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```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?

http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-question.html

```
 0
Reply bisht88.divya (73) 12/12/2011 11:57:56 AM

See related articles to this posting

```On Dec 12, 11:57=A0am, divya bisht <bisht88.di...@gmail.com> wrote:
> Toughest Probability Question
> (Source :http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-=
quest...
> )
>
> 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 belowhttp://hardest-puzzle.blogspot.com/2011/12/=
toughest-probability-quest...

Two. Invite a pair of twins - they're almost certain to have the same
birthday.
```
 0
Reply gw7rib (471) 12/12/2011 8:57:17 PM

```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
```
 0
Reply pfiland (6614) 12/13/2011 12:29:50 AM

```On Dec 12, 12:57=A0pm, divya bisht <bisht88.di...@gmail.com> wrote:
> Toughest Probability Question
> (Source :http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-=
quest...
> )
>
> 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?
>

What this has to do with C I don't know.  There are 365 (or 366)
choose N ways to arrange N people on the calendar with no collisions,
ignoring order. Considering order it's N! times 365 choose N, which is
365!/(365-N)!  On the other hand there are 365^N total ways N people
can be placed on the calendar.  We want the probability of a collision
to be less than half. So the answer is the smallest N such that

( 365! / (365 - N)! ) / 365^N < 1/2

This happens at N=3D23, which you can verify with a suitable C
program ;-)
```
 0
Reply gene.ressler (563) 12/13/2011 12:49:28 AM

```On Dec 13, 7:57=A0am, Paul N <gw7...@aol.com> wrote:
> On Dec 12, 11:57=A0am, divya bisht <bisht88.di...@gmail.com> 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?
>
> Two. Invite a pair of twins...

Or just invite someone else with the same birthday as you.

--
Peter
```
 0
Reply airia (1802) 12/13/2011 3:49:26 AM

```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.
```
 0
Reply usenet-nospam (2309) 12/13/2011 4:46:57 AM

```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.

```
 0
Reply geoff745 (472) 12/13/2011 6:19:35 AM

```On 2011-12-13, 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.

This kind of assumption is reasonable for this type of academic puzzle problem.

"People in a room", not qualified in a way, is easily recognizeable
"exam-speak" denoting a "random sample from a population".

Any bias in the selection of the people introduces more information into the
problem, and therefore makes the solution harder.

If the examiner posing this problem to you wanted you to work harder, he would

I.e. "don't make it harder than it is." If a key assumption is not stated which
you think should be, just write it down and proceed.  Assume that the coin
being tossed is fair, that the people in the room are randomly selected, that
events are independent even if the question forgot to say, etc. If you think
a key assumption should have been made explicit, write it down and keep going.
If you're wrong, you may still get partial marks.

Examsmanship is not exactly the same thing as real-world problem solving,
where you are usually right to avoid jumping to conclusions and investigate to
get as much information as you can, if you have time.
```
 0
Reply kaz15 (1143) 12/13/2011 6:55:50 AM

```On Dec 12, 11:57=A0am, divya bisht <bisht88.di...@gmail.com> wrote:
> Toughest Probability Question
> (Source :http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-=
quest...
> )
>
> 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 belowhttp://hardest-puzzle.blogspot.com/2011/12/=
toughest-probability-quest...

5s of googling would have found the answer...
Even if you hadn't seen the problem before it's not particularly hard.
```
 0
Reply nick_keighley_nospam (4575) 12/13/2011 8:40:54 AM

```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.
```
 0
Reply usenet-nospam (2309) 12/13/2011 9:19:10 AM

```On 2011-12-13, Kaz Kylheku <kaz@kylheku.com> wrote:
> This kind of assumption is reasonable for this type of academic
> puzzle problem.

That's quite likely, but...

> "People in a room", not qualified in a way, is easily recognizeable
> "exam-speak" denoting a "random sample from a population".

Yes.  But it's not what the questioner asked, and one of the things I look
for when I poke at brain teasers is people who question the "obvious"
assumptions -- as after all, that's one of the things that makes for a
good brain teaser.

I had a great deal of fun once with a very elaborate liars-and-truth-tellers
puzzle.  The question was "Can you, by asking a single non-compound question
of one person, find out reliably which path is safe?"

The answer, of course, was "no."

> Any bias in the selection of the people introduces more information into the
> problem, and therefore makes the solution harder.

Or possibly makes it completely trivial.  :)

> Examsmanship is not exactly the same thing as real-world problem solving,
> where you are usually right to avoid jumping to conclusions and investigate to
> get as much information as you can, if you have time.

Yes.  And neither is the same as answering brain teasers.  :)

-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.
```
 0
Reply usenet-nospam (2309) 12/13/2011 9:19:11 AM

```"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

```
 0
Reply bc (2337) 12/13/2011 1:43:16 PM

```Seebs 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?

I made a mistake in the program.

This line:

printf("The probability that %d "
"random birthdays are all different is %f\n",
people, probability);

should have been written as:

printf("The probability that %d "
"random birthdays are all different is %f\n",
people, 1.0 - probability);

and the output should have been :

The probability that 23 random birthdays are all different is 0.507297

--
pete
```
 0
Reply pfiland (6614) 12/13/2011 5:15:35 PM

```pete wrote:
>
> Seebs 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?
>
> I made a mistake in the program.

I had it right the first time,
but I rewrote the whole thing anyway.

/* BEGIN output from new.c */

The probability that out of 23 random birthdays,
that at least two are the same is 0.507297

/* END output from new.c */

/* BEGIN new.c */

#include <stdio.h>

#define YEAR                365  /* 366 */
#define PROB_SOME_SAME      0.5
#define PROB_ALL_DIFFERENT  (1 - PROB_SOME_SAME)

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 > PROB_ALL_DIFFERENT);
printf("The probability that out of %d "
"random birthdays,\n"
"that at least two are the same is %f\n",
people, 1 - probability);
puts("\n/* END output from new.c */");
return 0;
}

/* END new.c */

--
pete
```
 0
Reply pfiland (6614) 12/13/2011 6:20:07 PM

```"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"
```
 0
Reply kst-u (21963) 12/13/2011 6:27:35 PM

```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?
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.
```
 0
Reply usenet-nospam (2309) 12/13/2011 7:27:39 PM

```
"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

```
 0
Reply bc (2337) 12/13/2011 7:52:18 PM

```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.
```
 0
Reply usenet-nospam (2309) 12/13/2011 8:06:46 PM

```"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
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"
```
 0
Reply kst-u (21963) 12/13/2011 8:14:36 PM

```"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

```
 0
Reply bc (2337) 12/13/2011 11:34:51 PM

```"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

```
 0
Reply bc (2337) 12/13/2011 11:56:59 PM

```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.
```
 0
Reply usenet-nospam (2309) 12/14/2011 12:24:35 AM

```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.
```
 0
Reply usenet-nospam (2309) 12/14/2011 12:24:35 AM

```On Tue, 13 Dec 2011 23:56:59 -0000, "BartC" <bc@freeuk.com> wrote:

>"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?

To answer that question you must first calculate the probability that
the No Two Birthdays Alike Club (NTBAC) exists and then calculate the
probability that 23 members of it would be congregated for this
experiment.

The fact of the matter is that since the problem was expressed in the
category of probability and statistics and the implicit assumption is
that the congregation is a random sample of the general population of
"people" the special case need not be considered. Seebs is merely
trying to score points against what he admitted at the outset was a
spammer.

The probability equation applicable to the question won't change. The
correct answer is you can be certain the probability that any two
people in a gathering of 23 people or more will have the same birthday
is greater than 50% because the equation holds true regardless of the
outlying probability that your sample is skewed. The probability of
the NTBAC existing would merely add another term to the equation and
since the probability of the existence of the NTBAC is vanishingly
small it's influence on the problem is near zero.

It's important to note that the certainty is applied to the
probability being greater than 50%, not to the probability of there
actually being a birthday match. The vanishingly small probability
that you have a congregation of a subset of the NTBAC would only make
the number 24 instead of 23, at best.

Pr = 1-(365!)/(365^n (365-n)!) * (1 - Ps)
where
Pr is probability at least two with the same birthday
n is number of people
Ps is probability of sample being NTBAC members (skewed)
Ps is vanishingly small
```
 0
Reply geoff745 (472) 12/14/2011 12:35:38 AM

```"BartC" <bc@freeuk.com> writes:
> "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?

There is insufficient information to answer the question without
making further assumptions, in particular without knowing how the
23 people were selected.

If I were actually asked this question (say, in the course of a
job interview), I'd make any such assumptions explicit: "Assuming
that their birthdays are randomly distributed, and ignoring leap
years ...".  I might also discuss alternative answers; for example,
the probability is 100% if you have two people selected to have
the same birthday, and 0% if you have 366 people selected to have
distinct birthdays.

A random uniform distribution is no more valid than any other
distribution, and there are plenty of real-world circumstances in
which a uniform distribution is highly unlikely.

--
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"
```
 0
Reply kst-u (21963) 12/14/2011 12:44:09 AM

```On 2011-12-14, Geoff <geoff@invalid.invalid> wrote:
> Seebs is merely
> trying to score points against what he admitted at the outset was a
> spammer.

Not really; just pointing out that the way the question is phrased admits
a different interpretation of what the question is.  In particular, "certain"
is a rather strong claim.

> It's important to note that the certainty is applied to the
> probability being greater than 50%, not to the probability of there
> actually being a birthday match. The vanishingly small probability
> that you have a congregation of a subset of the NTBAC would only make
> the number 24 instead of 23, at best.

That depends on what you mean by "certain".  I would interpret it as meaning
"probability exactly unity".  In which case, the theoretical possibility
that the NTBAC exists is sufficient to make it "uncertain".  :)

-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.
```
 0
Reply usenet-nospam (2309) 12/14/2011 1:20:35 AM

```On 2011-12-14, Keith Thompson <kst-u@mib.org> wrote:
> A random uniform distribution is no more valid than any other
> distribution, and there are plenty of real-world circumstances in
> which a uniform distribution is highly unlikely.

Consider, for instance, the distribution of digits in numbers.  A lot of
people are aware that the distributions of first digits are uneven; for
instance, if you took a list of "the populations of all cities in an
almanac", you would find that the first digits 1-4 were more heavily
represented than 5-9 by a fair margin.  On the other hand, if you looked
in the tens place of the same numbers, you'd probably find a very even
distribution of 0-9.

Now consider the question of the overall distribution of digits in a
large body of material.  Say, every digit anywhere in said almanac.
It should be obvious that some digits will be more common than others,
with 0 significantly less common than 1.

.... Now, given that:  Take the pool of all the digits in a given almanac,
including repeated copies as separate entries in the pool.  Select two of
those digits at random.  What is the probability that they're the same?
It's not 1/10.

Real world sets are, quite often, NOT random.  In fact, random sets are
sort of an anomaly.  :)

-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.
```
 0
Reply usenet-nospam (2309) 12/14/2011 1:20:36 AM

```> pete wrote:
> I had it right the first time,
> but I rewrote the whole thing anyway.
> ...
> #define YEAR =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0365 =A0/* 366 */

To nitpick, this is limiting people to those only born in non
leap years, or those only born in leap years. The general figure
is somewhere in between...

#include <stdio.h>

double P2(int f29, int oth, int n)
{
double p =3D 0;
if (n =3D=3D 0) return 1;
if (oth < 365)
p +=3D (303. / 400 + 97. / 400 * 365 / 366)
* (365 - oth) / 365 * P2(f29, oth + 1, n - 1);
if (f29 < 1)
p +=3D 97. / 400 * 1 / 366 * P2(f29 + 1, oth, n - 1);
return p;
}

double P(int n) { return P2(0, 0, n); }

int main(void)
{
int n =3D 0;
double p;

do {
p =3D P(++n);
printf("%3d: %7.3f%%\n", n, 100 * (1 - p));
}  while (1 - p < .5);

return 0;
}

--
Peter
```
 0
Reply airia (1802) 12/14/2011 2:15:33 AM

```Peter Nilsson wrote:
>
> > pete wrote:
> > I had it right the first time,
> > but I rewrote the whole thing anyway.
> > ...
> > #define YEAR                365  /* 366 */
>
> To nitpick, this is limiting people to those only born in non
> leap years, or those only born in leap years. The general figure
> is somewhere in between...
>
>   #include <stdio.h>
>
>   double P2(int f29, int oth, int n)
>   {
>     double p = 0;
>     if (n == 0) return 1;
>     if (oth < 365)
>       p += (303. / 400 + 97. / 400 * 365 / 366)
>          * (365 - oth) / 365 * P2(f29, oth + 1, n - 1);
>     if (f29 < 1)
>       p += 97. / 400 * 1 / 366 * P2(f29 + 1, oth, n - 1);
>     return p;
>   }
>
>   double P(int n) { return P2(0, 0, n); }
>
>   int main(void)
>   {
>     int n = 0;
>     double p;
>
>     do {
>       p = P(++n);
>       printf("%3d: %7.3f%%\n", n, 100 * (1 - p));
>     }  while (1 - p < .5);
>
>     return 0;
>   }

I was wondering about how to deal with that.

--
pete
```
 0
Reply pfiland (6614) 12/14/2011 2:52:11 AM

```"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

```
 0
Reply bc (2337) 12/14/2011 12:12:16 PM

```On Dec 13, 4:44=A0pm, Keith Thompson <ks...@mib.org> wrote:
> "BartC" <b...@freeuk.com> writes:
> > "Seebs" <usenet-nos...@seebs.net> wrote in message
> >news:slrnjef9k1.30f0.usenet-nospam@guild.seebs.net...
> >> On 2011-12-13, BartC <b...@freeuk.com> wrote:
> >>> and note the word "chance". With 23 people, the chance IS greater tha=
n
> >>> 50%,
> >>> for certain.
>
> >> Since it is obviously possible to come up with a sample space *selecte=
d
> >> 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 yo=
u
> >> 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 h=
ave
> >> 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?
>
> There is insufficient information to answer the question without
> making further assumptions, in particular without knowing how the
> 23 people were selected.
>
The actual problem with you answering this question is that you
don't know how to solve it...

> If I were actually asked this question (say, in the course of a
> job interview), I'd make any such assumptions explicit: "Assuming
> that their birthdays are randomly distributed, and ignoring leap
> years ...".

To obfuscate the simple fact you don't know how to solve the
problem...your life strategy is "if you can't dazzle them with
brilliance,
baffle them with BS"...

> I might also discuss alternative answers; for example,
> the probability is 100% if you have two people selected to have
> the same birthday, and 0% if you have 366 people selected to have
> distinct birthdays.
>
Since the word "chance" was in the question, obviously the
question was how to perform stochastic calculations on an
assumed random distribution.  Since you don't know how to do
that, you waste the interviewer's time, just like you waste
the readers of this group's time...

> A random uniform distribution is no more valid than any other
> distribution, and there are plenty of real-world circumstances in
> which a uniform distribution is highly unlikely.
>
What is the chance that a comp.lang.c regular will waste
everybody's time by not answering a simple question but rather
re-iterate the return value of main()?  Hint: it is the same
chance that any two randomly-selected US race horses will
share the exact same official birthday...

> Keith Thompson (The_Other_Keith) ks...@mib.org =A0<http://www.ghoti.net/~=
kst>
> =A0 =A0 Will write code for food.

Liar.  You will argue the chrome off a trailer hitch for food, and
maybe in the past there was a special place and enough spare money
to pay you to do that, and who knows, maybe in the future there will
continue to be some type of welfare system to support people like
you.
But just like the word "chance", the common-sense interpretation of
the word "code" implies that the "code" will do something useful
for mankind, and I honestly don't believe you have the capability
to perform useful work.

---
William Ernest Reid
```
 0
Reply hormelfree1 (57) 12/14/2011 3:07:41 PM

```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.

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.
```
 0
Reply usenet-nospam (2309) 12/14/2011 5:47:02 PM

```"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"
```
 0
Reply kst-u (21963) 12/14/2011 9:12:49 PM

```BartC wrote:
) 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%).

Note that the original question was something like:
When can you be *certain* that the chance is above 50% ?

Here's a probability question for you:

Suppose you pick a restaurant that seats exactly 23 people.  It's so
popular that every evening, it is full (with 23 people).

For every evening of an entire year, you interview the 23 people that are
there that evening, and you note if there are two that have the same
birthday.

So now you have 365 cases of a group of 23 people.  You would say that in
about half of those cases, there would be a double birthday, right?

Now here's the question: In the given scenario, what is the probability
that there was a double birthday in less than 150 cases?

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
```
 0
Reply willem6 (255) 12/14/2011 9:53:41 PM

```
> BartC wrote:
> ) 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%).
>
> Note that the original question was something like:
> When can you be *certain* that the chance is above 50% ?
>
>
> Here's a probability question for you:
>
> Suppose you pick a restaurant that seats exactly 23 people.  It's so
> popular that every evening, it is full (with 23 people).
>
> For every evening of an entire year, you interview the 23 people that are
> there that evening, and you note if there are two that have the same
> birthday.
>
> So now you have 365 cases of a group of 23 people.  You would say that in
> about half of those cases, there would be a double birthday, right?
>
> Now here's the question: In the given scenario, what is the probability
> that there was a double birthday in less than 150 cases?

I've no idea how to work these things out myself. But I've done a rough
simulation, and in 10,000 years, there were only 4 years where there was a
double birthday on less than 150 nights, and none on less that 140. So
roughly 0.04%.

Now I suppose you're going to say there was something funny about the
restaurant, that it's used exclusively by people with the same birthday, or
all different, or there's something that's going to be contrary to the
theoretical probabilities.

--
Bartc

```
 0
Reply bc (2337) 12/14/2011 10:35:45 PM

```BartC wrote:
) I've no idea how to work these things out myself. But I've done a rough
) simulation, and in 10,000 years, there were only 4 years where there was a
) double birthday on less than 150 nights, and none on less that 140. So
) roughly 0.04%.
)
) Now I suppose you're going to say there was something funny about the
) restaurant, that it's used exclusively by people with the same birthday, or
) all different, or there's something that's going to be contrary to the
) theoretical probabilities.

I'll just say that you proved that even with a random distribution, you
can't be *certain* that the outcome of a given number of experiments will
be the number you expected by probability calculus.

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
```
 0
Reply willem6 (255) 12/15/2011 9:07:19 AM

```It's often overlooked that probability fundamentally
involves *ignorance* (ignoring some interpretations
of quantum mechanics).  For example, suppose I bet on the
Pass Line at a Craps Table and roll two dice.
Alice can see only a Five and will judge that my Passing is
very likely (6 or 2 on the other die win at once).
Bob can see only an Ace and judges Pass unlikely
(1 or 2 on the other die lose at once).  Carl and Donna
see neither and both dice respectively, and state other
probabilities.  All are correct -- all reflect their respective
ignorances.

Thus a question about the "certainty" of a probability
is poorly phrased.

On Dec 14, 8:20=A0am, Seebs <usenet-nos...@seebs.net> wrote:
> ... if you took a list of "the populations of all cities in an
> almanac", you would find that the first digits 1-4 were more heavily
> represented than 5-9 by a fair margin.

This is called Benford's law.

> On the other hand, if you looked
> in the tens place of the same numbers, you'd probably find a very even
> distribution of 0-9.

Nitpick:  '0' will be, by far, the most common ten's digit.
The published population of Casablanca is 3,027,000;
of Rangoon 4,350,000; Singapore 5,183,700, and so on.

James Dow Allen

```
 0
Reply jdallen2000 (495) 12/15/2011 10:33:52 AM

```On Dec 15, 10:33=A0am, James Dow Allen <jdallen2...@yahoo.com> wrote:
> It's often overlooked that probability fundamentally
> involves *ignorance*

Or, alternatively, information. And even then, all you're describing
is the notion of conditional probability - that one's estimate of
probability is based on one's a priori information.

The probability of a given event is fundamentally the same for
everyone, but one's estimate of that probability is based on one's own
information.

P(win at craps betting on the pass line) does NOT vary (given fair
dice, etc).
P(win at craps betting on the pass line | information known by A,B,C,D
etc) *does* vary when the level of information varies.
```
 0
Reply gwowen (551) 12/15/2011 1:38:11 PM

```> I'd disagree for at least two obvious cases:
> * Restaurants that specifically seek out birthday party traffic.

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.

```
 0

```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.
>
> 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."
```
 0
Reply superpollo (98) 12/15/2011 6:14:43 PM

```On 12/15/2011 6:38 AM, gwowen wrote:
> On Dec 15, 10:33 am, James Dow Allen<jdallen2...@yahoo.com>  wrote:
>> It's often overlooked that probability fundamentally
>> involves *ignorance*
....
> The probability of a given event is fundamentally the same for
> everyone, but one's estimate of that probability is based on one's own
> information.

Then the probability of having matching birthdays _in a given group_ is either 1
or 0, depending on the actual birthdays.  To be certain it is greater than 50%,
one must know the birthdays or know that there are enough people to guarantee it.

I agree with James that the question is poorly worded.  If the question were
instead "How many randomly selected people must be gathered together in a room
before there is a greater than 50/50 chance that at least two of
them have the same birthday?", the answer would be different.

```
 0

```In article <79664e6d-da9a-48ff-9666-f3f4740e0120
>
> On Dec 12, 11:57�am, divya bisht <bisht88.di...@gmail.com> wrote:
> > Toughest Probability Question
> > (Source :http://hardest-puzzle.blogspot.com/2011/12/toughest-probability-quest...
> > )
> >
> > 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 belowhttp://hardest-puzzle.blogspot.com/2011/12/toughest-probability-quest...
>
> Two. Invite a pair of twins - they're almost certain to have the same
> birthday.

The subject line is wrong.  This is a well known problem with a simple
solution:
http://en.wikipedia.org/wiki/Birthday_problem

The most difficult probability problem is:
What is the probability that someone who has taken an introductory level
probability course and yet has never heard of the birthday paradox?

Only slightly less difficult is:
What are the odds that a student of below average intelligence can get
news:comp.lang.c to perform his homework problem for him?
```
 0
Reply dcorbit (2698) 12/20/2011 12:55:33 AM

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