Letters from Readers to the Birdman

Which He Has Unfortunately Not Had the Time or Energy to Answer

**Note:**** ****Some of these letters have
been posted for awhile; others have been sitting in a cache which
I had forgotten about, and which are now posted below. Letters
are posted in date order as much as possible.**

**Letter from Mark GW**

**I was intrigued by the Paradox of
Irrelevant Information and think I can clear up a few problems
with it.
1. The "One/Other" Problem
You state that if you label one child as "one" and the
other as "the other"
that all of a sudden the statistical analysis yields a 50% chance
of the
woman's "other" child being a boy.
The "one/other" problem arises from this: that we label
"one" as a boy, but
don't alternate which one is "one!" To re-state: This
labeling leads us to
assume that "one" will always be a boy, when this
doesn't fit the woman's
situation. Let's label the older child "one" and the
younger child "the
other." If the older child is always a boy, then yes,
"the other" will be a
boy 50% of the time. This is analogous to the man's situation.
However,
what if the older child is a girl? Now, the child we have labeled
"one" is
in fact NOT a boy! In this case, "the other" is the boy
and "one" is a girl.
The English statement "one is a boy" holds in the sense
that the woman has a
boy, but the boy isn't one, he's the other!
I've never been formally schooled on probability, but I think I
can represent
this in an "event field."
One-B Other-B One-B Other-G One-G Other-B The event
"One-G Other-G" doesn't occur because the woman must
have at least one boy.
Since we have two boy/girl pairings and only one boy/boy pairing,
then the
odds still hold as one in three. The problem is that the English
statement
"one is a boy" leads us to label one of the children as
a constant boy. In
the woman's problem, it can be said that the
counter-intuitiveness sneaks in
through the back door by way of the fact that a girl could be
born first,
followed by a boy. This possibility is not allowed in the man's
situation.
The problem with the standard method of applying "one"
and "the other" is
much harder to spot, as the phrase "one is a boy" does
NOT rule out the fact
that "one" is a girl.
2. The overall counter-intuitiveness of the woman's situation
It seems to me that a better way of thinking of these sorts of
problems would
be to phrase them as "If a woman has X children and at least
one is a boy,
what are the odds that all of her children are boys?"
The total event field will always be 2^X, since each child has
two possible
sexes. Since all of the children must be boys, there will be only
one
possibility in the event field to provide the result we're
seeking (all
boys). Therefore, the numerator in the probability will always be
one.
Since there is only one possibility in the event field which
doesn't fulfill
the premises we've been given (that at least one be a
boy…therefore all girls
is the only one to be eliminated) then the denominator will
always be '2^X'
minus one. This leads to a chance of 1/3 for 2 children, 1/7 for
3 children,
etc.
It may seem like an invalid comparison, but if you think of it
for a minute,
it is entirely valid. Just as we ran into problems by using the
terms "one"
and "the other" to think of the two children, we run
into just as many
problems by thinking of it using the terms "one" and
"both." As we've seen,
the statistical analysis always leads us to the right conclusion,
but it
seems counter-intuitive because we immediately assume that the
only variable
is the sex of one individual child and therefore the answer must
be 50%. The
only way to set your brain on the right track is to use the terms
"at least
one" and "all." The image which comes immediately
to my mind is a little 2
by 2 Punnett square that introductory biology students use to
predict
possible genetic outcomes (dominant/recessive). Let's say that a
plant has 2
alleles for a trait, dominant and recessive, which each occur 50%
of the time
(this is obviously not a scientific example, but we need it to
match the
boy/girl, heads/tails scenario). If I were to say, "For a
given trait, a
plant has at least one dominant gene. What are the odds that both
of its
genes are dominant?" In your mind's eye, you could very
easily picture the
recessive/recessive corner of your square being blacked out,
leaving you with
3 possible squares, only one of which is dominant/dominant. The
answer is
intuitively 1/3. It seems to me that this phrasing alleviates
some of the
counter-intuitiveness. The fact that we're no longer thinking of
coin
flipping and childbirth no doubt also makes it easier to accept
that the
probability is not 50%. To put it back into the terms of the
woman, we could
eliminate some of the counter-intuitiveness by asking, "What
are the odds
that a woman's two children are both boys, assuming that she has
at least one
boy." Our base to work off of is immediately 1/4, as the
"both boys" has
put this into our heads. Soon after this, we realize the two
girls
possibility must be eliminated from the event field as it doesn't
fit the
premises. This immediately gets us back to 1/3. These are both
rather
obvious examples, but I'm just trying to show that the situation
isn't
necessarily counter-intuitive. It's all in the phrasing.
The phrases in question read as this: A Man has exactly two
children, the
older a boy. What is the probability of the Man's younger child
being a boy ?
A Woman has exactly 2 children, one of whom is a boy. What is the
probability
of the Woman's "other" child being a boy?
In fact, the man's problem is the one that contains irrelevant
information.
The man's younger child will be a boy 50% of the time, regardless
of whether
or not the older child was a boy. The inclusion of an older
brother doesn't
change this fact. The information contained in the woman's
problem is,
however, entirely relevant.
I hope these thoughts will prove helpful and I look forward to
receiving your
future weekly letters.
Sincerely,
Mark **

**Letter from Jeff P**

**I don't know anything about probability or
"Standard Analysis," but here
is what common sense tells me.
The age is certainly irrelevant.
I think the problem would be more clearly stated thusly:
A man has two children. The first we will consider is a boy. This
first
child happens to be older. The second child is either a boy or a
girl.
A woman has two children. The first we will consider is a boy.
The
second child is either a girl or a boy.
So, in each case, the possibilities are:
(BG,BB)
By introducing the category of age, the reader wants to consider
the
older child first. But there is no reason to do this. So he moves
on to
the next and, assuming that age is relevant, comes up with three
posibilities (BB, BG, GB), two of which are the same (BG and GB).
I don't care what SA or any other anylsis says, this is correct.
Surly
SA has a means of qualifying categories?
I would further state that the known boy is irrelevant
information as
well, making the problem look like this:
A man has a child. The child is either a girl or a boy.
A woman has a child. The child is either a girl or a boy.
What are the odds that each person's child is a boy?
There is an equal probability of the child being a girl or a boy.
Answer: 50%
duh.
-Jeff.**

**Letter #1 from Mark G:**

**Dear John,**** **** About the people who
say the probability of the woman's other child being .66 for a
girl, here's my questions:****
****
1. Will they actually pay off at 2-1 or even 3-2 if I
bet manchild for big $ each time on a population pool of 1000
pregnant women who each already have a son? **** ****
and**** **** 2. Are any
of them local to us? I'd like to organize a friendly game
of poker. This includes dos Savant if she's signed up to
the standard analysis.**** ****Mark**

**Letter #2 from Mark G:**

**John,**** **** After thinking while
driving let me restate the problem. The instance I
gave with the preggo with a boy is really a restatement of
the first case of a man with a known first son. What's
needed is a test where age order is unknown. Zo....**** **** 1. Take the USA Y2K
Census results. 2. Print out tickets for
every woman who reported two children and at least one boy.
Stack these face down. I therefore don't know the age
ordering.**** **** 3. I will
bet $1,000 on every flipover for the unknown 2d child being a boy
so long as dos Savant & Company give me 2-1 or even 3-2.**** ****
4. Alternately, if any of those blowhards can fnd me a
syndicate willing to book that game, I'll give them 50% of my
winnings.**** **** Mark**** **

**Letter from Mike S:**

**read your section on the Paradox of
Irrelevant Information and, with your indulgence, I would like to
throw in my two cents. Unfortunately, I don't believe that the SA
leads to a contradiction, but rather that the seeming paradox
resolves itself if you examine closely the nature of the
conditioning information laid out in the two cases. I will
attempt to elucidate exactly how the conditions are different in
each case, and exactly how they affect the probabilities.
Let's restate the standard situation here for convenience: A Man
has exactly two children, the older a boy. A Woman has exactly 2
children, one of whom is a boy. What is the probability of the
Man's younger child being a boy, and what is the probability of
the Woman's "other" child being a boy?
The standard situation lays out two scenarios, one for a Man and
one for a Woman. Each has two children. Naturally, we will assume
that the general probability that a child is a boy is 1/2, and
occurs independently. Now, the key to resolving the paradox is
analyzing the language of the paragraph in the light of two
questions. First, to what event exactly are we trying to assign a
probability? Secondly, exactly what conditioning information is
given in each case? When we have the answers to these questions,
we will be able to use the definition of conditional probability
to resolve the paradox.
In regards to the first question: in both cases, despite the
vague language of "other," it can be seen that the
event to which we are really trying to assign a probability is
not whether the "other" child is a boy, but rather
whether BOTH children are boys. In fact, in the second case,
asking whether the "other" child is a boy is nonsense
because "other" is not defined (more on this later).
What we mean to ask is whether they are both boys.
In regards to the second question: if we had no conditioning
information at all, then our (unconditioned) probability for the
event (x1=boy and x2=boy) would simply be 1/4. However, in the
two cases in question, we have (relevant) conditioning
information that requires us to adjust the probability we would
assign. Let's examine each case in turn.
In the Man's case, we have "peeked" at exactly ONE
child and established that HE is indeed a boy. We did not look at
the other child, and have drawn no information from him or her.
Incidentally, HOW we selected the child to peek at, whether by
oldness, slimness, "firstness", or whatever, is
completely irrelevant, as common sense would dictate. The
information gleaned from peeking at exactly one child increases
the assigned probability from the expected 1/4 to 1/2 -- in
effect, to calculate the desired probability we could simplify
our event space to the simpler space of whether the un-peeked-at
child is in fact a boy. This simplification is actually a
mathematical step in which the conditioning information cancels
the peeked-at child out of the probability equation and leaves
the un-peeked-at child in. Thus, in the Man's case, there IS in
fact an "other" child that we can talk about
meaningfully. This other child is independent of the conditioning
information.
Noting the _definition of conditional probability_:
Prob (A given B) = Prob (A and B) divided by Prob (B)
Which can be read:
"The conditional probability of A given B is the ratio of
unconditional probabilities: the numerator is the probability of
the intersection of the two events (A and B), and the denominator
is the probability of B,"
The man's case can be described as follows:
Prob (x1=boy and x2=boy, given x1=boy)
= Prob ((x1= boy and x2=boy) and x1=boy) divided by Prob (x1=boy)
= Prob (x1= boy and x2=boy) divided by Prob (x1=boy)
= Prob (x1= boy) times Prob (x2=boy), divided by Prob (x1=boy)
= Prob (x2=boy)
= .50
Note: Prob (x1= boy and x2=boy) reduces to Prob (x1= boy) times
Prob (x2=boy), because of the independence of each child's sex.
In the Woman's case, we have, or perhaps someone else has for us,
"peeked" at BOTH children and established that AT LEAST
ONE is a boy. This increases the assigned probability from the
expected 1/4 to only 1/3. We CANNOT simplify our event space to
the simpler space of whether the un-peeked-at "other"
child is in fact a boy because there IS NO un-peeked-at
"other" child. Our conditioning information is a
function of both children, and does not allow us meaningfully to
consider "one" or the "other" child.
Here is the appropriate calculation for the Woman's case:
Prob (x1=boy and x2=boy, given x1=boy or x2=boy)
= Prob ((x1= boy and x2=boy) and (x1=boy or x2=boy)) divided by
Prob (x1=boy or x2=boy)
= Prob (x1= boy and x2=boy) divided by Prob (x1=boy or x2=boy)
= .25 divided by .75
= .33
The crux of my defense of the SA is that the conditioning
information in the Man's case is not equivalent to that of the
Woman's. The Man's information is a datum about a single child.
Here it makes sense to talk about the "other child".
The Woman's information is an abstraction of both children,
obtained from applying the logical-OR function. In the process of
applying this function, details are lost, and the information is
distilled into a general statement about both children rather
than a statement about one child. This is why it is impossible to
assign a probability to the "other" child. We are
forced to consider both children and assign a probability to that
event.
For a moment, let's try to explain in common sense terms why the
conditioning information in the Woman's case does not result in
as much an increase in probability as in the Man's case. It is
because, in the Woman's case, by peeking at both children we
increased the likelihood that we would find at least one boy for
our conditioning information. That makes the woman's conditioning
information somewhat less significant; after all, three out of
four times we can expect to find at least one boy between two
children. In the Man's case, we had to look at only one child to
discover "at least" one boy; we can expect this to
happen only one out of two times. Thus in the Man's case we are
that much closer to the outcome we are testing for.
I believe there are two factors that obscure the important
distinction between the two cases. The first factor is the
phraseology used in the standard situation. The phrase "one
of whom is a boy" is very misleading. It suggests that there
is a particular child in mind and thereby evokes the same mental
picture that the phrase "the older a boy" does, despite
being information of a different sort. Such a mental picture
erroneously implies that we are able to consider the probability
of an actual "other" child. To be more precise, the
phrase in the Woman's case should be, "AT LEAST one of whom
is a boy."
The second factor that obscures the distinction is the relative
rareness with which we encounter the Woman's case in real life;
it's difficult to imagine a parent discussing his children in
such a manner. The conditioning information here is filtered
through Boolean logic: the phrase "one of whom is a
boy" (actually meaning "at least one of whom is a
boy") is the same as saying "either one or the other of
whom is a boy, and perhaps both of whom are boys." Humans
seldom think or relay information like this, at least not about
their children. In real life, if a parent mentions the gender of
one child, usually the statement will not imply anything about
the other child. For example, "I have two children. (My boy)
Bill likes to play hockey." conveys no information about the
other child. Such a statement provides conditioning information
equivalent to the Man's case. On the other hand, in common
English usage, the phrase "one of whom" will probably
more often mean "exactly one of whom," than it will
"at least one of whom." So when a person says something
like "I have two children, one of whom is a boy,"
common English usage strongly (but not definitively) implies that
exactly one child is a boy and the other child is a girl. A
follow-up like, "and the other of whom is also a boy"
sounds odd; we wonder why the parent did not say, "both of
whom are boys" in the first place. Thus, while it is
certainly possible to imagine a situation in which we are given
the conditional information of the Woman's case, in real life I
think if occurs relatively infrequently.
In summary, the two cases have different conditioning
information, and therefore we shouldn't expect the same
conditional probability. Careful examination of the conditioning
information resolves the paradox.
Respectfully,
Mike S**

**Letter from Stephen:**

**Believe it or not, I've
thought about this problem off and on for the last many years. I
have a degree in math, and fancied myself fairly competent in the
field. But, I have to admit, this particular problem, despite
being so simple in appearance, has given me fits.****
****I really do
not want to go into any deep analysis . This is because I find it
difficult to express my thoughts on it. That is because I find
writing difficult and also, to be quite frank, I'm not completely
comfortable with every possible detail concerning my
"solution", for lack of a better word.****
****Anyway, let
me just say this. Rather than go into detail, I'll offer a
particular point of view. **** ****A woman has 2 children. At least
one of them is a boy. What is the probability the other is a boy?
It depends on how you found out that info. For example, let's say
you were talking to a woman on the phone. She tells you she has 2
children. And they are both home right now. Then, you hear a
child's voice in the backround and it's a male. Then I would say
that the probablity that the other child is a male is 1/3.****
****If, however, before
you heard anything, you told the lady to put the younger child on
the phone. You hear it is a male's voice. Then the probability
that the other child is a boy would be 1/2. If, going back to the
first scenario, after hearing the childs (male)voice, you aks the
lady, "is that the younger child", and she says
"yes", I would say the probability that the other child
is male is still 1/3.**** ****It is similar, though not quite the
same as, to the following problem. Remember the Let's Make a Deal
show? There are 3 doors. One contains the Grand Prize. The
contestant picks Door #1. Monty then shows that the prize is not
behind Door #2. The he offers the contestant a chance to change
his pick to door #3. The question is should he change to #3 or
does it not matter. Well, it turns out, he SHOULD change to door
#3. (I will not provide the proof of this. I assume you are
familiar with the problem already, and it is not hard to show in
any event.) The points to ponder in this is that Monty KNOWS what
door the Grand Prize is behind. So, his decision to display Door
#2 was not random. Again, this is not the same problem, but may
have similar ideas about it to the 2 children problem.****
****With that in
mind, let's go back to the lady with the 2 children, not both
female. It certainly seems that the probability that both
children are male, if all you know is that the sex of both is NOT
female, is 1/3. However, many are claiming that if you knew that
the youngest was a male then the probability the the other is
male becomes 1/2. Well, I disagree. (Unless that knowledge came
about in some non-arbitrary way, as mentioned above.) ****
****Let's put it this
way. You are on the phone. You know she has 2 kids and they are
both home. You hear a male voice. You then ask if the voice came
from the youngest. She says yes. Then many are claiming that the
probability that the other is male becomes 1/2. Well, if she had
said the voice came from the oldest, then a similar line of
reasoning would also have one arrive at the conclusion that the
probability of the other child being male was 1/2. But, come on,
the voice you hear MUST be one or the other! (older or youner) In
other words, in this scenario, knowing the relative age of the
known male seems to give no added information. But again, if you
had first asked "what is the sex of the youngest?". If
she says "male", then It seems that the probability of
the other being male is indeed 1/2. ****
****I hope you see my
point here. Not to say you may be incapable, but rather I hope
I've expressed myself ok. And presented a relevent analysis.**

**Stephen**

**Letter from WCG:**

**A Man has
exactly three children, the oldest a boy. A Woman has exactly 3
children, one of whom is a boy. What is the probability of the
Man's younger
children being boys, and what is the probability of the
Woman's "other" children being boys?
Add
A man has exactly three children, the fattest a boy. What
is the probability
of the Man's skinnier children being boys?
Knowing the Man's oldest child is a boy of (BBB BBG BGG BGB, GGG
GGB GBB GBG)
we can eliminate (GGG GGB GBB GBG) knowing that one of the
woman's children
is a boy we can eliminate (GGG) knowing the man's fattest child
is a boy we
can only eliminate (GGG).
There is no paradox, older boy is a relevant piece of
information, by
establishing the first child is not a girl, which is the crux of
the problem.
Fatter, taller, smarter boy, does not establish anything
but that one of the
children is a boy and all the children are not girls, just as
saying one of
whom is a boy does.
There is no flaw in the standard analysis, only in your
reasoning.**

**Letter from Bryce L:**

**Dear John,****
****As you may
remember, we had a discussion via e-mail a while back about this
and that. For no good reason, I finally visited your
website, and read the Paradox of Irrelevant Information.
Alas, I hadn't been a member of Mensa at the time the issue came
about, but do have an answer that should clarify things once and
for all.**** ****Statistics and probability are two of the
things I hold nearest and dearest to my heart: I am fascinated by
them, and so, of course, seeing what the Paradox consisted of, I
had to analyze it.**** ****I'm afraid Ms. Vos Savant was
correct with this one (as she often is when developing problems
of probability,) in that the probability of the other child being
a boy if we know the older child is a boy, is 1/2, and if we know
that one of them is a boy, it is 1/3, and only 1/3.****
****You stated
that because a distinction is drawn based on an arbitrary
characteristic (say older/younger,) one could substitute any
characteristic (fatter/skinnier, taller/shorter, etc) with the
same result. This would include "one/the other,"
which you said isn't a category and would yield no
information. Yet, in fact, it does. Whether we
know that "the older" is a boy, or "this one"
is a boy, it infers the same specificity that excludes the other
from being that specific boy. ****
****To recapitulate,
let's list the possibilities: boy/boy, boy/girl, girl/boy,
girl/girl: or B/B B/G G/B G/G. Case 1: We know one is B,
but not which one. Thus G/G is not a possibility, leaving
B/B, B/G, G/B. The odds of the other being B is clearly 1
in 3.**** ****Case 2: We know one is B, and also which
one. This gives B/B and B/G. (Included in this case
is the scenario "If a person's oldest/fattest/stupidest
child. . ." or, as you pointed out, "If a person's
child" (this one, I said, pointing to them) "is a boy,
what are the odds of the other being a boy?" And
this is the distinction of one/the other. If one,
specifically, is the boy to which we refer, the other cannot be
that boy.) Thus, if we know that one is specifically a boy,
we can have B/B and B/G, ****but not G/B, ****since
we know it is that specific one (B/X, in this case) that is
the boy.**** ****This is analogous to a hypothetical coin
flipping game, in that, in case 1, I flip two coins and cover
them with my hand, and tell you, truthfully, that "at least
one coin is heads," (similarly, "at least one child is
a boy"). Well, they could both be heads, or one of
them could be with the other tails, and the other could be, with
the first one heads, but they could not both be tails. So
the probability of them both being heads is 1/3. In case 2,
I flip one coin so you can see that it's heads. The odds of
flipping the other as heads is 1 in 2.****
****The error you
made failing to recognize the distinction between: the inclusive
"one," in "one of them is a boy." Well,
it could be the first one, or the second one, the fat one, or the
short one: and the particular "one," as in "that
one is a boy." There it is the first one, the fat one,
etc.**** ****Regards,****
****Bryce **

**Letter from Dirk Van M:**

**A comment on a rather old
problem...
The Paradox of Irrelevant Information (October 26, 1997)
The question is "is the fact that the girl is named Mary is
relevant
information or not" ?
Once again, the answer comes from Reverend Bayes himself : it is
relevant if
we BELIEVE it is.
Otherwise it is not... We must use a priori knowledge in order to
answer
this question.
In order to prove this, draw the complete tree of events :
- having a first child (prob boy = prob girl = .5)
- if it is a girl, what wil be her name ? (prob MARY = p, prob
any other
name = 1-p)
- having a second child (prob boy = prob girl = .5)
- if it is a girl, what wil be her name ? for simplicity, admit
that if the
first child was a girl and if
her name is MARY, it is not allowed to name the second girl also
MARY ;
thus : prob MARY = p if the first
child was a boy or a girl not named Mary, prob MARY = 0 if the
first child
was a girl named Mary)
Developping the complete tree we find :
B & B : 1/4
B & G=M : 1/4 p
B & G!=M : 1/4 (1-p)
G=M & B : 1/4 p
G=M & G : 1/4 p
G!=M & B : 1/4 (1-p)
G!=M & G=M : 1/4 (1-p) p
G!=M & G!=M : 1/4 (1-p)(1-p)
And sum of prob = 1 as must be... Eliminating all cases where
there isn't a
girl named MARY we find :
B & G=M : 1/(4-p)
G=M & B : 1/(4-p)
G=M & G : 1/(4-p)
G!=M & G=M : (1-p) / (4-p)
Hence, prob that the other child is a boy : 2 / (4-p) !!
If we believe that MARY is a very ordinary name and it is higly
likely they
gave at least one of the girls
this name (p ~ 1), then P = 2/3. If on the other hand we believe
Mary is a
highly unlikely name (p ~ 0) than
P = 1/2.
In the first case, because we BELIEVE Mary is an ordinary name,
we gained NO
information from this fact ;
in the second case, we gained some information about it...
There is no unique answer because it all depends on the unknown A
PRIORI
probability p... And that is what
Bayesian analysis is all about.
Of course, one can make a more detailled analysis by allowing
parents to
give identical names to their
girls or estimate that if they rejected Mary as the name of their
first
child there is a chance p' < p that they will choose it for
their second
child, but all this is irrelevant. What's important is that you
cannot
do a posteriori analysis without a priori probabilities and most
of the time
these are just based on believes or common sense.
Hope this settles the question once and for all...
Dirk Van M
PS : another nice question :
A says : yesterday I went to X and I saw a yellow frog.
B says : that's funny, last summer I went to Y and I too saw a
yellow frog.
Where would you go if you wanted to meet a yellow frog ?**

**Letter from Zarko B:**

**I just came across your page and,
pardon my arrogance, it looks to me that
all you guys overcomplicated the matter.
In this particular case, the fact that question referes merely to
"other"
child means that there is no labeling - you are not dealing wiht
ordered
pairs but sets. So the event space (assembly - set of possible
configurations) is {{B,B}, {B,G}} - a set of two multisets (multi
since you
can have repeated element), and that is why probablility is 1/2.
Q.E.D.
A matter of wrong thatement of the event space at the beginging
which in the
heat of debate noone thought about checking.
If the question was about the probablility of women't older child
being boy
it would be a compound probability then with completely different
analysis.
There is a little thing constnatly used in quantum statistics
called
"indistinguishability of particles" i.e. when you have
item which don't have
any labeling relevant to your problem they are indistinguishable
for the
purpose of probabilistic reasoning and you end up working with
sets and not
ordered n-tuples which reduces number of states.
Regards,
Zarko B**

**Letter from Renato L**

**I read your essay on logical paradoxes
( Irrelevant Information Paradox) and I
enjoyed it.
I think I "saw the light" after reading Tom's tale
about the tricky hotel room
costing 30$.... that sort of kicked me into really understanding
what we are
talking about, instead jumping straight away to try to solve it.
As I see it, both the man and the woman ALREADY have a son.
So the probability that the oter is also a son is 1/2 in both
cases,
and there is no paradox. I try to explicate:
1) the man's older is a male. That is part of the data of the
problem,
that is, it has already happened. What is the chance that the
younger
is also a male? 1/2
2) the woman has two siblings, of which one is a male. This also
is
already set. the "other" can be male or female. the
prob. that he is
a boy is also 1/2
Easy!.... may be too easy?
Best regards
Renato L**

**Letter from Raphael P:**

**I just read your article you said you
gave up on in 2001 on irrelevant
material or something like that. You are correct in saying that
the
probability of a boy as the second or other child is 1/2 in
either case
and it is so simple to show I can't understand how Tom and Tony
and the
rest of Mensa could make it so complicated. They probably did the
same
thing with the study they did and got irrelevant results. The
sample
space is much simpler than they want to make it for both cases
and it
doesn't matter if the man's father had his wisdom teeth taken out
or if
the woman had her poodle groomed on New Years Day. For the man
whose
first child was a boy the sample space does not include that boy.
The
sample space includes only the possibilities for the second
child, which
is either (B or G), each with a probability of 1/2. For the
woman; to
say "at least one child is a boy" is the same as saying
"one child is a
boy". That leaves one child to choose the sample space for,
which is (B
or G), again, each having a probability of 1/2. The other boy
does not
fit into the sample space in either case.
Sincerely,
Raphael P**

**Letter from Ray B:**

**>This excerpt is from your paradox
page:
>
>The standard situation:
>A Man has exactly two children, the older a boy.
>A Woman has exactly two children, one of whom is a boy.
>What is the probability of the Man's younger child being a
boy? What is
>the probability of the Woman's "other" child being
a boy?
>
>My comments:
>
>The answer to both questions, as they are stated, is 50%.
>This is because the wording of the questions clearly treat
the gender
>event as an independent trial. An independent trial is not
affected by
>any other event past,present or future.
>
>The title of your page "The Paradox of Irrelevant
Information" is
>unfit.
>
>There is no paradox.
>
>A Better title would be "The Obfuscatory Power of
Irrelevant
>Information"
>
>The extra information invites the simple mind to shift into
permutation
>mode as evidenced by all the "irrelevant"
discussion on the page.
>
>Most of the discussion on the page seems to assume that the
questions
>were:
>
>1) What is the probability that a man has a boy and then
another boy?
>2) What is the probability that a woman has two boys?
>
>Again, there is no paradox here.
>
>It is more like a trick question on a college final.
>
>Please change the title of your page at once.
>
>Ray B**

**[Birdman responds:]**

**As you can see if you bother to read
the material thoroughly, there are
intelligent people who hold rather differing opinions on this
subject.
And since not all of us are as brilliant as you, we apprehend
paradox
where you do not. So I guess you will just have to forgive us.**

**[Ray replies:]**

**John,
...didn't mean to offend ... I omitted the usual LOL! for effect
... I
write gaming software professionally where statistical errors
become
immediately apparent when the game begins bleeding cash.
I was attracted by the word paradox in the page title and after
reading
the page and discovering no paradox I had to respond.
Again, this text, as is, contains no paradox:
-------
A Man has exactly two children, the older a boy.
A Woman has exactly two children, one of whom is a boy.
What is the probability of the Man's younger child being a boy?
What is the probability of the Woman's "other" child
being a boy?
-------
Anyone presenting an analysis that argues forward from this text
and
asserts paradox is either assuming something not present in the
text or
they have argued incorrectly. And many disagreements end up being
about
the correct set of assumptions. Numbers don't lie but words can
be
confusing.
Here's an example where the information does make a difference in
the
odds (but there is no paradox)
The Let's Make A Deal Game
--------------------------
There is a million in cash behind one of three doors.
You are invited to try and select the million dollar door.
You get the million if you guess correctly.
So you pick door #1.
But before door #1 is opened Monty Hall opens door #2 to reveal
Carol
Merrill on a donkey ... Everyone laughs.
He then informs you that you may now pick door #3 if you want or
you may
stick with door #1.
So what do you do?
-------------------
In this example, there is useful information and irrelevant
information,
but there is no paradox even though the probabilities are
different for
each of the remaining doors.
One of your respondents also feels that 'Paradox' is the
incorrect word.
The better adjective is Obfuscatory.
That was really the only point I was trying to make.
Ray.**

**Letter from Ben R:**

**Hello John, reguarding your analysis of
the second sibling paradox, I
believe that you are correct, but that all of the arguments are
much more
complicated than they need to be. The SA basically boils down to
a set
of {BB, BG, GB, GG}, with GG being eliminated since we know that
one of
them is a boy. As you put it, flipping 2 coins, but eliminating 2
tails.
This is where I think you go wrong. Given that there is 1 boy
with
irrelevant age, there is no need to flip 2 coins and end with a
2:1
ratio. Only 1 child's sex is in question... only 1 coin needs to
be
flipped. I don't accept the {BB, BG, GB} as the possible result
set.
It should be {BB, BG}, BG and GB are equivolent, since age is
irrelevant
here. 50/50, as it should be.
Ben**

**PS: Whenever you do get around to it, I
would also like to pose a second
angle on the question. If people do consider the age to be
relevant,
then we have 2 distict possibilities and the {BB, BG, GB} result
set
is still irrelevant. If the known boy is the older child, then
the
possible results are {BB, BG}. If he's the younger, then the set
is
{BB, GB}. Still 50/50, either way.**

**Letter from Chris:**

**Hey Birdman,****
****It looks
like this subject has been shelved for some time, but since
it is still on your website I'd like to comment upon the problem.****
****After
glossing over the paradox and its subsequent analyses by Tony,
Tom and yourself, I got this weird feeling something wasn't
right. My common sense wouldn't allow be to be taken in by
the logic of the situation. Suddenly something clicked in
my brain and I was left with this: ****NOIR,****
****- the
abbreviation I learned in 7th-grade Pre-Algebra.****
****The
N (Nominal) and the O (Ordinal) seemed to explain
it all for me. Whether the child is the second or the
other, the child should be a boy half the time.
OK, that makes sense. It doesn't make ****sense
****the probability of the Woman's other child
to be a boy 1/3 of the time. The crux of the problem is
this:**** ****The Man's situation implies order, while the
Woman's does not. To address the Woman's situation in light
of order throws askew the **__fact__**
that her kid will be a boy 1/2 the time.****
****(Ordinal)
Man: [BB, BG] <-- The 'B' is firmly fixed in the left position
because of its primacy.**** ****(Nominal) Woman: [BB, BG, GB]
<-- We make the mistake of IMPOSING ORDER on one/other.****
****BG and GB are
the same thing concerning the Woman. We unconsciously order
them with the way we compose these problems. (Left is first,
right is second) **** ****The Boy[B] and Girl[G] represent
two distinct entities, and to symbolize these on paper or in your
mind, we seem to be restricted to the theater of Event
Space. B is isolated and unconnected to G in the one/other
scope. How is this to be represented? ****
****Top-bottom,
Left-right, Front-back, Now-later. They ALL reflect Order.****
****B
BG
G or
GB**** ****Relationships are by their nature
ordered. Saying ONE and OTHER implies NO
relationship. **** ****The paradox only solidifies my
prejudice about Logic being less valuable than Common
Sense. Also, these exercises seem useless in context to the
Decay of Civilization. Though they do provide an escape
without the numbing effect of television.****
****Keep up the
Free Speech,**** ****Chris **** **

**PS. Here's a little brain
teaser for you: Given the equation, 5 + 5 + 5 = 550, place
a one (1) in the equation to make it true or equal. ****Hint:
Truth and Equality aren't always the same thing.****
**

**Letter from Luc
- comment on letter from Stephen:**

**Letter from Stephen :
For example, let's say you were talking to a woman on the phone.
She tells you she has 2 children. And they are both home right
now. Then, you hear a child's voice in the backround and it's a
male. Then I would say that the probablity that the other child
is a male is 1/3. If, however, before you heard anything, you
told the lady to put the younger child on the phone. You hear it
is a male's voice. Then the probability that the other child is a
boy would be 1/2. If, going back to the first scenario, after
hearing the childs (male)voice, you aks the lady, "is that
the younger child", and she says "yes", I would
say the probability that the other child is male is still
1/3…..
With that in mind, let's go back to the lady with the 2 children,
not both female. It certainly seems that the probability that
both children are male, if all you know is that the sex of both
is NOT female, is 1/3. However, many are claiming that if you
knew that the youngest was a male then the probability the the
other is male becomes 1/2. Well, I disagree. (Unless that
knowledge came about in some non-arbitrary way, as mentioned
above.) Let's put it this way. You are on the phone. You know she
has 2 kids and they are both home. You hear a male voice. You
then ask if the voice came from the youngest. She says yes. Then
many are claiming that the probability that the other is male
becomes 1/2. Well, if she had said the voice came from the
oldest, then a similar line of reasoning would also have one
arrive at the conclusion that the probability of the other child
being male was 1/2. But, come on, the voice you hear MUST be one
or the other! (older or youner) In other words, in this scenario,
knowing the relative age of the known male seems to give no added
information. But again, if you had first asked "what is the
sex of the youngest?". If she says "male", then It
seems that the probability of the other being male is indeed 1/2.
I hope you see my point here.
“Then, you hear a child's voice in the backround and it's a
male. Then I would say that the probablity that the other child
is a male is 1/3”**

**Luc comments on the above
letter:**

**He is wrong; the probability is 1/2; to hear a
boy’s voice on the phone is not the same as to know that
there is a boy at home.
There are now 8 possibilities :
(the two children plus the child at the phone)
B1B2 B1(at the phone)
B1B2 B2
B1G2 B1
B1G2 G2
G1B2 G1
G1B2 B2
G1G2 G1
G1G2 G2
As it is a boy at the phone, the possibilities are :
B1B2 B1
B1B2 B2
B1G2 B1
G1B2 B2
And the probability of the second child being a boy is 1/2
Now the crux : if you tell the lady to put a child on the phone :
If she RANDOMLY puts a child(the fattest, the older, the younger,
or with a dice…), we have the situation as above:
B1B2 B1(at the phone)
B1B2 B2
B1G2 B1
B1G2 G2
G1B2 G1
G1B2 B2
G1G2 G1
G1G2 G2
Now, if she systematically puts a boy on the phone (when there is
one, of course) :
B1B2 B1 or B2
B1B2 B2 or B1
B1G2 B1
B1G2 B1
G1B2 B2
G1B2 B2
G1G2 G1
G1G2 G2
So if you hear a boy :
B1B2 B1 or B2
B1B2 B2 or B1
B1G2 B1
B1G2 B1
G1B2 B2
G1B2 B2
The probability for a second boy is 1/3
Well, when you know that the younger of older of fattest or
“this child” is a boy, you hear a child at random, when
you know that “one of whom is a boy”, you always hear
the boy.
So, you may use any comparative (or not) criterion except
"to be a boy"
(please pardon my poor english and correct my mistakes...)**

**[End of Luc comment]**

**Letter from Hoop:**

**Regarding the paradox of irrelvant information
One can not solve paradoxes by reasoning only by understanding!
My understanding is this.
The rule for the womans case, counter intuitive but soundly
logical, changes the probabilityof her older child being a boy
(2/3 instead of 1/2). For the younger child it remains the same
(1/2).Two boys gives 2/3 x 1/2 = 1/3.
So the paradox evaporates.
Confusion in this case (as in so many similar examples) has
arissen from
1) aplying probabilities to the past. Contrary to Mark Twains
opinion predicting the past is not only more difficult than
predicting the future but impossible.
2) mixing rules (i.e. fixed outcomes) and probabilities
kind regards
Hoop**

**Letter from DanielZG1**

**Irrefutable Proof That the****
Paradox**** of Irrelevant Information is
Indeed Irrelevant
Dear Birdman,
For the past couple of weeks, I've been thinking, on and off,
about
what you claim is the paradox of irrelevant information. The more
I
thought about it the more frustrated I became and I was just
about
ready to concede that you did indeed find a paradoxical flaw in
standard analysis. But then, just the other day, I sat down on a
subway
and the answer hit me like a brick.
I looked through some of the **

were definitely right in saying that they simply overcomplicated the

issue and that the explanations they give are completely opaque. The

matter is really quite simple. I will explain it as clear and simply

that I possibly can, in hopes of convincing you that you were wrong in

claiming that this is a paradox. I will briefly review the matter at

hand and point out where the flaw was in your logic.

Lets start with the man. The man has 2 kids, the older being a boy.

Lets call this older boy kid Kid A and his other kid Kid B. This gives

us 2 possibilities;

1) Kid A is a boy who is older than Kid B, who is a boy

2) Kid A is a boy who is older than Kid B, who is a girl.

Because both of these are equally probable to happen, each of these

events has a 50% chance to happen. Therefore, the odds of the younger

kid (kid B) being a boy is 50%.

So far so good. Now lets move on to the woman:

The woman has 2 kids, one of whom is a boy. What you say is that there

are three possibilities; 1. both kids are boys, 2. one is a boy and

one is a girl and the boy is older, and 3. one is a boy and one is a

girl and the boy is younger. Then you go on to say that since these

three possibilities are equally probable, the probability of each is

33%. And since in only one of these possibilities is the

boy, the odds of the other kid being a boy is 33%, which you say is

paradoxical because these are different odds then the man but the only

difference between the man and the woman was one piece of irrelevant

information. Here's the flaw in your reasoning: There aren't only three

possibilities with the woman, there are actually four. Here's why:

Lets call the kid that the woman has whom we know is a boy Kid A, and

lets call her other kid Kid B. Now lets examine all of the

possibilities.

1) Kid A is a boy who is older than Kid B, who is a boy.

2) Kid A is a boy who is younger than B, who is a boy.

3) Kid A is a boy who is older than Kid B, who is a girl.

4) Kid A is a boy who is younger than Kid B, who is a girl.

Since all of these are equally probable, the probability of each one

occurring is a 25% chance. Since two of these possibilities involve her

a 50% chance that she will have a boy for her second kid-the same odds

as the man.

Here's another way of wording it: when you were examining the woman's

possibilities, you took into account that since we do not know the ages

of her kids, there are 2 possibilities that involve her having a girl

and a boy; either that boy is older or the boy is younger. And this

was absolutely correct. However, what you didn't take into account is

that not knowing the age of the kids also gives us two possibilities

involving her having two boys; that one boy is older than the other,

and that the

Therefore, what you call the paradox of irrelevant information is not a

paradox at all, but rather just a riddle. A good riddle, I might add,

but a riddle nonetheless.

What makes it a good and clever riddle is that, when thinking about the

possibilities of the woman having a boy and a girl, it is easy for us

to fathom the two possibilities; either the boy is older or the girl is

older. It is easy because the fact that one is a boy and one is a girl

makes it easy to differentiate between them in our minds. With the

possibility of the woman having two boys, however, because all we know

about them is that the are both boys, it is tempting to blend their

situation as one possibility, when in fact this isn't true. In other

words, just because we can't differentiate between the two boys, it

doesn't mean that they aren't two different people who therefore have

two different possibilities about which one of them is older.

On a final note, I would like to add that although I realize that you

pretty much put the debate over this

I still decided to write to you because it seems to me that you still

believe that you were correct in calling this riddle a paradox, and I

hope to help you see the light so that you finally admit/realize that

you were wrong. Please give what I have said some serious thought and

get back to me.

**Leter from Scott in Orlando**

**John,
I don't know if you ever resolved "The Paradox of Irrelevant
Information" to
your satisfaction but I believe you are correct in your
disagreement with
the SA. The apparent paradox is due to inadequate use of the
known
information for the second case (i.e. the woman's case). This
arises from
an incorrect probability assigned to that information.
Supporters of the SA assign a probability of 66.7% to the known
information
when in actuality it is 100%. Tony, and others, argue that
statistical
analysis always proves the SA correct. They give examples such as
the coin
flipping situation, or a roulette wheel, as practical evidence.
These
examples are not equivalent to the standard situation associated
with the
woman. In coin flipping it can never be assured *prior to
flipping *that
one of the next two flips will end up heads. In the woman's
situation the
sex of one of her children *is known* beforehand.
As an example we can adapt Tony's bag-of-balls approach to
generate a truly
equivalent situation to the standard situation for the woman.
There are two
bags one of which is known to contain a black ball of unknown
diameter. In
another bag are two balls, one black and one white, also of
unknown
diameters. If one ball is picked from each bag the probability of
picking
two black balls is 50% *regardless of the diameters (or age,
oldness, or any
other "label").*
In effect, the SA uses only half the information available and in
doing so
reduces the number event possibilities from 4 to 3 rather than
down to 2.
The known information should be thought of as occupying both
"labels" at the
same time only in half amounts, analogous I suppose to aspects of
quantum
mechanics. (Note that the total probability remains 100%). Once a
ball is
chosen from the bag containing two balls the "label"
(i.e., position) of the
other is then known.
If all the woman participating in the study group recently had
a baby and then adopted a baby of unknown age from an agency
providing only
boys I'm sure all the supporters of the SA would agree that the
probability
of her having two boys is 50%.
Sincerely,
Scott in Orlando**

**Letter from David C**

**John, I have read your article on PII, and methinks
it'd have been more convenient to put my mind in a blender with a
little lemon juice. At least that way Jimmy Buffet may find some
use for it. My 2 cents on this subject was on my lowest priority
for today, behind full time school and work (not being a member
of mensa is a factor as well). So that may well be all it's
worth, 2 cents. Not to mention the wine I have been drinking: and
unfortunately, In Vino, Veritas does not apply to logical truths,
only known situational facts. These factors aside, I sincerely
hope you will read my input, and grill me if need be. **

**Concerning the opposing viewpoint. I agree, Tom and
Jerry's opinions and argument's against JBR Yant's opinion seems
to have very little to do with the Birdman's own position. I
further agree that I do not understand the opposing arguments, at
least not the point they want made. **

**Secondly, gender is determined by the MALE. So
biologically, the father(s) of the female's 2 children not being
given, the male has a higher biological chance of having a second
male heir, the gender of his first child being known. So in this
sense age CAN be a factor, as a father with 2 daughters is more
likely to have a (third, younger) daughter. Altho this factor may
be irrelevant; as this argument is not one of genetics or biology
but of probability and ir/relevant information. It is relevant
becoz, as convenient as it may be to compare a concrete situation
to an abstract idea, a flawed example/comparison can possibly
result in a flawed conclusion. **

**This last statement being said.... Is it possible that
we have not come up with a satisfactory concrete analogy to fully
represent the abstract philosophical issue at hand? Or that we
ever shall regarding this particular issue? This is the very
reason I avoided mathematics in my argument. **

**The tossing coins scenario won't work either: a
perfect machine tossing them in a precise manner will generally
yield some consistent result. At the opposite spectrum human
physical hand tossing would not be precise enuf to be
scientifically consistent. This, coupled with the fact that no
known human can conjure up true randomness to even make said
examples work theoretically even; leads me to ask if it is
possible that as yet no absolute truth may be known in this
argument? **

**Is it that a possibly superior analogy to this issue
will one day improve our (my?) understanding of it? Truly, **

**Yours David C only 21 years of age **

**Letter from Sber**

**I came across your paradox section and this problem -
The Paradox of Irrelevant Information - has kept nagging me for a
while so I decided to have my take on it. For me, the paradox is
not. It is just that a problem must be reduced to its simplest
and clearest expression by eliminating all the language
ambiguities. **

**The BOY-GIRL PARADOX : A QUALITATIVE APPROACH **

**If we rephrase a bit the problem in a logical
equivalent we get : **

**First we select our samples : **

**- Population A : Let's find a family with exactly 2
children where at least one is a boy and is also the older child
- Population B : Let's find a family with exactly 2 children
where at least one is a boy **

**Then we ask a question on the probability of a
particular event occurring in our samples : **

**- What is the probability, for each population, of the
other child being a boy ? **

**If we agree that the above is the logical equivalent
to the problem then it is rather obvious that Population A must
be smaller than Population B because the selection criteria is
more restrictive for A than for B. The boy must be the older
child for population A whereas no such restriction is imposed for
population B. Hence the resulting probability for the event
"the other child is a boy" is higher in A than in B. **

**In other words the more we put restrictions, the
smaller gets the sample size and the higher gets the probability
of one particular event as is shown in the progression below with
the standard notation : **

**- No restriction : Sample = {BB,BG,GB,GG}. Probability
for BB = 1/4 - At least one is a boy : Sample = {BB,BG,GB}.
Probability for BB = 1/3 - At least one is a boy and is also the
older child : Sample = {BB,BG}. Probability for BB = 1/2 And if I
didn't make myself clear, for the insane ;-) : - At least one is
a boy, is the older one and has a brother : Sample ={BB},
probability for BB = 1 **

**Thanks for your time. I love your site. Keep it going
! **

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