At Least One Girl

The probability of rolling a 7 with a pair of ordinary dice is 
1/6, because of the 36 possible (and presumably equally likely) 
combinations, six of them yield the sum 7.  Now, suppose the dice 
are rolled out of our sight, and some honest person who can see
the results tells us that at least one of the die came up 6.  This
restricts the total number of possible combinations to the following
eleven pairs

 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (6,1) (6,2) (6,3) (6,4) (6,5)

of which exactly two yield a sum of 7, so we would now assess the
probability of a 7 as 2/11, just slightly better than 1/6.  Of course,
the same analysis would give a probability of 2/11 if our honest 
friend had reported "at least one 5" instead of "at least one 6".
In fact, we would arrive at the same probability if he reported
"at least one n" for ANY value of n.  This sometimes strikes people
as paradoxical, because obviously we have "at least one n" for SOME
value of n, so why not just assume this without waiting for our 
friend to tell us?  It wold seem that this magically improves our
a_priori odds of rolling a seven from 1/6 to 2/11.  

This is the kind of connundrum that often puzzles beginning students 
of probability.  It's actually a 6x6 version of the classical 2x2 
problem that's usually expressed in terms of boy/girl children.  If we 
regard a man's two children as a pair of 2-sided dice, each of which 
is equally likely to come up as either a "1" (boy) or a "2" (girl), 
then what is the probability of "total = 3"?  Here the obvious answer 
is 1/2 (just as the probability of rolling a 7 with two 6-sided dice 
is 1/6).

Now we alter the problem by having the father provide us with one of 
two specific pieces of (factual) information.  He either tells us "At 
least one is a boy" or he tells us "At least one is a girl".  Given one 
of these pieces of information, what is the probability of "total = 3"?

Strictly speaking, the problem is underspecified in this form, because 
we don't know how the father decides which piece of information to 
give us if he has a choice (i.e., if he has one child of each gender).
For example, suppose the father tells us "At least one girl" whenever 
possible, and only tells us "At least one boy" if he has no choice.  
Then obviously if he tells us "At least one boy" we know the 
probability of "total = 3" is zero.

The key point here is the difference between

     Condition 1:  There is at least one boy.
     Condition 2:  There is at least one girl.
and
     Condition 3:  We are told that there is at least one boy.
     Condition 4:  We are told that there is at least one girl.

When assessing the conditional probability of "total=3" we are not 
given C1 or C2; we are given C3 or C4.  C1 and C2 are each true 
exactly 3/4 of the time, and half the time they are both true 
simultaneously.  In contrast, C3 and C4 are mutually exclusive and 
complementary conditions, but we don't really know what fraction 
of the time either of them is true.  Therefore, the problem is 
indeterminate.

However, by the "unwritten rule of probability riddles" (the UROPR), 
let's assume that all specified choices are to be made randomly from 
the available options.  Thus we assume that if the man has one child 
of each gender he is equally likely to report "At least one boy" as 
to report "At least one girl".  It follows that each of these reports 
has a probability of 1/2.  For clarity, let's denote C1 as [boys>=1], 
and C3 as [told:boys>=1].  On this basis we can use Bayes' formula 
to compute the conditional probability as follows

                                   Pr{ [told:boys>=1] AND total=3 }
 Pr{ total=3 | [told:boys>=1] } =  ---------------------------------
                                            Pr{ told:boys>=1 }

The denominator is known to be 1/2, and the numerator is 1/4 because 
the total equals 3 exactly half the time, and (by assumption) the 
father reports "At least one boy" exactly half of those times.  Thus 
the conditional probability of "total=3" given that the father tells 
us "At least one boy" is exactly 1/2.

Similarly in the 6x6 case we know that  Pr{total=7} = 1/6, and by the
UROPR we have

             Pr{ [told:(n's)>=1] AND total=7} = 1/36       

for any n from 1 to 6, so Bayes formula gives the conditional 
probability

                                              1/36
       Pr{ total=7 | [told:(n's)>=1] }   =   -------  =  1/6 
                                               1/6

for any n from 1 to 6.  This in no way conflicts with the fact that

              Pr{total=7 | [(n's)>=1] }    =   2/11

because the conditions are different.  Let me try to express this 
distinction verbally for the 2x2 case.

Suppose there are 100 fathers in an auditorium, and each is the father 
of two children.  Each father is instructed to tell you (truthfully) 
if at least one of his children is a boy.  This will apply to about 
75 of the fathers.  Now, of those 75 Dads, 2/3 (i.e., 50) have a 
daughter, and 1/3 (i.e., 25) have two sons.  Thus, if you want to 
guess the gender of their "other" child, the chances are 2/3 that 
it is a girl.  (Of course, for the remaining 25 fathers - those 
who did not report at least one son - you know immediately they 
have two daughters.)

However, suppose instead that all 100 fathers were instructed to tell 
you either (a) "At least one of my children is a boy" or (b) "At least 
one of my children is a girl".  Based on what each father tells you, 
you try to guess the gender of his "other" child.  Strictly speaking 
this problem is indeterminate, but if it's also stipulated that fathers 
with both a son and a daughter should flip a coin to decide what to 
tell you, then the probability that the "other" child is of the 
opposite gender is exactly 1/2.  The breakdown is

    25 have two sons, and they report at least one son
    25 have a son and daughter, and report at least one son

    25 have a son and daughter, and report at least one daughter
    25 have two daughters, and they report at least one daughter

Thus, regardless of what a particular father reports, you have only a
50% chance of correctly guessing the gender of his "other" child.