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Re: Toughest Probability Question

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  Re: Toughest Probability Question pete <pfiland@mindspring.com> - 2011-12-12 19:29 -0500
    Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-13 04:46 +0000
      Re: Toughest Probability Question Geoff <geoff@invalid.invalid> - 2011-12-12 22:19 -0800
        Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-13 09:19 +0000
          Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-13 13:43 +0000
            Re: Toughest Probability Question Keith Thompson <kst-u@mib.org> - 2011-12-13 10:27 -0800
              Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-13 19:52 +0000
                Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-13 20:06 +0000
                Re: Toughest Probability Question Keith Thompson <kst-u@mib.org> - 2011-12-13 12:14 -0800
                  Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-13 23:34 +0000
                    Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-14 00:24 +0000
            Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-13 19:27 +0000
              Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-13 23:56 +0000
                Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-14 00:24 +0000
                  Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-14 12:12 +0000
                    Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-14 17:47 +0000
                      Re: Toughest Probability Question gordonb.qunma@burditt.org (Gordon Burditt) - 2011-12-15 08:22 -0600
                        Re: Toughest Probability Question Robert Wessel <robertwessel2@yahoo.com> - 2011-12-15 14:41 -0600
                      Re: Toughest Probability Question superpollo <superpollo@tznvy.pbz> - 2011-12-15 19:14 +0100
                    Re: Toughest Probability Question Keith Thompson <kst-u@mib.org> - 2011-12-14 13:12 -0800
                    Re: Toughest Probability Question Willem <willem@toad.stack.nl> - 2011-12-14 21:53 +0000
                      Re: Toughest Probability Question "BartC" <bc@freeuk.com> - 2011-12-14 22:35 +0000
                        Re: Toughest Probability Question Willem <willem@toad.stack.nl> - 2011-12-15 09:07 +0000
                          Re: Toughest Probability Question Geoff <geoff@invalid.invalid> - 2011-12-15 02:07 -0800
                            Re: Toughest Probability Question Keith Thompson <kst-u@mib.org> - 2011-12-15 12:13 -0800
                              Re: Toughest Probability Question Geoff <geoff@invalid.invalid> - 2011-12-15 13:03 -0800
                                Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-15 21:36 +0000
                Re: Toughest Probability Question Geoff <geoff@invalid.invalid> - 2011-12-13 16:35 -0800
                  Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-14 01:20 +0000
                Re: Toughest Probability Question Keith Thompson <kst-u@mib.org> - 2011-12-13 16:44 -0800
                  Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-14 01:20 +0000
                    Re: Toughest Probability Question James Dow Allen <jdallen2000@yahoo.com> - 2011-12-15 02:33 -0800
                      Re: Toughest Probability Question gwowen <gwowen@gmail.com> - 2011-12-15 05:38 -0800
                        Re: Toughest Probability Question Thad Smith <ThadSmith@acm.org> - 2011-12-18 23:34 -0700
                  Re: Toughest Probability Question Bill Reid <hormelfree@gmail.com> - 2011-12-14 07:07 -0800
      Re: Toughest Probability Question Kaz Kylheku <kaz@kylheku.com> - 2011-12-13 06:55 +0000
        Re: Toughest Probability Question Seebs <usenet-nospam@seebs.net> - 2011-12-13 09:19 +0000
      Re: Toughest Probability Question pete <pfiland@mindspring.com> - 2011-12-13 12:15 -0500
        Re: Toughest Probability Question pete <pfiland@mindspring.com> - 2011-12-13 13:20 -0500
          Re: Toughest Probability Question Peter Nilsson <airia@acay.com.au> - 2011-12-13 18:15 -0800
            Re: Toughest Probability Question pete <pfiland@mindspring.com> - 2011-12-13 21:52 -0500

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#14536 — Re: Toughest Probability Question

divya bisht wrote:
> 
> Toughest Probability Question
> (Source : http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-question.html
> )
> 
> How many people must be gathered together in a room, before you can be
> certain that there is a greater than 50/50 chance that at least two of
> them have the same birthday?
> 
> Discuss Solution in link below

/* BEGIN output from new.c */

The probability that 23 random birthdays are all different is 0.492703

/* END output from new.c */



/* BEGIN new.c */

#include <stdio.h>

#define YEAR 365 /* 366 */

int 
main(void)
{
    int days = YEAR;
    double probability = 1;
    int people = 1;
    puts("\n/* BEGIN output from new.c */\n");
    do {
        ++people;
        --days;
        probability *= 1.0 * days / YEAR;
    } while(probability > 0.5);
    printf("The probability that %d "
        "random birthdays are all different is %f\n", 
        people, probability);
    puts("\n/* END output from new.c */");
    return 0;
}

/* END new.c */



-- 
pete

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#14540

On 2011-12-13, pete <pfiland@mindspring.com> wrote:
> divya bisht wrote:
>> How many people must be gathered together in a room, before you can be
>> certain that there is a greater than 50/50 chance that at least two of
>> them have the same birthday?

>> Discuss Solution in link below

> The probability that 23 random birthdays are all different is 0.492703

So?

First off, I'm not sure why so many people respond to the spammer.  Secondly,
consider the wording:  Not before you would reasonably *assume* that there
is a greater than 50/50 chance, but until you can be *certain* that there is a
greater than 50/50 chance.

The answer, of course, is 367.  With 366 people in a room, while it seems
*extremely likely* that the chances are >50% that two share a birthday, that
is not the case if it is the annual gathering of the No Two Birthdays Alike
club.  The question does not promise that the people will be randomly
selected with regards to birthdays.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14541

On 13 Dec 2011 04:46:57 GMT, Seebs <usenet-nospam@seebs.net> wrote:

>On 2011-12-13, pete <pfiland@mindspring.com> wrote:
>> divya bisht wrote:
>>> How many people must be gathered together in a room, before you can be
>>> certain that there is a greater than 50/50 chance that at least two of
>>> them have the same birthday?
>
>>> Discuss Solution in link below
>
>> The probability that 23 random birthdays are all different is 0.492703
>
>So?
>
>First off, I'm not sure why so many people respond to the spammer.  Secondly,
>consider the wording:  Not before you would reasonably *assume* that there
>is a greater than 50/50 chance, but until you can be *certain* that there is a
>greater than 50/50 chance.
>
>The answer, of course, is 367.  With 366 people in a room, while it seems
>*extremely likely* that the chances are >50% that two share a birthday, that
>is not the case if it is the annual gathering of the No Two Birthdays Alike
>club.  The question does not promise that the people will be randomly
>selected with regards to birthdays.
>
>-s

Seebs,

You proved you don't know anything about statistics and probability.

http://www.wolframalpha.com/input/?i=birthday

The probability of AT LEAST two of them having the same birthday is
greater than 50% if the number of people is at least 23.

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#14550

On 2011-12-13, Geoff <geoff@invalid.invalid> wrote:
> You proved you don't know anything about statistics and probability.

No, I didn't.

> The probability of AT LEAST two of them having the same birthday is
> greater than 50% if the number of people is at least 23.

Read my post again, please.  Not the part where I point out that the problem
as stated *does not* specify that the distribution is at all random.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14556

"Seebs" <usenet-nospam@seebs.net> wrote in message 
news:slrnjee4qh.2irh.usenet-nospam@guild.seebs.net...
> On 2011-12-13, Geoff <geoff@invalid.invalid> wrote:

>> The probability of AT LEAST two of them having the same birthday is
>> greater than 50% if the number of people is at least 23.
>
> Read my post again, please.  Not the part where I point out that the 
> problem
> as stated *does not* specify that the distribution is at all random.

And read the OP's post again:

"before you can be certain that there is a greater than 50/50 chance"

and note the word "chance". With 23 people, the chance IS greater than 50%, 
for certain. What is not for certain is that any two of those people WILL 
have the same birthday.

-- 
Bartc 

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#14565

"BartC" <bc@freeuk.com> writes:
> "Seebs" <usenet-nospam@seebs.net> wrote in message
> news:slrnjee4qh.2irh.usenet-nospam@guild.seebs.net...
>> On 2011-12-13, Geoff <geoff@invalid.invalid> wrote:
>
>>> The probability of AT LEAST two of them having the same birthday is
>>> greater than 50% if the number of people is at least 23.
>>
>> Read my post again, please.  Not the part where I point out that the
>> problem
>> as stated *does not* specify that the distribution is at all random.
>
> And read the OP's post again:
>
> "before you can be certain that there is a greater than 50/50 chance"
>
> and note the word "chance". With 23 people, the chance IS greater than
> 50%, for certain. What is not for certain is that any two of those
> people WILL have the same birthday.

An annual gathering of the No Two Birthdays Alike club, which has
366 members (and lives up to its name), satisfies the conditions
stated in the problem and results in a 0 probability that at least
two of them have the same birthday.

The probability for 23 people is greater than 50% *if* their
birthdays are random.  The author of the problem almost certainly
*intended* to be random, but didn't actually say so.

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
    Will write code for food.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

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#14569

"Keith Thompson" <kst-u@mib.org> wrote in message 
news:ln39co8gyw.fsf@nuthaus.mib.org...
> "BartC" <bc@freeuk.com> writes:
>> "Seebs" <usenet-nospam@seebs.net> wrote in message
>> news:slrnjee4qh.2irh.usenet-nospam@guild.seebs.net...
>>> On 2011-12-13, Geoff <geoff@invalid.invalid> wrote:
>>
>>>> The probability of AT LEAST two of them having the same birthday is
>>>> greater than 50% if the number of people is at least 23.
>>>
>>> Read my post again, please.  Not the part where I point out that the
>>> problem
>>> as stated *does not* specify that the distribution is at all random.
>>
>> And read the OP's post again:
>>
>> "before you can be certain that there is a greater than 50/50 chance"
>>
>> and note the word "chance". With 23 people, the chance IS greater than
>> 50%, for certain. What is not for certain is that any two of those
>> people WILL have the same birthday.
>
> An annual gathering of the No Two Birthdays Alike club, which has
> 366 members (and lives up to its name), satisfies the conditions
> stated in the problem and results in a 0 probability that at least
> two of them have the same birthday.

That's true. But it doesn't say anything about what kind of people are in 
the room, therefore we can't speculate.

The probability of just over 50%, for 23 people people, means that's what 
you might expect when you examine large numbers of such gatherings; half the 
time, a birthday will be shared. Other times, in the case of your annual 
gathering for example, it won't be.

If you had to place a bet, in the absence of any extra knowledge, the odds 
would have to be about even.

-- 
Bartc

 

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#14571

On 2011-12-13, BartC <bc@freeuk.com> wrote:
> That's true. But it doesn't say anything about what kind of people are in 
> the room, therefore we can't speculate.

No, therefore we can't rule out possible weirdnesses.

> If you had to place a bet, in the absence of any extra knowledge, the odds 
> would have to be about even.

Yes.  But he didn't say "when would you make a bet", but "when would you
be certain".  And I can't be certain that the probability exceeds 50% until
I know what the sample space is.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14572

"BartC" <bc@freeuk.com> writes:
> "Keith Thompson" <kst-u@mib.org> wrote in message 
> news:ln39co8gyw.fsf@nuthaus.mib.org...
>> "BartC" <bc@freeuk.com> writes:
>>> "Seebs" <usenet-nospam@seebs.net> wrote in message
>>> news:slrnjee4qh.2irh.usenet-nospam@guild.seebs.net...
>>>> On 2011-12-13, Geoff <geoff@invalid.invalid> wrote:
>>>
>>>>> The probability of AT LEAST two of them having the same birthday is
>>>>> greater than 50% if the number of people is at least 23.
>>>>
>>>> Read my post again, please.  Not the part where I point out that the
>>>> problem
>>>> as stated *does not* specify that the distribution is at all random.
>>>
>>> And read the OP's post again:
>>>
>>> "before you can be certain that there is a greater than 50/50 chance"
>>>
>>> and note the word "chance". With 23 people, the chance IS greater than
>>> 50%, for certain. What is not for certain is that any two of those
>>> people WILL have the same birthday.
>>
>> An annual gathering of the No Two Birthdays Alike club, which has
>> 366 members (and lives up to its name), satisfies the conditions
>> stated in the problem and results in a 0 probability that at least
>> two of them have the same birthday.
>
> That's true. But it doesn't say anything about what kind of people are in 
> the room, therefore we can't speculate.

Agreed.

And yet you *are* speculating.  Specifically, you're speculating
that the birthdays are randomly distributed.

You cannot be *certain* that there's a greater than 50/50 chance
(as stated in the original question) without more information,
unless there are at least 367 people.

[snip]

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
    Will write code for food.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

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#14585

"Keith Thompson" <kst-u@mib.org> wrote in message
news:lnpqfs6xg3.fsf@nuthaus.mib.org...
> "BartC" <bc@freeuk.com> writes:
>> "Keith Thompson" <kst-u@mib.org> wrote in message

>>> An annual gathering of the No Two Birthdays Alike club, which has
>>> 366 members (and lives up to its name), satisfies the conditions
>>> stated in the problem and results in a 0 probability that at least
>>> two of them have the same birthday.
>>
>> That's true. But it doesn't say anything about what kind of people are in
>> the room, therefore we can't speculate.
>
> Agreed.
>
> And yet you *are* speculating.  Specifically, you're speculating
> that the birthdays are randomly distributed.

I'm not speculating at all. That's why I have to come up with a probability
that is not 0% or 100%.

The probability can't be 100% unless there are 367 people gathered, as you
say, and it cannot be 0% unless you have a group of one person or less.

What you're saying is that with a group size of 2 to 366 people, any
probability you can come up with (for at least two of the group to share a
birthday) is meaningless. But I think that figure does have a practical use.

-- 
bartc 

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#14587

On 2011-12-13, BartC <bc@freeuk.com> wrote:
> I'm not speculating at all.

Sure you are.  You're speculating that the population is random.

> That's why I have to come up with a probability
> that is not 0% or 100%.

There's two different layers of speculation under consideration.

One is speculating as to the exact set of people in the room.  The other is
speculating as to the sample space from which those people were selected.

Consider a maternity ward; the chances of duplicate birthdays are MUCH
higher there than they are in the rest of the world.

> The probability can't be 100% unless there are 367 people gathered, as you
> say, and it cannot be 0% unless you have a group of one person or less.

Not so!

> What you're saying is that with a group size of 2 to 366 people, any
> probability you can come up with (for at least two of the group to share a
> birthday) is meaningless. But I think that figure does have a practical use.

No.  I'm saying that we can give a general answer that's true for random
distributions, and we can qualify it for some specific non-random
distributions.

If you ask me what I think the probability is that, given 23 people, two of
them have the same birthday, I'll say "a bit over 50%".  But if you ask me
how *sure* I am, I'll have to admit that I'm not very sure, because I have
*assumed* that the 23 people are selected randomly without regard to their
birthdays.  Since that might not be the case, the probability might not really
be 50%.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14568

On 2011-12-13, BartC <bc@freeuk.com> wrote:
> And read the OP's post again:

I did.

> "before you can be certain that there is a greater than 50/50 chance"

> and note the word "chance". With 23 people, the chance IS greater than 50%, 
> for certain.

Only if they're randomly selected.

Okay, I'm gonna have to be clearer.

My point is that "be *certain* that there is a greater than 50/50 chance"
(emphasis mine) could be read as implying not merely casual back of the
envelope probability math, but a rather higher standard, which is checking
for unusual circumstances and the like, such as *non-random distributions*.

Consider the question "how many people *can* be in a room and you can
still be certain there is a less than 50/50 chance...".  Obviously, 22,
right?

What if the room is the Department of Motor Vehicles license office?
Driver's licenses around here expire on your birthday, meaning that there
is a slight bias towards people whose birthdays are near.  You might find
that, in a DMV office, the chances of duplicate birthdays are about 50%
at only 16 people.

Or, for the original problem, what about a Chuck E. Cheese pizza restaurant,
which specializes in holding birthday parties for children?  The chances
will be 50% of two identical birthdays at lower than 23, probably --
specifically, today.

Since it is obviously possible to come up with a sample space *selected for*
lack of duplicate birthdays, it is easy to have a room containing 366
people selected from a sample such that, no matter which 366 people you
pick, the chances are 0% that any have the same birthday, because if they'd
had the same birthday as another member of the sample, they wouldn't have
been in the sample.

To be *certain* of a chance is not just to do the math with common
assumptions.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14586

"Seebs" <usenet-nospam@seebs.net> wrote in message 
news:slrnjef9k1.30f0.usenet-nospam@guild.seebs.net...
> On 2011-12-13, BartC <bc@freeuk.com> wrote:

>> and note the word "chance". With 23 people, the chance IS greater than 
>> 50%,
>> for certain.

> Since it is obviously possible to come up with a sample space *selected 
> for*
> lack of duplicate birthdays, it is easy to have a room containing 366
> people selected from a sample such that, no matter which 366 people you
> pick, the chances are 0% that any have the same birthday, because if 
> they'd
> had the same birthday as another member of the sample, they wouldn't have
> been in the sample.
>
> To be *certain* of a chance is not just to do the math with common
> assumptions.

So what *is* the probability that out of a group of 23 people, at least two 
will share a birthday? Or any group of people smaller than 367? And how 
certain are you of that figure?

-- 
bartc 

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#14588

On 2011-12-13, BartC <bc@freeuk.com> wrote:
> So what *is* the probability that out of a group of 23 people, at least two 
> will share a birthday?

Depends on what sample space those 23 people were selected from.

> Or any group of people smaller than 367? And how 
> certain are you of that figure?

Not very certain at all unless someone tells me the sample space.

There are a whole lot of ways to select random people which produce an uneven
distribution of birthdays.  Consider the birthdays of successful professional
sports players, for instance.  Do you think that's going to be an even
representative sample?  It's not, at least in the US.

Why, you ask?  Well.  Professional sports players tend to be people who did
well in sports as kids, yes?  That group, it turns out, has a bias; because
of the way the cutoffs for entering school are calculated, there exists a pair
of adjacent days such that kids born on those two days will actually be just
shy of a full year different in age (counted as number of days since birth,
not rounded-off years) when they enter 9th grade.  Meaning one of them has
almost a full year's growth on the other.  And that turns out to correlate
noticably with performance, and thus with motivation to train harder and
pursue sports actively.

See, for instance:

	http://www.socialproblemindex.ualberta.ca/relage.htm#Elite

So what I'm saying is, in the absence of qualifiers, I can give an answer
which is correct for a random distribution, but not for a non-random
distribution.

What's the chance of getting 18 by rolling three 6-sided dice and adding
them?  PROBABLY one in 216, but loaded dice are available for purchase, and
with a set of loaded dice, it's much closer to unity.  If you stipulate fair
dice, it's 1 in 216.  If you take dice which are known to be loaded, it's
obviously not.  But what if I don't *know* whether the dice are fair or not?
Then I don't really know the probability, or have enough information to allow
me to calculate it.

Consider it a philosophical point -- but perhaps a significant one.  People
who have studied probability but not thought carefully about the assumptions
make a LOT of mistakes in evaluating the probabilities of events in the
real world.

To bring this vaguely on topic:

You have a pointer which contains stack garbage.  Your process has a
4MB memory footprint in a 4GB address space.  What are the chances that the
pointer will point into your memory footprint?

Hint:  They're almost certainly not 1/1024.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14600

"Seebs" <usenet-nospam@seebs.net> wrote in message 
news:slrnjefr5m.7q7.usenet-nospam@guild.seebs.net...
> On 2011-12-13, BartC <bc@freeuk.com> wrote:
>> So what *is* the probability that out of a group of 23 people, at least 
>> two
>> will share a birthday?

> Consider it a philosophical point -- but perhaps a significant one. 
> People
> who have studied probability but not thought carefully about the 
> assumptions
> make a LOT of mistakes in evaluating the probabilities of events in the
> real world.

Probability is a useful mathematical tool. There will be discrepancies when 
applied to real-life situations, but that in itself can be useful, when you 
investigate why.

In the case of real-life groupings of two dozen or more people that you 
might typically find (even if you have no idea why any particular gathering 
has occurred) I don't think the theoretical probabilities are going to that 
different from actual results.

So if you're at a party with some 60 people, I believe even you would take 
on a bet against someone who didn't think anyone would share a birthday. (I 
believe the theoretical probability is about 99%; someone who hasn't thought 
it about it too carefully, and might be also be drunk, might think it was 
16%).

>
> To bring this vaguely on topic:
>
> You have a pointer which contains stack garbage.  Your process has a
> 4MB memory footprint in a 4GB address space.  What are the chances that 
> the
> pointer will point into your memory footprint?
>
> Hint:  They're almost certainly not 1/1024.

I've no idea. The answer could be anything. However I wouldn't try 
dereferencing such a pointer, even if the probability was 100%.

-- 
Bartc 

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#14603

On 2011-12-14, BartC <bc@freeuk.com> wrote:
> Probability is a useful mathematical tool. There will be discrepancies when 
> applied to real-life situations, but that in itself can be useful, when you 
> investigate why.

Yes.

> In the case of real-life groupings of two dozen or more people that you 
> might typically find (even if you have no idea why any particular gathering 
> has occurred) I don't think the theoretical probabilities are going to that 
> different from actual results.

I'd disagree for at least two obvious cases:
* Restaurants that specifically seek out birthday party traffic.
* License bureaus in places where licenses expire on birthdays.

Both of these, I'd guess that at 24 people the probability would be noticably
higher than 50%, not just very marginally higher.

> So if you're at a party with some 60 people, I believe even you would take 
> on a bet against someone who didn't think anyone would share a birthday. (I 
> believe the theoretical probability is about 99%; someone who hasn't thought 
> it about it too carefully, and might be also be drunk, might think it was 
> 16%).

Oh, sure, I'd take the bet -- but I wouldn't claim to be *certain*.

> I've no idea. The answer could be anything. However I wouldn't try 
> dereferencing such a pointer, even if the probability was 100%.

Of course not.  But it might be useful to have an estimate when trying
to figure out why something is or isn't going wrong with code that you're
suspicious of.

-s
-- 
Copyright 2011, all wrongs reversed.  Peter Seebach / usenet-nospam@seebs.net
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
I am not speaking for my employer, although they do rent some of my opinions.

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#14628

> I'd disagree for at least two obvious cases:
> * Restaurants that specifically seek out birthday party traffic.
> * License bureaus in places where licenses expire on birthdays.

And another case:  US war veterans (from the Vietnam war) born in
the range 1944-1956.  (For those of you not familiar with the draft
lottery in the USA, based on birthdays, see http://www.sss.gov/lotter1.htm
).  Now, granted, some different birth years used a different set
of drawings, although 1944-1950 used the first drawing, which is
more than half the range.  I do not know whether that bias is enough
to skew the birthday distribution of meetings of retired veterans
(any war), or perhaps occupants of veterans hospitals significantly.

Any organization where membership is based on being a multiple birth
may have seriously skewed birthday distribution at meetings.  Multiple
births run in families, so certain family reunions may have more
multiple births than would be expected.

Do humans even have an even distribution of birthdays across seasons?
Many animals don't.  There are known instances of spikes in human
births 9 months after blackouts.  Is there a more general spike in
births 9 months after, say, winter in cold climates?  Or how about
9 or 10 months after June weddings?

Question:  given that a person at a gathering of people is a twin,
what is the probability that this person's twin is also at the
meeting?  There are also statistics available on the percentage of
multiple births.

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#14637

On Thu, 15 Dec 2011 08:22:29 -0600, gordonb.qunma@burditt.org (Gordon
Burditt) wrote:

>> I'd disagree for at least two obvious cases:
>> * Restaurants that specifically seek out birthday party traffic.
>> * License bureaus in places where licenses expire on birthdays.
>
>(...)
>
>Do humans even have an even distribution of birthdays across seasons?
>Many animals don't.  There are known instances of spikes in human
>births 9 months after blackouts.  Is there a more general spike in
>births 9 months after, say, winter in cold climates?  Or how about
>9 or 10 months after June weddings?


There's actually considerable literature on that point.  These days,
European and American populations have a minor birth rate peak in the
spring, and a major one in the fall.  This is also changeable - in the
first half of the twentieth century, the European (but not American)
peaks were reversed (big peak in spring, small peak in fall).  The
size of the peaks also appears to correlate with latitude.

Humans do not have estrous tied to the seasons as many animal do,
which makes this more choice for us than nature.  Polar bears, as an
example, mate in April and May, so that their cubs are born during the
next hibernation cycles, and are mature enough to emerge from the den
at the end of that winter with a good chance of surviving the
(relatively) mild summer and maturing enough to deal with the
following winter.  A polar bear giving birth in (say) August, would
have few if any surviving offspring.

The reasons are debated endlessly, but one consideration (at least in
the US) appears to be school schedules - most schools have a birthday
cutoff at the beginning of the school year (or sometimes a month or
two earlier).  So if you time your child's birth for September,
they'll be amongst the oldest in their class, while the August babies
in the same class will be almost a year younger.  That too has changed
- when I entered school, the common cutoff was December 1st, and I was
one of the youngest (my birthday being just a few days before that).

At some point someone gathering these statistics may have used a C
program...

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#14631

Seebs ha scritto:
> On 2011-12-14, BartC <bc@freeuk.com> wrote:
>> Probability is a useful mathematical tool. There will be discrepancies when 
>> applied to real-life situations, but that in itself can be useful, when you 
>> investigate why.
> 
> Yes.
> 
>> In the case of real-life groupings of two dozen or more people that you 
>> might typically find (even if you have no idea why any particular gathering 
>> has occurred) I don't think the theoretical probabilities are going to that 
>> different from actual results.
> 
> I'd disagree for at least two obvious cases:
> * Restaurants that specifically seek out birthday party traffic.
> * License bureaus in places where licenses expire on birthdays.
> 
> Both of these, I'd guess that at 24 people the probability would be noticably
> higher than 50%, not just very marginally higher.
> 
>> So if you're at a party with some 60 people, I believe even you would take 
>> on a bet against someone who didn't think anyone would share a birthday. (I 
>> believe the theoretical probability is about 99%; someone who hasn't thought 
>> it about it too carefully, and might be also be drunk, might think it was 
>> 16%).
> 
> Oh, sure, I'd take the bet -- but I wouldn't claim to be *certain*.

but you know, probability is all about taking bets...

bye

-- 
"Remember when teachers, public employees, Planned Parenthood,
NPR and PBS crashed the stock market, wiped out half of our 401Ks,
took trillions in TARP money, spilled oil in the Gulf of Mexico,
gave themselves billions in bonuses, and paid no taxes? Yeah,
me neither."

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#14605

"BartC" <bc@freeuk.com> writes:
> "Seebs" <usenet-nospam@seebs.net> wrote in message 
> news:slrnjefr5m.7q7.usenet-nospam@guild.seebs.net...
>> On 2011-12-13, BartC <bc@freeuk.com> wrote:
>>> So what *is* the probability that out of a group of 23 people, at least 
>>> two
>>> will share a birthday?
>
>> Consider it a philosophical point -- but perhaps a significant one. 
>> People
>> who have studied probability but not thought carefully about the 
>> assumptions
>> make a LOT of mistakes in evaluating the probabilities of events in the
>> real world.
>
> Probability is a useful mathematical tool. There will be discrepancies when 
> applied to real-life situations, but that in itself can be useful, when you 
> investigate why.
>
> In the case of real-life groupings of two dozen or more people that you 
> might typically find (even if you have no idea why any particular gathering 
> has occurred) I don't think the theoretical probabilities are going to that 
> different from actual results.
>
> So if you're at a party with some 60 people, I believe even you would take 
> on a bet against someone who didn't think anyone would share a birthday. (I 
> believe the theoretical probability is about 99%; someone who hasn't thought 
> it about it too carefully, and might be also be drunk, might think it was 
> 16%).

If I'm at a party with 60 people, I probably know something about them,
in particular that they weren't selected for their birthdays.

The problem, as stated, asked for *certainty* that the probability of a
match is greater than 50%.  And it didn't say anything about a party.

And this has nothing to do with C.

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
    Will write code for food.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

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