There is a room. It is initially empty. You have full control over
who enters and leaves the room, with one exception: if you know a
person's birthday or anything about them that would give you
knowledge relating to their birthday, then you aren't allowed to
let them in the room.
Your task is to create a situation such that, with at least a 50%
probability, there will be at least two people in the room with
the same birthday.
What is the smallest number of people you can send into the room to
achieve this?
Hint: it is not what you are probably thinking.
-- Mark Brader | "I thought it was a big joke.
Toronto | Dr. Brader is known for joking around a lot."
m...@vex.net | --Matthew McKnight
The
original
post
is
being
preserved
as
spoiler
space
for
those
of
you
with
big
monitors.
On Tue, 24 Apr 2012 16:52:09 -0500, Mark Brader wrote:
> There is a room. It is initially empty. You have full control over
> who enters and leaves the room, with one exception: if you know a
> person's birthday or anything about them that would give you
> knowledge relating to their birthday, then you aren't allowed to
> let them in the room.
> Your task is to create a situation such that, with at least a 50%
> probability, there will be at least two people in the room with
> the same birthday.
> What is the smallest number of people you can send into the room to
> achieve this?
> Hint: it is not what you are probably thinking.
Let in a pair of conjoined twins? Although I suppose you could argue
that it means you have "knowledge relating to" the second twin's
birthday.
> The
> original
> post
> is
> being
> preserved
> as
> spoiler
> space
> for
> those
> of
> you
> with
> big
> monitors.
> On Tue, 24 Apr 2012 16:52:09 -0500, Mark Brader wrote:
>> There is a room. It is initially empty. You have full control over
>> who enters and leaves the room, with one exception: if you know a
>> person's birthday or anything about them that would give you
>> knowledge relating to their birthday, then you aren't allowed to
>> let them in the room.
>> Your task is to create a situation such that, with at least a 50%
>> probability, there will be at least two people in the room with
>> the same birthday.
>> What is the smallest number of people you can send into the room to
>> achieve this?
>> Hint: it is not what you are probably thinking.
> Let in a pair of conjoined twins? Although I suppose you could argue
> that it means you have "knowledge relating to" the second twin's
> birthday.
Smullyan relates an anecdote in which a student victimized him on
just this point. He'd explained the Birthday Paradox to his class, and
how you needed twenty-mumble people to get a fifty-mumble percent chance
of a birthday collision. One student objected, claiming that collisions
could be far more likely than the BP predicted, and that even in the
small class at hand there was very likely to be a collision. He denied
knowing anybody's birthday other than his own, so Smullyan began to
poll the students -- until he got to the twins, whose birthday was
unknown to all present except each other.
Let's see: You might place your faith in people's basic honesty
(in the absence of apparent incentive). As each person approaches,
say "Step inside if you were born on February 11;" you can issue this
instruction without prior knowledge of anybody's birthday, and you
gain no certain knowledge while the person heads for the door (he
might be lying, or mistaken, or simply baffled as George Washington
might have been). Yet once there are two people inside, you could be
reasonably confident of a birthday collision.
Ted Schuerzinger writes [spoiler converted to rot13]:
> Yrg va n cnve bs pbawbvarq gjvaf?
Good idea, but wrong.
-- Mark Brader | "It is impractical for the standard to attempt to
Toronto | constrain the behavior of code that does not obey
m...@vex.net | the constraints of the standard." -- Doug Gwyn
> Yrg'f frr: Lbh zvtug cynpr lbhe snvgu va crbcyr'f onfvp ubarfgl
> (va gur nofrapr bs nccnerag vapragvir). Nf rnpu crefba nccebnpurf,
> fnl "Fgrc vafvqr vs lbh jrer obea ba Sroehnel 11;"...
Nope. This still violates the "knowledge relating to their birthday"
rule.
-- Mark Brader | "Courtly love-poetry may first have been written
Toronto | during long periods of abstinence on the Crusades,
m...@vex.net | but it would not have flourished in the cold of
| northern Europe without some help from the chimney."
| -- James Burke
> Eric Sosman [spoiler converted to rot13]:
>> Yrg'f frr: Lbh zvtug cynpr lbhe snvgu va crbcyr'f onfvp ubarfgl
>> (va gur nofrapr bs nccnerag vapragvir). Nf rnpu crefba nccebnpurf,
>> fnl "Fgrc vafvqr vs lbh jrer obea ba Sroehnel 11;"...
> Nope. This still violates the "knowledge relating to their birthday"
> rule.
Perhaps you should elucidate the rule. I may say "Mark, glad
to see you, please enter if you were born on February 11," and I
can do so with no knowledge whatever[*] about your birthday. It is
entirely up to you whether you enter or not; if I have influenced
your choice of action it is only by my personal charm and not by
my use of knowledge.
[*] Rigid interpretation: You cannot admit anyone at all into
the room, because you cannot escape having "knowledge relating to
their birthday," however slight. For example, mid-August has more
births than late April (<http://www.panix.com/~murphy/bday.html>),
so I'm already in possession of a smidgen of a bit of information
about your likely birth date. More blatantly, I can assert that
your birthday is more likely to be September 15 than February 29.
Perhaps epidemiological data might tell me that infant mortality
is higher in some seasons than in others, in which case observing
that you have survived infancy provides a hint. None of these gives
me a whole lot of prior knowledge, but the rule as stated requires
the absolutist "anything."
> > This still violates the "knowledge relating to their birthday" rule.
Eric Sosman:
> Perhaps you should elucidate the rule. I may say "Mark, glad
> to see you, please enter if you were born on February 11," and I
> can do so with no knowledge whatever[*] about your birthday. It is
> entirely up to you whether you enter or not...
But if I start to enter, you now have knowledge and have to stop me.
> [*] Rigid interpretation: You cannot admit anyone at all into
> the room, because you cannot escape having "knowledge relating to
> their birthday," however slight. For example, mid-August has more
> births than late April...
Good point. I'll allow you to ignore general knowledge of that type.
-- Mark Brader, Toronto | "...everything else in [the] list is wrong;
m...@vex.net | why should [this] be correct?" -- Rob Novak
You could always let only infants....those who are obviously less than 1 year old....into the room. That would seriously increase the odds of someone being born in the latter half of the preceding year.
On Wednesday, April 25, 2012 5:52:09 AM UTC+8, Mark Brader wrote:
> There is a room. It is initially empty. You have full control over
> who enters and leaves the room, with one exception: if you know a
> person's birthday or anything about them that would give you
> knowledge relating to their birthday, then you aren't allowed to
> let them in the room.
> Your task is to create a situation such that, with at least a 50%
> probability, there will be at least two people in the room with
> the same birthday.
> What is the smallest number of people you can send into the room to
> achieve this?
> Hint: it is not what you are probably thinking.
> -- > Mark Brader | "I thought it was a big joke.
> Toronto | Dr. Brader is known for joking around a lot."
> m...@vex.net | --Matthew McKnight
> My text in this article is in the public domain.
Find a hospital with maternity facilities. Admit two people about to deliver to a shared delivery/recovery room. You did not admit the expected children.
(This is not an adequate solution because, I'd really expect you to also send in medical people.)
> There is a room. It is initially empty. You have full control over
> who enters and leaves the room, with one exception: if you know a
> person's birthday or anything about them that would give you
> knowledge relating to their birthday, then you aren't allowed to
> let them in the room.
> Your task is to create a situation such that, with at least a 50%
> probability, there will be at least two people in the room with
> the same birthday.
> What is the smallest number of people you can send into the room to
> achieve this?
On Wednesday, April 25, 2012 6:26:45 PM UTC+8, Alexander Thesoso wrote:
> Find a hospital with maternity facilities. Admit two people about to > deliver to a shared delivery/recovery room. You did not admit the > expected children.
> (This is not an adequate solution because, I'd really expect you to also > send in medical people.)
> On 4/24/2012 5:52 PM, Mark Brader wrote:
> > There is a room. It is initially empty. You have full control over
> > who enters and leaves the room, with one exception: if you know a
> > person's birthday or anything about them that would give you
> > knowledge relating to their birthday, then you aren't allowed to
> > let them in the room.
> > Your task is to create a situation such that, with at least a 50%
> > probability, there will be at least two people in the room with
> > the same birthday.
> > What is the smallest number of people you can send into the room to
> > achieve this?
> > Hint: it is not what you are probably thinking.
Or only admit people who declare that they are carrying twins, but be explicit that they cannot tell you anything about the due date.
> Or only admit people who declare that they are carrying twins,
Bingo! That's the intended answer -- only one person needs to be
admitted to produce the desired conditions.
> but be explicit that they cannot tell you anything about the due date.
Not a requirement -- I didn't say anything about knowledge of birthdays
of future people. For the convenience of the one person admitted, it
would be ideal if she was already in labor.
-- Mark Brader "In general, it is safe and legal to
Toronto kill your children and their children."
m...@vex.net -- POSIX manual, quoted by Thomas Koenig
> On Wednesday, April 25, 2012 6:26:45 PM UTC+8, Alexander Thesoso wrote:
>> On 4/24/2012 5:52 PM, Mark Brader wrote:
>>> There is a room. It is initially empty. You have full control over
>>> who enters and leaves the room, with one exception: if you know a
>>> person's birthday or anything about them that would give you
>>> knowledge relating to their birthday, then you aren't allowed to
>>> let them in the room.
>>> Your task is to create a situation such that, with at least a 50%
>>> probability, there will be at least two people in the room with
>>> the same birthday.
>>> What is the smallest number of people you can send into the room to
>>> achieve this?
>>> Hint: it is not what you are probably thinking.
>> Find a hospital with maternity facilities. Admit two people about to
>> deliver to a shared delivery/recovery room. You did not admit the
>> expected children.
>> (This is not an adequate solution because, I'd really expect you to also
>> send in medical people.)
> Or only admit people who declare that they are carrying twins, but be
> explicit that they cannot tell you anything about the due date.
(quotes reordered)
If unborn babies don't count as "people" yet for the purpose of this
puzzle, then this works even if they tell you the due date, because
(again) you admitted them while they weren't "people" yet. If they do
count as "people" already, then this doesn't work (you know the due
date is some time in the next nine months or so) and neither does
Alexander's version.
m...@vex.net (Mark Brader) writes:
> Myron Buck:
>> Or only admit people who declare that they are carrying twins,
> Bingo! That's the intended answer -- only one person needs to be
> admitted to produce the desired conditions.
>> but be explicit that they cannot tell you anything about the due date.
> Not a requirement -- I didn't say anything about knowledge of birthdays
> of future people. For the convenience of the one person admitted, it
> would be ideal if she was already in labor.
You also have to arrange that one baby isn't born before 24.00 and the
second one after 24.00 :)
) Myron Buck:
)> Or only admit people who declare that they are carrying twins,
)
) Bingo! That's the intended answer -- only one person needs to be
) admitted to produce the desired conditions.
Disagree.
You stated that there had to be two "people" with the same birthday,
and you also stated that you could not let "people" into the room
wnen you know anything about their birthday.
If you let someone in that is carrying twins, then either those twins
do not count as "people", in which case they don't count towards the
probability of there being two with the same birthday, or they *do*
count as "people", in which case you let them into the room while you
had knowledge about their birthdays, which is not allowed.
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
>>> Or only admit people who declare that they are carrying twins,
Mark Brader:
>> Bingo! That's the intended answer -- only one person needs to be
>> admitted to produce the desired conditions.
"Willem":
> If you let someone in that is carrying twins, then either those twins
> do not count as "people", in which case they don't count towards the
> probability of there being two with the same birthday, or they *do*
> count as "people", in which case you let them into the room while you
> had knowledge about their birthdays, which is not allowed.
The trick is, even though you control who enters the room, there's
still another way for people to get in there, viz. birth. When the
pregnant woman enters, you set up a situation where there will *later*
be three people in the room, satisfying the conditions given.
-- Mark Brader | "It is refreshing to have Republican presidential
Toronto | candidates we can believe about *something*.
m...@vex.net | I believe what Bush says about Dole...
| And... what Dole says about Bush." --Craig B. Leman
) The trick is, even though you control who enters the room, there's
) still another way for people to get in there, viz. birth. When the
) pregnant woman enters, you set up a situation where there will *later*
) be three people in the room, satisfying the conditions given.
Ah, I see.
But "pro-life" advocates would disagree with you! :P
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
>> The trick is, even though you control who enters the room, there's
>> still another way for people to get in there, viz. birth. When the
>> pregnant woman enters, you set up a situation where there will *later*
>> be three people in the room, satisfying the conditions given.
"Willem":
> Ah, I see.
> But "pro-life" advocates would disagree with you! :P
Ah, but they're talking about the way they would *like* the law to
be, and I'm talking about the way it *is*. Anyway, I didn't want to
suppress posting the puzzle on those grounds.
-- Mark Brader "A clarification is not to make oneself clear.
Toronto It is to PUT oneself IN the clear."
m...@vex.net -- Lynn & Jay, "Yes, Prime Minister"
On Thursday, April 26, 2012 3:38:58 AM UTC+8, Willem wrote:
> Mark Brader wrote:
> ) Myron Buck:
> )> Or only admit people who declare that they are carrying twins,
> )
> ) Bingo! That's the intended answer -- only one person needs to be
> ) admitted to produce the desired conditions.
> Disagree.
> You stated that there had to be two "people" with the same birthday,
> and you also stated that you could not let "people" into the room
> wnen you know anything about their birthday.
> If you let someone in that is carrying twins, then either those twins
> do not count as "people", in which case they don't count towards the
> probability of there being two with the same birthday, or they *do*
> count as "people", in which case you let them into the room while you
> had knowledge about their birthdays, which is not allowed.
> 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
How can you have knowledge about their birthday, when they don't have one yet?
mbuck <mb...@hkis.edu.hk> writes:
> On Thursday, April 26, 2012 3:38:58 AM UTC+8, Willem wrote:
>> Mark Brader wrote:
>> ) Myron Buck:
>> )> Or only admit people who declare that they are carrying twins,
>> )
>> ) Bingo! That's the intended answer -- only one person needs to be
>> ) admitted to produce the desired conditions.
>> Disagree.
>> You stated that there had to be two "people" with the same birthday,
>> and you also stated that you could not let "people" into the room
>> wnen you know anything about their birthday.
>> If you let someone in that is carrying twins, then either those twins
>> do not count as "people", in which case they don't count towards the
>> probability of there being two with the same birthday, or they *do*
>> count as "people", in which case you let them into the room while you
>> had knowledge about their birthdays, which is not allowed.
> How can you have knowledge about their birthday, when they don't have one yet?
You've got a lot more than zero knowledge - you can place it within a
month or so. That meets the requirement.
-- Online waterways route planner | http://canalplan.eu Plan trips, see photos, check facilities | http://canalplan.org.uk
On Wednesday, April 25, 2012 5:52:09 AM UTC+8, Mark Brader wrote:
> There is a room. It is initially empty. You have full control over
> who enters and leaves the room, with one exception: if you know a
> person's birthday or anything about them that would give you
> knowledge relating to their birthday, then you aren't allowed to
> let them in the room.
> Your task is to create a situation such that, with at least a 50%
> probability, there will be at least two people in the room with
> the same birthday.
> What is the smallest number of people you can send into the room to
> achieve this?
> Hint: it is not what you are probably thinking.
> -- > Mark Brader | "I thought it was a big joke.
> Toronto | Dr. Brader is known for joking around a lot."
> m...@vex.net | --Matthew McKnight
> My text in this article is in the public domain.
You miss my point. They don't HAVE a birthday. Whatever you 'know' is purely theoretical.
:-)
