**October 26, 1997 **

**To: Fellow Mensans, Fellow Libertarians and other
Highly Intelligent People **

**From: Mensa's Resident Iconoclast (Tom Elliott's
words: You did see Tom's review of my books in the October
Mensa Bulletin, did you not?), aka John Bryant **

**Subject: An invitation to cast your vote on the
Paradox of Irrelevant Information **

**For the last several weeks, about 20 of us (mostly
Mensans, but with a scattering of Promethians and Libertarians)
have been having a lively (and often acrimonious) exchange over a
problem recently addressed by IQ- superstar Marilyn vos Savant, a
problem which I have named the Paradox of Irrelevant Information
(PII). Because I and most of my opponents disagree with each
other just as strongly as we did at the outset, at my invitation
(to all opponents), two of them have agreed to write refutations
(or, in my opinion, "refutations") of my position, with
the view that the Mensa email membership should have an
opportunity to read our essays and then vote on whether they
thought my position was right, or whether that of my opponents
(including Marilyn, as it happens) is. You are hereby invited to
read these three essays and cast your vote accordingly. **

**It may be noted that Marilyn took up this problem for
a third time in her Parade Magazine column of 19 Oct,
due apparently to the particularly large amount of mail (and
interest) generated by this topic. Marilyn, as it turns out, did
not have the benefit of my input, or she might have responded
differently (she basically just restated the Standard Analysis
(SA), given below.) She did, however, report the result of a
statistical survey which seemed to support her analysis. In view
of this, I should make it clear that I am not arguing that the
statistics she cites are "wrong"; rather I am arguing
that the SA leads to a contradiction. What that means is that the
SA may give the right answers sometimes, but that
because it leads to a contradiction, its results must always be
suspect. But if I am right about the SA being wrong, then that
fact would seem to shoot a big hole in probability theory. For
me, the point in arguing my case is simply to find either a
confirmation that I am right or a convincing argument that I am
wrong, by presenting my arguments to the smartest and best
informed people I can find. My thought is that if Mensans find my
arguments convincing, perhaps the Mensa Bulletin editor
will run a feature to let all Mensans hear the
arguments, and maybe even Her Royal High-IQ-Ness Marilyn will
deign to confront them. **

**As an aside, it may be noted that the explanation of
the PII given in my first letter to the Mensa email membership
has undergone considerable evolution due to the arguments thrown
at me by opponents. But useful as those arguments have been in
helping me clarify my thinking, they have nonetheless convinced
me even more strongly that my position is basically right, since
these counterarguments have been -- in my view, at least -- so
totally ineffective in putting any real dent in my position. But
whatever your view on the subject, the essays in this letter
ought to convince you that the PII is not a simple problem which
is answerable -- as HRH Marilyn would have us believe -- in a
single paragraph or column. **

**So don't forget: After you finish reading, remember to
send me your vote! Comments and criticisms -- to say nothing of a
solution -- will also be appreciated. **

**The Paradox of Irrelevant Information **

**The standard situation: 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 SA: We assume that the probability of a
boy (or girl) at any given birth is 1/2. This means that the
probability of the Man's younger child being a boy is 1/2.
Another way to characterize this is to say that the "event
space" for the Man is the set of equally probable events
{BB, BG}, and the probability of BB is the ratio of the number of
"successes" (BB) to the number of possible events (BB,
BG) in the event space, namely, 1/2. **

**As for the Woman's case, because we do not know
whether it is the Woman's older or younger child who is the boy,
the event space is the set of equally probable events {BB, BG,
GB}, and the probability of the event "both boys" is
the ratio of successes (BB) to possibilities (BB, BG, GB), ie
1/3. **

**Explication of the SA in terms of
coin-flipping: Coin-flipping can be made to model the
Man/Woman situation as follows: (1) Represent the event of the
Man's younger child as a coin flip, with H for boy, T for girl,
so that the event space becomes {HH, HT}; (2) represent the
Woman's event space similarly as flipping two coins
simultaneously, and eliminating all {TT} events from
consideration, so that the event space becomes {HH, HT, TH}.
Everyone agrees that the probabilities represented by the coin
flips are the same as in the SA for the Man/Woman situation. **

**The intuitive objection to the SA: (Note:
This is not a precise objection, but rather what is known in
mathematics as a "motivation".) Consider the following
statements: (1) Of x's 2 children, the older is a boy (2) Of x's
2 children, one is a boy The SA says in effect that (1) implies a
higher probability for x's second (other) child being a boy than
does (2). This, however, is intuitively unsettling, because
olderness is not a quality which impacts on the probability of
sex (ie, sex and age are independent), and hence olderness is
irrelevant to the probability of being a boy. The supporters of
SA have two basic responses to this: (a) Statement (1) -- and in
particular, the assertion of olderness -- gives "more
information" (b) Any category matters, hence is not
irrelevant. (To be understood, this point needs explanation; but
because it is irrelevant to our arguments below, we merely state
it here for the sake of completeness.) Because the above
arguments are all somewhat indefinite, they settle nothing. They
merely point to the field of battle. **

**Using an additional category with SA: Note
that virtually any comparative quality whatsoever (eg, fatter,
smarter) could be substituted for "older" and the SA
would not be essentially changed. Now let us suppose that we know
a comparative quality of the Woman's boy (or one of them), eg,
suppose we know that the Woman's fatter child is a boy.
Then we can use this information in the SA to show that the
probability of the Woman's second child being a boy is 1/2, not
1/3, in exactly the same way we use olderness to show the
probability that the Man's younger child is a boy equals 1/2. We
can use the two bits of information (olderness, fatness)
separately or simultaneously: If we use them simultaneously, the
event space for the Man -- using F/S for fatter/slimmer and O/Y
for older/younger -- can be represented as (2 boys) {BOF, BYS} +
{BOS, BYF} + (1 boy, 1 girl) {BOF, GYS}, {BOS, GYF}; while the
Woman's event space can be represented as (2 boys) {BOF, BYS} +
{BOS, BYF} + (1 boy, 1 girl) {BOF, GYS} + {BYF, GOS}. (Note that
the Man's case for (1 boy, one girl) excludes the events {BYF}
GOS}, {BYS, GOF}; while the Woman's case for the same excludes
{BOS, GYF}, {BYS, GOF}.) This event space yields a probability of
1/2 for the second boy for both the Man and the Woman, just as
does using the bits of information (O/Y, F/S) individually. **

**Why the SA is contradictory: We have already
noted how virtually any comparative category will yield a higher
probability for the second child being a boy. While this is
counterintuitive against the SA, the situation is actually much
worse, for rather than a comparative category (eg, older/younger,
fatter/slimmer, smarter/dumber) we can simply use the category
"one/the other". This is counterintuitive to the
contention that the SA is valid not only because this new
category gives no new information whatsoever (recall that one
justification for SA was that the older/younger category gave
"new information") but it is actually a non-category
category. Or to put it another way, we can raise the probability
of the Woman's other child being a boy without any new
information whatsoever by simply recognizing that
"one/the other" may be used in SA as a category. Which
means that SA has two possible interpretations, one of which
yields a probability of 1/2 for the Woman, and one
which yields a probability of 1/3; ie, SA is
contradictory! **

**Arguments against using "one/the
other": SA supporters will say that "one/the
other" does not make a unique designation for the children,
unlike the other comparative qualities (fatter/slimmer, etc),
since either child could be "one" (or "the
other"). But as far as I can see, this carries no more
weight (and in fact, much less weight) than the objection
against the SA that using irrelevant information to
determine probabilities is illegitimate. But in any event, as I
noted in earlier letters, the nature of using comparative
qualities in the context of the SA constitutes nothing more than
slapping an arbitrary label on the objects of the event space (in
the present case, the children), so "one/the other"
does quite as well for labeling as anything else, altho it is
different from the other labels in the sense that we can only
imagine that they uniquely designate the children,
since we cannot know "which" child each designates.
However, it is my contention that this fact doesn't matter,
because even if we knew which was "one" and which
"the other", it would have absolutely no effect on the
analysis. The above response is reinforced by a completely
different approach, namely, the fact that we know that there
always exists a comparative quality Q (and in fact there exist
infinitely many) which uniquely distinguishes each
child, eg, "noisier/quieter at time t". Thus if Q
exists (and we know it does), then it is -- unlike "one/the
other" -- a real quality; hence this overcomes the potential
objection against "one/the other" that it is empty of
content. In fairness, it should be noted that, in general, we do
not know what Q is; but since we know it exists, this seems
adequate for use in SA. **

**Responding to a final objection: The major
argument in the minds of several of my correspondents in favor of
SA appears to be that, because the coin- flipping situation
described earlier seems to represent the Man/Woman problem
precisely, and since the SA seems to correctly describe the coin-
flipping situation, it follows SA must also correctly describe
the Man/Woman problem. To this I admit that I do not have a
conclusive response, except in the sense that I have discovered a
contradiction in the SA. Perhaps the problem is that the
coin-flipping does not properly represent the Man/Woman
situation, as suggested by my discussion of labels. If all else
fails, we can simply say that this is the crux of the PII,
namely, that we seem -- like Buridan's ass -- to be drawn in two
opposite directions at the same time. To which the supporters of
SA will undoubtedly respond that when an ass is drawn in two
opposite directions, the result can only be the exposure of an
asshole. **

**Additional reinforcement: Libertarian Eric
Carroll came up with an independent analysis of the PII which
reaches the same conclusion as mine. While he has been known to
flip-flop on his positions, I looked his work over and it seems
right to me, and in fact quite elegant. However, the last time I
heard, he had begun to waffle. All I can say is, Eric, stiffen
that backbone! **

**Final remarks: It is my suspicion that the
resolution to the PII is in having better criteria for setting up
problems. But because I am not well- enuf acquainted with
probability theory to know what impact my work might have, I can
only speculate on this point. But whatever implications the PII
carries for probability theory, it is clear from our discussion
that there are some rather confusing elements in this area which
no one has yet succeeded in properly clearing up, Marilyn and my
correspondents included. Which means that there is hope, after
all, that Las Vegas can be beaten. **

**What follows below are two essays by Mensans who
believe I am mistaken and who have agreed to write responses to
my essay above for publication. The first is by Anthony C.
Patzelt (acp@pipeline.com); the second by Tom Serb
(TSerb@aol.com). I have not changed the text of either of their
essays except to indicate with "note 1", "note
2" etc the places in the respective essays to which my
comments at the end of each essay apply. My general reaction to
these essays is that they are illogical and do not prove their
point. Specific problems with these essays are noted in my
comments following the essays. **

********************** **

**Tony's essay: **

**I have attempted to relay this to John Bryant in
private correspondence to allow him to correct himself. I have
tried various illustrations in an attempt to point out his
logical flaw. He seems to be ignoring this information while
continuing to publicly bluster. **

**John seems to believe that by introducing what he
terms "irrelevant information" into a probabilistic
Standard Analysis that you generate a "paradox." I
think "paradox" is far too strong a term and I might
feel more comfortable with "counter-intuitive." **

**Now, we know that the statistics, the experimental
data prove John wrong. So either we can believe that John has
discovered a "paradox" that shakes the foundations of
probability theory, or more likely there is a flaw in John's
logic. **

**His logic is basically: "If SA predicts blah,
blah using irrelevant information" **

**Well, lets stop right there. John has been very crafty
in concocting this puzzle. He storms quickly through the SA, and
guess what, he is using it incorrectly. **

**John has tried to bamboozle us into believing that the
SA is based on THE NUMBER OF POSSIBLE OUTCOMES as opposed to the
correct, NUMBER OF ALL POSSIBLE WAYS TO REACH ALL POSSIBLE
OUTCOMES. **

**I will illustrate this with the more classic
presentation of this problem, which contains no potentially
"irrelevant" information. (note 1) **

**You have a black, velvet sack. Inside the sack are two
blue balls and one green ball. The balls were all manufactured in
the same way, so there is not way for you to tell which color a
ball is when you reach into the bag to pull one out. So it is
equally probable that you pick any ball. **

**The equivalent of the Man situation: **

**You reach into the bag and pull out a blue ball, what
is the probability that the next ball you pick will be blue? Note
that this is predicated on pulling out a blue ball first. If you
pull out a green ball first, you have to try again. **

**John's (incorrect) analysis: **

**Event Space: **

**First Ball Second Ball **

**Blue Green Blue Blue **

**So, if the first ball you pick is Blue, the
probability of the second ball being blue is 1/2 or 50% **

**Note that for the correct analysis we need to label
the blue balls Blue(1) and Blue(2) to avoid the contradiction of
retrieving Blue(1) first and then Blue(1) second or the
contradiction of retrieving Blue(2) first and Blue(2) second.
(note 2) **

**The correct analysis: **

**Event Space: **

**First Ball Second Ball **

**Blue(1) Blue(2) Blue(2) Blue(1) Blue(1) Green Blue(2)
Green **

**So, if the first ball you choose is blue there is a
2/4 or 50% chance that the second ball will be blue. **

**The equivalent of the Women situation: **

**You reach into the bag and simultaneously extract two
balls, what is the probability of both balls being blue? **

**John's (incorrect) analysis: **

**Event Space: **

**One Ball The other Ball **

**Blue Blue Green Blue **

**Therefore the probability of retrieving two blue balls
is 1/2 or 50%. **

**Again, for the correct analysis, we need to label the
blue balls Blue(1) and Blue(2) to avoid the contradiction of
selecting Blue(1) and Blue(1) simultaneously or the contradiction
of selecting Blue(2) and Blue(2) simultaneously. **

**The correct analysis: **

**Event Space: **

**One Ball The other Ball **

**Blue(1) Blue(2) Blue(2) Blue(1) Green Blue(1) Green
Blue(2) Blue(1) Green Blue(2) Green **

**(note 2a) So the probability of you picking two blue
balls 2/6 or one in three. **

**I know that John will argue that the labels on the
blue balls are "irrelevant" because this lets him slide
past the problem of THE NUMBER OF POSSIBLE OUTCOMES versus THE
NUMBER OF ALL THE WAYS TO REACH ALL POSSIBLE OUTCOMES. (note 3) **

**Unfortunately, that is not "irrelevant."
This leads to the contradictions highlighted above. **

**So, not surprisingly, John's misapplication of the
Standard Analysis has led to his supposed "paradox." **

******************** **

**The following are my comments on Tony's
essay: **

**General remark: As is obvious, Tony did not actually
bother to respond to my arguments except in an indirect way. You
may judge for yourself as to whether this means he cannot
understand them, or simply prefers to avoid dealing with them. In
any event, I find his essay completely unconvincing. **

**(note 1) Using the balls-in-a-sack situation to
represent the children situation is dubious at best, so any
conclusion drawn from it is dubious. Furthermore, it adds
unneeded complexity to a situation which can use all the
simplification it can get. **

**(note 2) Tony blithely decides to slap labels on the
balls without any consideration of the implications which this
has for his analysis. It is the act of changing objects from
unlabeled to labeled which I used to show SA to be contradictory,
so if Tony justifies his labeling, how does he avoid my
contradiction? **

**(note 2a) Unless Tony is using "One Ball"
and "The Other Ball" as labels for the balls in
addition to 1 and 2, then lines 1&2, 3&5 and 4&6
represent the same event, since the balls they refer to are
supposedly not ordered in being drawn from the bag. **

**(note 3) I did not so argue: see note 2. I also find
Tony's apparently- crucial distinction between possible outcomes
and possible ways to reach them opaque. **

******************* **

**The following is Tom's essay: **

**When I was a child, my grandfather told me a riddle:
Three men walk into a hotel and ask for a room. The manager tells
them that the room costs $30, so each man puts in $10. After they
leave for the room, the manager realizes he gave them the key to
a $25 room, and he sends the bellboy after them with the $5
overpayment. **

**The bellboy decides to pocket some of the money; he
gives each of the men $1, and keeps $2 for himself. Now each man
paid $10, and got $1 back, so each really paid $9. $9 times three
men is $27, and the bellboy kept two, which makes $29. Where did
the other dollar go? **

**This was somewhat confusing to me as a lad, until I
wrote it out, and saw the error: the correct expression would be
($9 * 3) - 2 = $25, the cost of the room. Funny how words can get
in the way of understanding. **

**With that in mind, let me propose my proof exposing
the 'paradox' posed by John Bryant. John has challenged others to
prove the validity of standard analysis in response to his
approaches; I'll not do that -- it should be sufficient to prove
he is wrong. **

**I. THE PROBLEM: **

**A man has two children, and the older one is a boy. A
woman also has two children, and at least one is a boy. What are
the probabilities that each of them has two boys, given that the
likelihood of a boy or girl for any given birth is equal? **

**II. THE PROOF **

**1. Probability may be expressed as a fraction: the
number of desired outcomes divided by the number of total
outcomes. 2. The chance of any given child at random being a boy
is 0.5 3. The chance of any two unrelated events occurring
together can be found by multiplying their probabilities
together; in this case, the chance of two boys is (0.5) * (0.5),
or 0.25. This becomes the numerator of the fraction, the outcome
to be found in either case. **

**For the man: 4. The denominator will be all events
which result in an older boy (1.0 by definition) times the events
that result in a boy at random second birth (0.5). 5. This can be
expressed: (0.5 * 0.5) / (1.0 * 0.5) which reduces to 1/2 **

**For the woman: 4. The number of total outcomes that
will satisfy her parameters is the sum of probabilities of
two-boy occurrences (shown to be 0.25) plus the chance of a mixed
boy-girl occurrence. 5. Mixed boy-girl probability can be found
by taking all births (1.0) and subtracting the probability of
two-boy and two-girl births. 6. The probability of two boys is
expressed: (0.5 * 0.5) / ((0.5 * 0.5) + (1.0 - (0.5 * 0.5) - (0.5
* 0.5)) which reduces to 1/3. **

**III. ILLUSTRATION **

**John argues that this is counter-intuitive, because
"this boy", "fatter boy", etc. can be
substituted. Let's take three children, A= a 10 year old boy, B =
an 8 year old girl, and C= a six year old girl (note 1). These
result in the following combinations, with either child selected
first; columns note whose parameters are satisfied, and mark
those resulting in two boys with asterisks: **

**Man's? Woman's? AB YES YES AC YES* YES* BA YES YES BC
NO YES CA YES* YES* CB NO YES **

**Note that of all possible outcomes, the man will have
two boys 2/4, or 50% of the time; the woman will have two boys
2/6, or 33% of the time. **

**IV. JOHN'S ERRORS **

**A. THE HIGHER PROBABILITY FOR A SECOND BOY FOR THE
WOMAN IS COUNTER- INTUITIVE. **

**John argues that Standard analysis implies a higher
probability for the man's second child being a boy than it does
for the woman. Of course they do; so are her odds of having an
older boy less than those of the man (0.67 versus 1.0). He finds
this unsettling because olderness does not impact the probability
of sex. (note 2) **

**This is certainly true in the man's case. However, if
we accept the fact that the woman's older child may be a boy 50%
of the time, we find the following: Boy-boy = 25% (50% of 50%)
Boy-girl = 25% (50% of 50%) Girl-boy = 50% (100% of 50%) (note 3)
This is by nature counter-intuitive. We now have twice the
likelihood that a mixed-sex set will consist of an older girl.
(note 4) **

**In the case of the woman, the second child's sex will
be a dependent variable if the first child is a girl. John
correctly defined an event space as one in which each outcome is
equally likely; if we accept the fact that a boy may be older 50%
of the time in the woman's case, we may not create a Standard
Analysis event space to solve the problem: by definition, it will
not work. Returning to the mathematical expression, the chance of
either child at random being a boy is 2 in 3, and we get a
correctly weighted space: Boy-Boy = 33% (1/2 of 2/3) Boy-Girl =
33% (1/2 of the other 2/3 of first boy) Girl-Boy = 33% (the
remainder; note that independent of birth order, B/G and G/B must
be equally likely) **

**The error here is the assumption that if all birth
possibilities are equal, then the first child has an equal chance
of being a boy or a girl. This is only true if we include the
girl-girl possibility, which has been eliminated in the problem
definition. **

**B. CHANGING THE LABELS WILL CHANGE THE RESULT. **

**Will it? Let's restate this and say that one of the
woman's children is fatter, and that child is known to be a boy.
The fatter child will be shown in caps. **

**We now have, using the same A,B,C possibilities shown
above: **

**Ab, Ac, bA, bC, Ca, Cb, all of which fulfill her
parameters. 50% of the time the fatter child will be the older,
and 50% the younger. What John is actually doing is finding that
events (Ab, bA, and aC, Ca) are pairs of identical events, and
therefore reduce the universe of possibilities. **

**For this we need to look again at weighting: in any
given set of random boy-girl births, considered in order, any one
ordered pair will occur 25 percent of the time. If we exclude Ab
and bA as being identical, we have to ask why we are excluding
the possibility of an existing Bc set (a boy- girl set with a
fatter girl). John's addition of this label does in fact reduce
the universe of possibilities; he has re-stated the problem in
terms other than those originally proposed. **

**Recognizing that fat and chosen are independent
identifiers, we would find: a random 25% chance of two boys with
either fatter; a 12.5% chance of a boy-girl with fatter boy (the
other 12.5% of the time the girl will be fatter), and a 12.5%
chance of girl-boy with fatter boy. Once again, the outcomes he
proposes are not suited to an event-space analysis, as they are
either not equally likely. **

**To state that one child is fatter than the other makes
no change to the numerator; however, associating this quality
with sex WILL change the denominator. That's exactly why the
man's and woman's probabilities differed in the first place. **

**C. NON-UNIQUE IDENTIFIERS HAVE THE SAME RESULT **

**To substitute "This child", and say that
we've chosen "this child" from the woman's two, and
found it a boy, we can easily see that 50% of the time the boy
will be the older child, and 50% the younger. The odds don't
change: although if the boy is the older one, 1/2 of the time he
will have a sister, and 1/2 of the time a brother, the same is
true of finding if we find a younger boy first. If we treat each
case individually and then combine them, we must eliminate the
duplicates from the resulting set: AC,CA,AB,BA + AC,CA,BC,CB =
AC,CA,AB,BA,BC,CB **

**Considering each case separately, and then eliminating
the true duplicates, will give us a proper solution for an event
space analysis, but both possibilities must be considered
simultaneously, and we must recognize those that are in fact the
same event. **

**IV. CONCLUSION **

**Any information attached to the sex of a child will
change the odds. For the information to be truly 'irrelevant', it
cannot restrict the variable to be found: in this case, sex. To
state that the woman has at least one boy, and at least one fat
child, will have no impact; to state that the woman has a fatter
boy will change the nature of the beast. **

***************** **

**My comments on Tom's essay: **

**General remarks: While Tom is slightly better in
attempting to address my actual argument, I nonetheless find his
argument opaque, so that while there might be something to it, I
am unable to figure out what it is. **

**(note 1) Like Tony, Tom cannot seem to avoid the
temptation to complexify a situation which needs to be simplified
(eg, why mention ages?). He also seems to have made a mistake in
substituting "girl" for "boy" in referring to
the six-year old. **

**(note 2) I'm not sure what Tom is saying, but it does
not seem to properly represent anything I have said. **

**(note 3) I do not understand the basis for this
statement **

**(note 4) The latter statement is irrelevant to the
matter at hand as far as I can see. **

**Note (Jan 2001): Three years ago I burned out on this
particular subject. However, I am posting the final essasy, and
if there is some interest from the cybercommunity, I may actually
get back into it.**

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