You meet a stranger on the street, and ask how many children he has.

Additional material

[ Editor: a 1996 rec.puzzles thread, "Equiprobability, oldest-girl,
gold/silver coins, and Monty Hall", followed by a reply from the
rec.puzzles oracle. ]

There is a fundamental conceptual and philosophical link between: (1)
the question of whether partitions or permutations are equally probable
in chance set-ups; (2) the oldest girl puzzle (i.e., what are the odds
that the second of two children is a girl, given that first [oldest or
not] is a girl; (3) the gold/silver coin problem (i.e., one bag
contains two silver coins, one bag contains two gold coins, and one bag
contains one gold and one silver coin; what are the odds the second
coin in a bag is gold, given that the first is gold); and (4) the Monty
Hall problem (which I assume needs no explication).  This link raises
several questions, which are the puzzles contained in this post.

In a recent thread on the first topic above, it was agreed that one
must learn by experience whether partitions or permutations are
equiprobable in dice.  [Partitions are insensitive to which die shows
which number; permutations take ordering into account.  Thus, there is
only one partition and one permutation for making 3 points using three
dice (1-1-1), whereas there is one partition for making four points
(1-1-2) but three permutations (1-1-2, 1-2-1, 2-1-1).]

The question of ordering is obviously relevant to the oldest girl
puzzle, because to arrive at the "correct" answer one must recall that
the single partition (boy-girl) can be split into two permutations
(boy-girl; girl-boy); whereas both boy-boy and girl-girl have only one
permutation each.  Thus, if one knows (only) that the oldest (this
really should be older) of two children is a girl; there is a 50-50
chance that the younger is also a girl.  Specifically, one is
indifferent between the girl-girl and girl-boy permutations, which are
the only two still in the running.  But if one knows only that at least
one of two children is a girl, only the boy-boy permutation is
eliminated, leaving both girl-girl, girl-boy, and boy-girl as
equiprobable contenders.  Thus in this case the chances are only one in
three that the second child is a girl.

On its face the gold/silver coin puzzle would seem essentially
identical to the oldest girl puzzle.  However, when this problem was
discussed at length in rec.puzzles a year ago or so, the consensus was
that given a gold coin selected at random from one of the the three
bags (also selected at random) the probability that the second coin in
that bag was gold is 2/3.  The best explanation I saw to explain this
ran something like this.  "Given a gold coin selected, there is now
zero likelihood that the selected bag was silver-silver.  But it is
also _relatively_ unlikely that the bag is gold-silver because in that
case a silver coin might just as well have been selected (but wasn't).
But if it is the gold-gold bag, the (randomly selected) coin _had_ to
be gold.  Thus it is twice as likely that a gold coin was selected from
the gold-gold bag as the gold-silver bag, and therefore the second coin
in the bag is twice as likely to gold as silver."

This reasoning is identical to the so-called "rule of restricted
choice" in cards, especially bridge.  If a player plays a card (say, a
jack) of equivalent value to one or more unplayed cards (say, a queen),
the odds are against him holding those unplayed cards, because if he
held them he might have played them.  If he didn't/doesn't hold them,
however, his choice was _restricted_ to the card he played.  The same
theme underlies the Monty Hall problem.  Speaking in the past tense,
Monty was twice as likely to have showed the door he did because he
_had_ to do so (the car being behind the door you didn't select and
that he therefore can't open given the rules) than that he _happened_
to select that door when he _could have selected_ another one equally
well.  So the probability that the car is behind the door he didn't
open (and that you didn't select) is twice that of the (a priori)
probability that it was behind the door he did open, namely 1/3, which
is the same as the [unchanged] probability that it is behind the door
you did select.  This is getting at the right answer to the problem
from a different perspective than the usual one, a very useful one, I
think.

Returning to the gold/silver coin problem, naturally it doesn't matter
what order the coins were but in the mixed-coin bag (i.e.,
permutations) for the above reasoning to hold true.  The question
immediately arises, therefore, why is ordering relevant in the
oldest-girl puzzle but not in the gold/silver coin puzzle?  Clearly if
we had three bags; one containing two boys, one containing two girls,
and one containing one of each, and a girl was randomly selected from
one of the bags, the probability that the other child in the bag is a
girl is 2/3 in view of the above argument.

Of course the reason for the apparent discrepancy has something to do
with how one comes to know about the one girl, exactly, and this
apparently influences whether it is partitions or permutations that are
equally probable.  This is seen by the fact that we now have three
different (personal) probabilities for the proposition that the second
of two children is a girl, given that one is a girl.  These are 1/2,
1/3, and 2/3, depending on whether we know, specifically and
respectively, that (1) the older child is a girl, (2) at least one is a
girl, and (3) a randomly selected child is a girl.  What has happened
(perhaps among other things) is that permutations (girl-boy vs
boy-girl) have ceased to matter once random selection is employed.  In
the random case it is partitions that are equally probable, whereas in
the other two set-ups it is partitions.  [ Editor: so in the original;
the second "partitions" was presumably meant to be "permutations". ]
Specifically, on the non-random set-ups girl-boy and boy-girl each hold
equal weight with girl-girl, whereas on the random set-up the total
probability of girl-boy/boy-girl is only half that of girl-girl (given
the evidence of one randomly selected girl).  Thus the girl-boy and
boy-girl permutations each appear to be worth on one-quarter of the
girl-girl permutation on this set-up.

What's going on here, exactly?  Why the difference in equiprobability
of partitions vs. permutations based on what (exactly) we know?  Are
the different probabilities (described above) derivable without
reference to experience?  It would seem so, perhaps, given the argument
based on restricted choice above and logical principles alone.  But if
this is true, what's different, exactly, about this chance set-up
versus dice?

Finally, consider the following Monty Hall variant.  Monty presents you
with three bags, containing pairs of gold or silver or mixed coins as
above.  The rules of the game are that you "randomly" pick a coin from
any bag, note its composition, and draw again from the same or a
different bag.  You win a car if the second coin has the same
composition as the first.  Between your two draws, Monty will pull out
and reveal one coin from each of the two bags you did'nt choose,
leaving one coin in each bag for you to choose among.  Monty will
pick/reveal a coin either randomly or with full knowledge of the
contents of all bags.  If he does know the composition, he will always
leave the situation ambiguous after he draws.  If he selects randomly,
sometimes he will give away the show by revealing a silver coin (say)
from Bags #2 and #3 after you have pulled a gold coin out of #1.

Assume then that you you are playing this game and have selected a
gold coin out of Bag #1.  Monty then reveals a gold coin from Bag #2
and a silver coin from Bag #3.  What is the winning strategy, and does
it matter what Monty knows?  [QUASI-SPOILER]  My (current) answer is
that nothing has changed in either case, and that one should always
draw the remaining coin from the originally selected bag.  If correct,
this answer differs from the classical Monty Hall problem, in which it
_does_ matter what Monty knows (as we've discussed ad nauseum
recently).  But why this difference, exactly?

Thanks for any help sorting all this out.  Again, I think there may be
something important going on in all of this with respect to
probability theory, statistical inference, and epistemology.  On the
other hand, maybe not.

-- David C. Hadorn <david.hadorn@vuw.ac.nz>

david.hadorn@vuw.ac.nz (David Hadorn) writes:
>In a recent thread on the first topic above, it was agreed that one
>must learn by experience whether partitions or permutations are
>equiprobable in dice.

Yes, but to qualify that agreement:  I don't believe the "experience"
for most of us consists of rolling multiple dice many times and
counting the occurrences of the partitions and permutations---it's very
hard for us even to identify the "permutations" of three dice rolled
simultaneously on a surface.

Rather, I think the experience we have is that macroscopic objects like
dice don't interfere with each other's state unless there are some
discernable forces acting between them, for example a mechanical
linkage, or magnets inside the objects.  So we reason that rolling N
identical dice simultaneously is equivalent to rolling the same die N
times---that is, the likelihood of any partition in the first case is
the same as in the second.  Alternatively, we may reason that painting
each of the N dice a different color---so that we can now tell the
"permutations" of the outcomes apart---does not affect the way any of
the dice rolls, so once again we should see the same partitions of
numbers rolled as we did when the dice were all the same color.

>The question of ordering is obviously relevant to the oldest girl
>puzzle, because to arrive at the `correct' answer one must recall that
>the single partition (boy-girl) can be split into two permutations
>(boy-girl; girl-boy);

And in fact those permutations are quite easy to identify in real life,
since in all known cases, *one* of the two children was born first, so
we may simply list that child's sex first.

But we have to justify why we assume a boy or girl is equally likely
for the first born, because there is no mechanical symmetry to justify
this conclusion.  In fact the assumption is only partially true: the
observed probabilities of male vs. female births are *not* precisely
equal, they are merely close enough that we can get away with
approximating them as 1/2.  And we have to justify why we assume the
first and second birth's sexes are independent.  We simply decide not
to consider identical twins (since they're relatively infrequent
anyway---if they accounted for most births, we'd have to change our
question or our analysis!), and we observe that the correlation between
the sexes of different children born to the same woman is so weak it
is extremely difficult to measure, so we decide to ignore it for the
purposes of our puzzle.

>This reasoning is identical to the so-called `rule of restricted
>choice' in cards, especially bridge.

It seems to me the rule of restricted choice is just a corollary of
Bayes' Theorem.  Moreover, its use in bridge and other games is based
on reasoning from experience (from which we conclude certain things
bout the distribution of hands of cards).

>Of course the reason for the apparent discrepancy has something to do
>with how one comes to know about the one girl, exactly, and this
>apparently influences whether it is partitions or permutations that
>are equally probable.

Not at all.  What makes the difference is quite simply that in one case
you chose the pair of children from a set in which boy-girl
permutations are known to be (at least approximately) equally likely,
and in the other you created a specially manipulated situation in which
partitions are equally likely instead.  It is not "how one comes to
know about the one girl" that makes the difference in this case, but
rather how one comes to choose a family from which to find out about a
child.

>This is seen by the fact that we now have
>three different (personal) probabilities for the proposition that the
>second of two children is a girl, given that one is a girl.  These
>are 1/2, 1/3, and 2/3, depending on whether we know, specifically and
>respectively, that (1) the older child is a girl, (2) at least one is
>a girl, and (3) a randomly selected child is a girl.

This is a very poor comparison, because the family in cases (1) or (2)
can be just any old two-child family in the general population, but to
get your answer for case (3) you must select the girl from one of your
carefully screened families.  This is comparing apples to oranges.

Let's compare apples to apples and oranges to oranges instead.  There
are two possible circumstances in which we may find ourselves:
  (A) choose any real two-child family at random, or
  (B) choose one from a selected set in which "partitions by sex"
      are equiprobable.
And we are considering three possible ways you could find out something
about the sex of the children:
  (1) you ask if the older child is a girl, and the answer is yes,
  (2) you ask if there are any girls, and the answer is yes, or
  (3) you choose a child randomly and find out it's a girl.

Here is the probability that both children are girls, given each of the
six possible setups:

                 Information about child
                  (1)     (2)     (3)
Family      (A)   1/2     1/3     1/2
selection   (B)   2/3(*)  1/2     2/3

(*) Case (B)(1) requires some extra assumptions about how you screened
your families to obtain equal probability of the partitions.  In
general, one of the children in any pair is older than the other
(especially if they actually are siblings, as we are given to assume in
the classic puzzle); but what was your selection criterion?  The answer
given (2/3) assumes your criterion did not either girl-eldest or
boy-eldest pairs over the other kind, but if you did favor one
permutation over the other, you can make the probability for question
(B)(1) be anywhere from 0 to 1.

>What has
>happened (perhaps among other things) is that permutations (girl-boy
>vs boy-girl) have ceased to matter once random selection is employed.
>In the random case it is partitions that are equally probable,

This is simply not correct.  Whether partitions or permutations are
equally likely has nothing to do with how you select the child, it has
everything to do with how you select the family.

>Finally, consider the following Monty Hall variant.  [...]
>Assume then that you you are playing this game and have selected a
>gold coin out of Bag #1.  Monty then reveals a gold coin from Bag #2
>and a silver coin from Bag #3.  What is the winning strategy, and does
>it matter what Monty knows?  [QUASI-SPOILER]  My (current) answer is
>that nothing has changed in either case, and that one should always
>draw the remaining coin from the originally selected bag.  If correct,
>this answer differs from the classical Monty Hall problem, in which it
>_does_ matter what Monty knows [...]

But the answer is not entirely correct.  Yes, bag #1 is still the most
likely place to find another gold coin, but if you know Monty really
chose the two coins at random, you should assign only 1/2 probability
to the proposition that the other coin in bag #1 is gold, rather than
2/3 probability, and in fact the remaining gold coin is just as likely
(probability 1/2) to be in bag #2 as in bag #1.

In fact the correct answer is exactly the *same* as the classical Monty
in important respects: if Monty always makes an ambiguous revelation,
you can assign 2/3 probability to one of your remaining options, but if
Monty's revelation is made at random, you wind up with two equally
likely options.  (In the bag game, one move, choosing the remaining
coin from bag #3, can logically be eliminated no matter which of the
two behaviors Monty follows.)

To me this sameness is far more surprising than your claimed difference
would have been, considering how radically you've changed the play of
the game.

-- David A. Karr <karr@cs.cornell.edu>

Just realized that the Monty Hall variant I proposed at the end of my
previous post was not set up correctly.  Arrgh!  This is confusing
stuff!  (At least for me.)  I was trying to construct a MH variant that
incorporates the flavor and lessons of the oldest-girl and gold/silver
coin puzzles, especially with respect to partitions and permutations.

Another (tentative) try: say that Monty pulls a coin out of Bag #2 or
Bag #3 after you've selected a coin from Bag #1.  Does this new
"evidence" change the probability that the other coin in Bag #1 is the
same composition as the first?  Clearly the answer is "no" if Monty
knows the three bags' contents (and acts according to standard rules)
and "yes" if Monty selects his coin at random.  Moreover, it seems to
me that in the latter case the correct solution might again involve
permutations, rather than partitions as in the standard gold/silver
coin problem.  I'm going to work on this again when I can; I just
wanted to get this erratum out quickly.

-- David C. Hadorn <david.hadorn@vuw.ac.nz>

karr@cs.cornell.edu (David Karr) wrote:

[Discussing whether experience is required to discern the probabilities
for different total points at dice.]

>Rather, I think the experience we have is that macroscopic objects
>like dice don't interfere with each other's state unless there are
>some discernable forces acting between them, for example a mechanical
>linkage, or magnets inside the objects.  So we reason that rolling N
>identical dice simultaneously is equivalent to rolling the same die N
>times---

I think it goes deeper than this.  In an earlier quote from Hacking's
book "The Emergence of Probability" he related how Leibnitz thought
that partitions were equiprobable in dice, whereas Galileo realized
that in fact permutations are equiprobable.  Surely Leibnitz knew that
dice throws are independent events!  Hacking's point is that _even
knowing_ this fact, experience was required to distinguish between
partitions and permutations as equiprobable.  Hacking:

"It seems natural to call this our first witness of the refutation of
a statistical hypothesis by 'long observation.'  We have the hypothesis
that permutations are equally probable versus the hypothesis that
partitions are equally probable; the latter is inconsistent with the
facts, while the former fits the facts perfectly. . . ."

Now compare the situation with respect to the eldest girl and
gold/silver coin puzzles.  It would appear that we can solve both of
these puzzles by reference only to "armchair" calculations of available
permutations (again, assuming independence of child gender in the
former puzzle).  Why can we do this, if we cannot figure out the dice
problem independent of experience?  And, to return to another issue
that interests me, granted that we can solve the two "simple" puzzles a
priori, does this mean that we then know something meaningful about how
the world works a priori?  Even if this knowledge were conditional on
the [empirically verifiable] assumption of independence of events, this
would still appear to be empirically significant.

>Whether partitions or permutations are equally likely has nothing
>to do with how you select the child, it has everything to do with
>how you select the family.

Yes, of course.  It's obvious when you lay it out like that.  Another
way to look at the situation is that in order to be comparable to the
oldest girl problem the gold/silver coin problem would have to have
_four bags_, not three, with two bags containing one gold/one silver
coin, not just one.  This parallels the fact that of all actual (or
potential) two-child families it _turns out_ that there are twice as
many mixed-sex as single-sex pairs.  This base rate is really what
matters, perhaps, not the number of permutations per se.  But again, it
seems that all of this is knowable a priori, but is it really?  And if
it is, again it would seem to pertain to how the world works.

What's funny about the oldest-girl puzzle is that the initial 1/2
probability that any _individual_ child is a girl appears to drop to
1/3 with the knowledge that at least one child of a pair of children is
a girl.  This seemed counterintuitive (at least to me) because it
seemed to entail the following three conditions:

1. P(both are girls | oldest is a girl) = 0.5
2. P(both are girls | youngest is a girl) = 0.5
3. P(oldest is a girl or youngest is a girl [or both]) = 1.0
   (Stipulated in the "at-least-one" scenario)

These three sentences seem to imply that if we know that (at least)
one child is a girl, the probability that "the other" child is a girl
would still be 0.5, just like it was before we were given the
information.  But I guess what is happening is that the phrase "the
other child" in the previous sentence is misleading, as it really
refers to both children in the pair, as does the "at least one child"
term.  And of the three possible permutations consistent with this
ambiguous reference, two contain boys.  But here again, the question
can be raised, why can we figure this out a priori, but not the dice
thing?

-- David C. Hadorn <david.hadorn@vuw.ac.nz>

david.hadorn@vuw.ac.nz (David Hadorn) writes:
>I think it goes deeper than this.  In an earlier quote from Hacking's
>book "The Emergence of Probability" he related how Leibnitz thought
>that partitions were equiprobable in dice, whereas Galileo realized
>that in fact permutations are equiprobable.  Surely Leibnitz knew
>that dice throws are independent events!

One certainly might expect that Leibniz would have admitted to this
knowledge---*if* you had put it to him this way.  But knowing a thing
and using that knowledge in one's calculations are two entirely
different things.  Just look at all the professional mathematicians who
surely know Bayes' Theorem---at least if specifically asked about
it---and yet fail to apply it in the classical Monty Hall puzzle and
therefore give an incorrect answer.

>Hacking:
>
>"It seems natural to call this our first witness of the refutation of
>a statistical hypothesis by 'long observation.'

When did this refutation occur, and on the basis of what observations
was the hypothesis refuted?  Without historical evidence, this
statement is no more than an "armchair" speculation by Hacking.  (I'm
not saying it didn't happen, just that there is no justification for
saying it *did* happen unless specific historical evidence can be
presented.)

>It would appear that we can solve both of these puzzles by reference
>only to "armchair" calculations of available permutations (again,
>assuming independence of child gender in the former puzzle).  Why can
>we do this, if we cannot figure out the dice problem independent of
>experience?

Ah, but we *can* figure out the dice problem independent of experience
(in the sense you intend here).  Galileo appears to have done it, and
he did his work *before* Leibniz came up with the incorrect answer.

Galileo is famous for being much better than his contemporaries at
constructing just the right thought experiments.  For example, he
"proved" that objects fall at a speed independent of their weight by
constructing a thought experiment in which two objects were linked
together before being dropped.  Did his contemporaries think that
linking two cannonballs with a length of chain would make them fall
faster?  Surely not, but apparently nobody had ever thought of applying
this concept to the "heavier falls faster" claim.

The history of mathematics, especially early mathematics, is full of
cases where people overlooked things early on that turned out later to
be important.  And sometimes, as with Galileo and Leibniz, someone
worked out the correct answer early on and then someone did it
incorrectly shortly afterward.  So why did the mistake happen with the
dice problem and not with the boy/girl problem?  Because Galileo and
Leibniz were more interested in dice than in the sex of children, I
suppose.  There are two obvious reasons why this should be so:

  1) People bet on the rolling of dice much more often than on the
     sex of children.
  2) It's a lot easier to convince oneself that the different faces
     of one die really are equiprobable (by symmetry) than that
     boys and girls are really equiprobable (after all, they
     aren't really!).

So the difference between the dice and the boy/girl problem is not that
the dice are harder to work out without extensive and detailed
observation.  In fact the dice are *easier* to figure out without such
observation, and that's part of the reason why they were addressed
earlier---at a time when the best ways to do this reasoning weren't as
widely known and people like Leibniz were more likely to make mistakes
about it.

>  And, to return
>to another issue that interests me, granted that we can solve the two
>"simple" puzzles a priori, does this mean that we then know something
>meaningful about how the world works a priori?  Even if this knowledge
>were conditional on the [empirically verifiable] assumption of
>independence of events, this would still appear to be empirically
>significant.

Why?  If we start with an empirically established notion of
independence, and it turns out logically to imply some other new
notion, what's the significance (other than that logic and mathematics
are not wholly worthless tools)?  Suppose I meet someone and notice
he's missing his left index finger but has all his other fingers, so I
conclude he has nine fingers without counting them.  There, I've just
reached a conclusion about the world without empirically verifying it
(e.g. by counting his fingers 1-2-3-4-5-6-7-8-9 the way a 3-year-old
might have to).  We do this sort of thing all the time.  What's the big
deal?

>Another way to look at the situation is that in order to be
>comparable to the oldest girl problem the gold/silver coin problem
>would have to have _four bags_, not three, [...]
>But again, it seems that all of this is knowable a priori, but is it
>really?  And if it is, again it would seem to pertain to how the world
>works.

Of course it does, but we do not really know it a priori.  Based on
observation, we think we know something about how babies are made, and
we think we have a pretty good idea of what the chances are for a girl
and how they correlate between successive pregnancies.  In fact the
notions we use in the "armchair" calculation are somewhat naive and
inaccurate, but not enough to merit discarding the calculation
entirely.  (It would be an entirely different matter if we had assumed
probability 1/2 of any given child being a girl and found out later the
probability was only 1/3.)

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: in January 1997 Peter Amundson <petera@scn.org> wrote to the
rec.puzzles oracle that his friends argued: "by saying 'I have two children; at least one is a
girl; what is the probability that the other is a girl?', you're really
asking 'What is the probability that both are girls?', since one is
already given to be a girl.  Therefore, since we know one is a girl (we
don't care whether she's older or younger), the question further
distills to 'What is the probability that the one remaining child is a
girl?', which is 1/2."  The reply: ]

Well, the enumeration of cases analysis is really the most convincing
ANALYSIS that I know of.  There are other analyses (e.g., Bayesian) but
they really amount to the same thing and are very unconvincing.
However, there is another kind of ARGUMENT that cannot be put in a FAQ,
but which you can use to convince your friend.  All you need is two
coins.  Flip both coins, and if either comes up heads, say to your
friend:

"At least one of my coins is a head."

Ask him to bet you that both coins are heads.  If he believes the odds
of this are 50-50, he should be willing to bet you at even odds.  since
the true odds are 1 in 3, after a while he'll get tired of this game,
he'll learn a valuable lesson in probability, and you'll have been paid
for delivering the lesson!

-- Chris Cole <chris@questrel.com>