10 mind-boggling paradoxes _ mental floss
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10 Mind-Boggling Paradoxes
A paradox is a statement or problem that either appears to produce two entirely contradictory(yet possible) outcomes, or provides proof for something that goes against what we intuitively
expect. Paradoxes have been a central part of philosophical thinking for centuries, and are
always ready to challenge our interpretation of otherwise simple situations, turning what we
might think to be true on its head and presenting us with provably plausible situations that are
in fact just as provably impossible. Confused? You should be.
1. ACHILLES AND THE TORTOISE
The Paradox of Achilles and the Tortoise is one of a number of theoretical discussions of
movement put forward by the Greek philosopher Zeno of Elea in the 5th century BC. It begins
with the great hero Achilles challenging a tortoise to a footrace. To keep things fair, he agrees
to give the tortoise a head start of, say, 500m. When the race begins, Achilles unsurprisingly
starts running at a speed much faster than the tortoise, so that by the time he has reached the
500m mark, the tortoise has only walked 50m further than him. But by the time Achilles has
reached the 550m mark, the tortoise has walked another 5m. And by the time he has reachedthe 555m mark, the tortoise has walked another 0.5m, then 0.25m, then 0.125m, and so on.
This process continues again and again over an infinite series of smaller and smaller
distances, with the tortoise alwaysmoving forwards while Achilles alwaysplays catch up.
Logically, this seems to prove that Achilles can never overtake the tortoisewhenever he
reaches somewhere the tortoise has been, he will always have some distance still left to go no
matter how small it might be. Except, of course, we know intuitively that he canovertake the
tortoise. The trick here is not to think of Zenos Achilles Paradox in terms of distances and
races, but rather as an example of how any finite value can always be divided an infinite
number of times, no matter how small its divisions might become.
2. THE BOOTSTRAP PARADOX
The Bootstrap Paradox is a paradox of time travel that questions how something that is taken
from the future and placed in the past could ever come into being in the first place. Its a
common trope used by science fiction writers and has inspired plotlines in everything from
Doctor Whoto the Bill and Tedmovies, but one of the most memorable and straightforward
examplesby Professor David Toomey of the University of Massachusetts and used in his
book The New Time Travellersinvolves an author and his manuscript.
Imagine that a time traveller buys a copy of Hamletfrom a bookstore, travels back in time to
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Elizabethan London, and hands the book to Shakespeare, who then copies it out and claims it
as his own work. Over the centuries that follow, Hamletis reprinted and reproduced countless
times until finally a copy of it ends up back in the same original bookstore, where the time
traveller finds it, buys it, and takes it back to Shakespeare. Who, then, wrote Hamlet?
3. THE BOY OR GIRL PARADOX
Imagine that a family has two children, one of whom we know to be a boy. What then is the
probability that the other child is a boy? The obvious answer is to say that the probability is 1/2
after all, the other child can only be eithera boy ora girl, and the chances of a baby being
born a boy or a girl are (essentially) equal. In a two-child family, however, there are actually
four possible combinations of children: two boys (MM), two girls (FF), an older boy and a
younger girl (MF), and an older girl and a younger boy (FM). We already know that one of thechildren is a boy, meaning we can eliminate the combination FF, but that leaves us with three
equally possible combinations of children in which at leastone is a boynamely MM, MF, and
FM. This means that the probability that the other child isa boyMMmust be 1/3, not 1/2.
4. THE CARD PARADOX
Imagine youre holding a postcard in your hand, on one side of which is written, The
statement on the other side of this card is true. Well call that Statement A. Turn the card
over, and the opposite side reads, The statement on the other side of this card is false
(Statement B). Trying to assign any truth to either Statement A or B, however, leads to a
paradox: if A is true then B must be as well, but for B to be true, A has to be false. Oppositely,
if A is false then B must be false too, which must ultimately make A true.
Invented by the British logician Philip Jourdain in the early 1900s, the Card Paradox is a simple
variation of what is known as a liar paradox, in which assigning truth values to statementsthat purport to be either true or false produces a contradiction. An even morecomplicated
variation of a liar paradox is the next entry on our list.
5. THE CROCODILE PARADOX
A crocodile snatches a young boy from a riverbank. His mother pleads with the crocodile to
return him, to which the crocodile replies that he will only return the boy safely if the mother
can guess correctly whether or not he will indeed return the boy. There is no problem if the
mother guesses that the crocodile willreturn himif she is right, he is returned; if she is
wrong, the crocodile keeps him. If she answers that the crocodile will notreturn him, however,
we end up with a paradox: if she is right and the crocodile never intended to return her child,
then the crocodile has to return him, but in doing so breaks his word and contradicts the
mothers answer. On the other hand, if she is wrong and the crocodile actually did intend to
return the boy, the crocodile must then keep him even though he intended not to, thereby also
breaking his word.
The Crocodile Paradox is such an ancient and enduring logic problem that in the Middle Ages
the word "crocodilite" came to be used to refer to any similarly brain-twisting dilemma where
you admit something that is later used against you, while "crocodility" is an equally ancient
word for captious or fallacious reasoning
6. THE DICHOTOMY PARADOX
Imagine that youre about to set off walking down a street. To reach the other end, youd first
have to walk half way there. And to walk half way there, youd first have to walk a quarter of the
way there. And to walk a quarter of the way there, youd first have to walk an eighth of the way
there. And before that a sixteenth of the way there, and then a thirty-second of the way there,
a sixty-fourth of the way there, and so on.
Ultimately, in order to perform even the simplest of tasks like walking down a street, youd
have to perform an infinite number of smaller taskssomething that, by definition, is utterly
impossible. Not only that, but no matter how small the first part of the journey is said to be, it
can always be halved to create another task; the only way in which it cannotbe halved would
be to consider the first part of the journey to be of absolutely no distance whatsoever, and in
order to complete the task of moving no distance whatsoever, you cant even start your
journey in the first place.
7. THE FLETCHERS PARADOX
Imagine a fletcher (i.e. an arrow-maker) has fired one of his arrows into the air. For the arrow
to be considered to be moving, it has to be continually repositioning itself from the place
where it is now to any place where it currently isnt. The Fletchers Paradox, however, states
that throughout its trajectory the arrow is actually not moving at all. At any given instant of no
real duration (in other words, a snapshot in time) during its flight, the arrow cannot move to
somewhere it isnt because there isnt time for it to do so. And it cant move to where it is now,
because its already there. So, for that instant in time, the arrow must be stationary. Butbecause all time is comprised entirely of instantsin every one of which the arrow must also
be stationarythen the arrow must in fact be stationary the entire time. Except, of course, it
isnt.
8. GALILEOS PARADOX OF THE INFINITE
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September 29, 2014 - 9:00am
In his final written work, Discourses and Mathematical Demonstrations Relating to Two New
Sciences(1638), the legendary Italian polymath Galileo Galilei proposed a mathematical
paradox based on the relationships between different sets of numbers. On the one hand, he
proposed, there are square numberslike 1, 4, 9, 16, 25, 36, and so on. On the other, there
are those numbers that are notsquareslike 2, 3, 5, 6, 7, 8, 10, and so on. Put these two
groups together, and surely there have to be more numbers in general than there arejust
square numbersor, to put it another way, the total number of square numbers must be less
than the total number of square andnon-square numbers together. However, because every
positive number has to have a corresponding square and every square number has to have a
positive number as its square root, there cannot possibly be more of one than the other.
Confused? Youre not the only one. In his discussion of his paradox, Galileo was left with no
alternative than to conclude that numerical concepts like more, less, or fewercan only be
applied to finite sets of numbers, and as there are an infinite number of square and non-
square numbers, these concepts simply cannot be used in this context.
9. THE POTATO PARADOX
Imagine that a farmer has a sack containing 100 lbs of potatoes. The potatoes, he discovers,
are comprised of 99% water and 1% solids, so he leaves them in the heat of the sun for a day
to let the amount of water in them reduce to 98%. But when he returns to them the day after,
he finds his 100 lb sack now weighs just 50 lbs. How can this be true? Well, if 99% of 100 lbs of
potatoes is water then the water must weigh 99 lbs. The 1% of solids must ultimately weigh
just 1 lb, giving a ratio of solids to liquids of 1:99. But if the potatoes are allowed to dehydrate
to 98% water, the solids must now account for 2% of the weighta ratio of 2:98, or 1:49
even though the solids must still only weigh 1lb. The water, ultimately, must now weigh 49lb,
giving a total weight of 50lbs despite just a 1% reduction in water content. Or must it?
Although not a true paradox in the strictest sense, the counterintuitive Potato Paradox is a
famous example of what is known as a veridical paradox, in which a basic theory is taken to a
logical but apparently absurd conclusion.
10. THE RAVEN PARADOX
Also known as Hempels Paradox, for the German logician who proposed it in the mid-1940s,
the Raven Paradox begins with the apparently straightforward and entirely true statement that
all ravens are black. This is matched by a logically contrapositive (i.e. negative and
contradictory) statement that everything that is notblack is nota ravenwhich, despite
seeming like a fairly unnecessary point to make, is also true given that we know all ravens are
black. Hempel argues that whenever we see a black raven, this provides evidence to support
the first statement. But by extension, whenever we see anything that is notblack, like an
apple, this too must be taken as evidence supporting the second statementafter all, an
apple is not black, and nor is it a raven.
The paradox here is that Hempel has apparently proved that seeing an apple provides us with
evidence, no matter how unrelated it may seem, that ravens are black. Its the equivalent of
saying that you live in New York is evidence that you dont live in L.A., or that saying you are 30
years old is evidence that you are not 29. Just how much information can one statement
actually imply anyway?
Paul Anthony Jones Follow @HaggardHawks
Paul Anthony Jones is a writer and musician from Newcastle upon Tyne. He is the
author of word origins guide Haggard Hawks and Paltry Poltroons, and he runs
@HaggardHawks.
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224 Comments
Brian
The only way that the 'Potato Paradox' is apparently absurd is if you are bad at math. It
is true, the only way to get the water content down to 98% (with 1 pound of solids) is by
having a total weight of 50 lbs.
The only way the answer appears absurd is if you confuse removing 1% of the weight
with changingthe ratio by 1%, which is just bad math.
Luka
I have to admit it did strike me as odd that a 1% reduction in one quality can
equate to a 50% reduction in another, but now you point it out I can totally see
how easily the outcome could be expected. Great article.
DrFlimFlam
I don't understand the boy or girl paradox. While creating the 1/3 conditions, you are
including the older girl/younger boy option, even though that option for a family with a
boy already could not possibly include that option. The only available options for a boy
with the M already in place are MF and MM. Cansomeone explain why this is a
paradox? The next birth is still roughly50/50; it isonly a stranger's guess, knowing that
one of the family's children is a boy (without knowing priority of birth, that is a 1/3 shot.
EDIT: I see my mistake. I took the "being born" toa conclusion the original statement
did not intend. I still don't see it as much of a paradox, but whatever.
Brian
The paradox is not about prediction, it is about the existing state. In a family that
has two children, the possible combinations are MM, FF, MF, FM. If you were
asked to guess what combination they had (without any other info), you would
have a 1 in 4 chance of being right. If you are told one is a boy, that only leaves
three possible combinations. Of those three combinations, two of them have
girls. Therefore, the chance that the other child is a girl is 2/3.
Bleus
You're falling for the same trap the author did: differentiating MF and FM
by introducing an outside component (time) that is NOT an element of
the question. The one sibling is Male, he can have EITHER a sister (who
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may e o er or younger, u we on care e can ave a ro er
(who also may be older or younger, and again, it doesn't matter). With
two choices, and all other factors not being considered (genetic or
societal predispositions for example), there's a 50/50 chance of his
sibling being male.
Tim
No. 50/50 would only work if the question was phrased as
follows:
'A couple has two children, the ELDER of which you know to be aboy. What are the odds that the other one is a boy?'
But you DON'T know this. Within the set of info you know about
the family, it's 1/3. You seem to be arguing that having extra info
wouldn't change the odds at all.
Bleus
LoL -- how about you first explain for us all, the exact mechanism
by which the gender of one sibling born to a couple impacts the
gender-selection process of another one?
I say this because, unless you can point to a specific mechanism
wherein the gender of one sibling (born before OR after)
somehow CHANGES the statistical likelihood of any OTHER
sibling being one gender or another, than the gender of ONE
child has no bearing whatsoever on the gender of another, and
thus the statistical probability of ANY child being "a boy" is always
50%...
Tim
You're mistaking a probability problem for a biology problem.
Bleus
Lmao -- then they asked the wrong question, because my facts
are EXACTLY correct for answering the question asked with theinformation provided. If they want to ask a DIFFERENT question
and/or provide other information, then my answer may be
different. Nevertheless, the PROBABILITY of the "other child"
being a boy IS 50%... PERIOD.
Tim
No. It becomes clearer when you put it in practical terms.
(Apologies for the clumsy & slightly archaic scenario.)
You meet your friends, Bob & Carol, on the street after an
absence of some years. They have a five-year-old boy with them.
'This is Billy,' says Carol. 'Our other kid's at a friend's place.'
Unfortunately, you get a phone call & have to dash before gettingfurther details.
The next day, you get a call from Bob. 'The kids' birthdays are
this week, and we were wondering if you'd like to come to their
party tomorrow,' he says. 'Are they twins?' you ask. 'No, one's a
year older--the older one was born on 7 January, the younger on
9 January. We celebrate their birthdays together.'
At the toy shop later that day to buy presents, you realise that
you didn't ask whether Billy's sibling was a boy or a girl. You try to
call Bob and Carol, but can't get through.
Tim
(Minor edit--at A friend's place, not at HIS friend's place.)
Bleus
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That's a wonderful story, but it's ENTIRELY USELESS for
answering the question posed with the information given.
First of all, you've added TONS of information that wasn't
provided in the scenario as posited (ages, and birth order being
the key ones). All you know in the situation provided is that there
are two children and that one of them is a boy (which itself is
irrelevant, since that fact DOES NOT impact the gender-selection
process of the other sibling IN ANY WAY!) Also, the "twins"
comment is another red-herring, even with twins one is always
the "oldest", even if only by a few minutes... Every child has a
~50% chance of being a boy. Unless/until you can demonstrate a
mechanism whereby the presence/absence/gender of the other
child CHANGES those odds, they remain constant FOREVER...
Tim
No, I haven't. The ages of the kids are irrelevant: the only reason
I added those was to make it resemble a real-life scenario. Also, I
purposely left birth order information out--read it again. None of
the scene-setting description in the story makes it functionally
different from the original problem.
Your 'twins' point is trivially true--you could delete all reference to
this, and the basic point would still be identical.
The scenario's quite similar to the Monty Hall Problem, which you
may be familiar with (quoting BBC website for this one). (EDIT: I
just replaced the BBC version with the clearer US one below.)
*
Imagine a TV game show not unlike Deal or No Deal in which
you choose one of three closed doors and win whatever is behind
it.
One door conceals a Cadillac - behind the other two doors are
Randall
Lol. I've seen people almost come to blows over the Monty Hall
problem. Sometimes the only way to convince someone is to do
a demonstration with cards or something.
Tim
I'm pretty sure there was a lot of screaming involved when I
heard it the first time, followed by grovelling apologies....
But it works quite similarly to the siblings problem above, right?
(i.e. it seems as if Event A is miraculously affecting the probability
of Event B, even though it isn't.)
Bleus
Mythbusters did a pretty good job of illustrating it on one of their
shows...
Tim
If you've seen Monty Hall explained on 'Mythbusters', how come
you can't grasp the functionally similar sibling problem?
Bleus
Because the question asked above is NOT a restatement of the
Monty Hall problem (although it WANTS to be, except they
worded it wrong).
To clarify for you, In the Monty Hall paradigm, there are THREE
initial choices, only one of which "pays", and there are two
opportunities to make a choice, AND the consequences/odds of
the second choice are directly impacted by having made the first
one...
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In the siblings issue, on the other hand, there are only two
possible outcomes, either of which is equally likely, and neither of
which has its end-state affected by that of the other...
Oh, and nobody gets to choose anything, you are only being
asked to state a probability...
Tim
OK, now we're getting somewhere. What would be the correctly
stated version, in your view?
Bleus
I edited my previous post to include why they are not the same
question, but you may have to refresh the page to get the update
:p
Tim
Wrong. The end-state (i.e. the configuration of objects behind the
doors) of the MH paradox isn't affected by the outcome of the
contestant's choices, either.
Bleus
Except that I didn't say it was: I said the ODDS of making a
correct choice are altered by getting to make the first choice...
Furthermore, there are no "choices" in the sibling problem,
nobody gets to "choose" anyone's gender, and nobody gets to
eliminate possibilities or alter the odds by making said choices...
There's a perfectly good (and short!) mathematical "proof" for the
sibling problem at the bottom of this thread (and I even used a
binary grid that time for poor old Brian's sake!)
Bob You are better off buying legos, because every kid loves those
and you aren't arbitrarily assigning gender roles to a child you've
never met. A
Brian
Nobody is claiming that the gender of one sibling affects the
other. What people are saying, but you seem to be unable to
grasp, is that given a set of two binary objects, if you know the
state of one (but not which one) you can make a guess as to the
state of the other.
A truth table is one of the most basic tools of logic. So let's use
one. Here are all the possible states of two binary objects:
00, 01, 10, 11
If I ask you what is the probability that at least one of two binary
objects is a 1, the answer is 3/4.
If I tell you that at least one of the objects is a 1, all you need to
do is count. There are three sets that have at least one 1. If I ask
what is the probability that the other one is also a 1, just count
again. Of the 3 sets possible, only 1 of them has two 1s.
Therefore, knowing that at least one of the objects is a 1, the
probability of the other one also being a 1 is 1/3. It has NOTHING
to do with one influencing the other or any such nonsense. If has
EVERYTHING to do with knowing all the possible states, then
eliminating the ones that don't apply,
Bleus
Math, using a Binary Grid:
1 = male, 0 = female.
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But in 2/3 of these scenarios, the "other" child is a girl.
If Jane is a boy, Joe is a boy in one and a girl in the other.
If Joe is a boy, Jane is a boy in one and a girl in the other.
Sure looks like if one is a boy, the other must be a boy or girl and
either is just as likely.
Bleus
You want math? Try this:
There are two children (C=[1,2]), one of which is a boy: Cn = (b).
What are the odds that the other one (which I'll define as Cn') is a boy?
i.e. What are the odds that Cn' = (b)
Given these conditions, the ONLY possible permutations ( where n =
[1,2] and n' = [2,1] ) are:
* Cn = (b) / Cn' = (b)
* Cn = (b) / Cn' = (g)
Thus, since there are only two possibilities for Cn' (i.e. b | g), the answer
to the question is 50%.
Furthermore, this result remains EXACTLY the same even if the grid is
constructed with Cn = (g) - - which means that the gender of Cn is
unrelated to the answer. This can be logically deduced/confirmed since
neither child's gender plays any role whatsoever in how the other child's
gender-selection process plays out during gestation and the base-
chance of *any* child being born a male is ~50%.
Does THAT feel any better?
Tom Burns
No, you are not considering that he may have an older brother, or a
younger brother, you are only considering "brother" as one option when
it is really 2 options. If we change the 1 that we know as a boy to X, it is
MX, XM, XF, FX = 50/50
Patrick
This is not a paradox. It doesn't assume that the boy we know of is the older
sibling. However, you're really left with 4 equally likely scenarios, not two. The
other unknown sibling can be an older brother, a younger brother, an older
sister, or a younger sister.
So the MM possibility is twice as likely as each of the other two (MF and FM).
You disappoint me Mental Floss.
Brian
Age has nothing to do with it. Consider there are two kids in the family,
named 'J' and 'M'. The possibilities are:
J=B,M=B/J=G,M=G/J=B,M=G/J=G,M=B.
We know ONE (but not WHICH one) is a boy. That leaves
J=B,M=B/J=G,M=B/J=B,M=G.
Make a random selection from that list. What are the odds that your
selection contains only boys? 1/3. Therefore, given the fact that you
know at least one of the children is a boy, the chances are 1/3 that the
other one is also a boy.
Your mistake is that you are saying that there is something known, when
there is not.
Eric
You are incorrect. There is something known. It is stated (known)
that one of the children is male.
The question is this:
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Picking the right combination of kids is not the same as the
likelihood of one gender of child or the other existing.
Mitchell
Beautifully done, thank you.
Tim
Yet wrong.
Tim
Actually, after wasting a large portion of the day thinking about it,
now I am not so sure. I think this may hinge on a semantic
interpretation of the problem. Well explained, anyway!
Eric
Actually, without knowing order of birth, it's 75%. 3 of the 4 possible outcomes
include at least one male. The odds of it being one male and one female is
50%, since 2 of the 4 starting possibilities give that combination. The "paradox"
given above is simply bad math. With the birth of the f irst child, they eliminated
1 of the 4 outcomes, when they should have eliminated 2 of them.
Brian
Order does not matter. The question is not 'if they have a boy, what are
the chances that the next one will be a boy?'. The only time a possibility
is eliminated is if the first child is a girl, then we know the other one won't
be a girl. If the first child is a boy, no possibilities are eliminated.
Eric
The point stands. Birth order is irrelevant no matter what.
It's basic statistics. With both children as unknowns, you have a
75% chance of it being any combination including at least one
boy. As soon as one of the genders is known, then 2 of the
original 4 possibilities are eliminated, leaving it at 50/50.
This is something that many statisticians get wrong. Once an
element is known, it changes the calculation.
This "paradox" is just bad math. That's all.
Brian
Sorry, but the statisticians are correct. Your mistake is in
assuming things that are not stated. You say 'one of the genders
is known', which is true, but you don't know which child is the boy.
Since you don't know which child is the boy, that leaves 2
possibiities right there.
Show us all the combinations that you think are possible (using
the information given), and I will show you your error.
Bleus
I don't even know what you're trying to say there... we know one
child is a boy, and we are asked what the odds are that the
*OTHER* child is also a boy. Leaving out semantic arguments of
genetic or societal predisposition, and hemaphrodity, the child
can be either male or female, giving a 50/50 chance. To put it
another way, please explain why you are making "girl-boy" and
"boy-girl" into separate results and not just two descriptions of the
same condition, worded differently?
Bort
This is a probability problem. The question is not: what is the
probability of any given child being a boy? That would be 50%.
The question is: given, that one child is known to be a boy, what
is the probability that both children are boys? Then it goes down
to 33%. The coin flip analogy from Brian is perfect and it can be
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Load more comments
a en o ano er eve or c ar y. a s e pro a y a w en
1000 coins are flipped that they are all heads? It is not 50%.
However, the chance of any given coin being heads is 50%. The
paradox seems counter intuitive because we are thinking
sequentially.
MontanaMade
I see your point, but in this case, that is exactly what the question
was- the question was NOT in what order are the children as to
age AND what is the sex of the unknown child? They asked quite
simply: What are the odds of the other child being a male?
period. nothing else. So now we are left with 2 choices: M or F.
period. Everything else is assumptions of information not known
or asked and creates a scenario never posed. The question of
whether the boy was older or younger was never introduced- just
assumed- which causes a false argument- an argument that
seems to keep going like the Energizer bunny!
Which brings up another question: If the Energizer bunny had
kids, what are the odds of them having batteries included??
Bleus
You're reading the question wrong. That we are told that one of
the children is male is a red-herring since the question
*specifically* asks, what are the odds that the OTHER child
[about whom we have been told NOTHING, other than that they
exist] is a male?" Since we have been told nothing else about
them, and the gender of the child we have been told something
about has no bearing on the answer to this question, there are
only two options, making the odds 50%. The question posed in
the article may as well have been, "What are the odds that one
particular child, about whom you know nothing, is male?".
Brian
The question is correct and there are no tricks or red herrings. It
is straightforward.
If there is a family of two children, and you don't have any more
information than that, what is the gender distribution?
25% 2 boys
25% 2 girls
50% one boy one girl
OK so far?
Now I add some information - one of the children is a boy. The
distribution is now
33% 2 boys
66% one boy one girl
(the two girls option is eliminated)
The question, when read correctly, is not what is the gender of
any individual child, but what are the chances of there being 2
boys. By giving you the 'one is a boy' hint I increased your
chances of being right from 25% to 33%.
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