 16-year-olds we would find that there continued to
be an excess of boys at that level, too.  Instead, we have a small
excess of females, enough to believe that:

       For populations of human beings, when they hit breeding age, it
       is desirable to have a sex ratio near 50/50 but with a small
       excess of females.

'Desirable' here just means that this is the answer that pops out when
you study real populations (or model them with computer simulations).

And to get this Desirable outcome, you have to produce many more male
babies than female ones.

Laura


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