I haven't received a single word of feedback on my comments to Marilyn
w.r.t. the litter of four puppies. Damn, I'm almost up to p. 100 in my
probability book. I never got that far before. Naturally, I am giddy
from the experience of possessing sooo much knowledge.
I could have made my points more succinctly: A *probability system* is
a triple of (i) the *sample space* S of all possible *outcomes* r_i , i =
1,2,3, ... , (ii) the *event class* E , the class of *events* A_j , j
= 1,2,3, ... such that the event A_k belongs to E provided that it is a
subset of S that satisfies a few non-demanding conditions (at least in
the finite case and which we can assume are satisfied by all the *events*
in this discussion, which I checked cursorily), and (iii) P, a
*probability measure* that satisfies normal requirements plus *countable
additivity*, which is fairly easy to verify too. So, a probability
system
X = {S,E,P} .
Now, Marilyn asks, given that your dog has alitter of four, "Is it most
likely that two are males and two are females?" There are 16 possible
outcomes; so, representing males by ones and females by zeros, I can
write S = {1111, 1110, 1101, 1100, 1011, 1010, 1001, 1000, 0111, 0110,
0101, 0100, 0011, 0010, 0001, 0000} . Now, there is nothing about
*simple events* and *complex events* in this theory, but it is not hard
to suppose that subsets formed by direct enumeration of outcomes are
simple events and events formed by taking unions, intersections,
complements, etc. are complex. Marilyn invites us to do this by offering
as a candidate the event A_1 formed by simply enumerating
every outcomes with two males and two females. A_1 = {1100, 1010, 1001,
0110, 0101, 0011} with six outcomes out of a possible (equally likely)
sixteen outcomes. So, P(A_1) = 3/8. There are two next largest simple
events, namely, A_2 is precisely three females: A_2 = {1000, 0100, 0010,
0001} and A_3 = {1110, 1101, 1011, 0111}. P(A_2) = P(A_3) = 1/4 . The
unwary player will not consider complex events like
A_2 UNION A_3. This is the event that precisely three puppies have the
same gender. This is Marilyn's solution. But all kinds of unions and
intersections can be formed and they all belong to E, which makes them
eligible to be "most likely" - unless the word "split", which is not used
in the problem statement, can be introduced into the theory without a
definition. I don't consider three females and one male the same
"split" as three males and one female; therefore, I shall assume that not
only is every split an event but *every event is a split*. I am back to
a term actually defined in the theory. I hope that Marilyn and others
find this acceptable.
Let A_4 be the event that at least two are female. Let A_5 be the event
that precisely two are female, A_6 precisely three, and P_7 precisely 4.
So, A_4 = UNION { A_k | k = 5,6,7}. This is an event. Is it a split?
I have defined it to be. Clearly, A_k, k = 5, 6, and 7 are splits by any
standard. Is the union of a finite number of splits a split? The theory
says nothing about that because split is undefined. But, three males
and one female should be a split and likewise three females and one male.
But, three of one gender and one of the other gender? Gender wasn't
even defined to be male or female or as anything really.
Anyway the split A_4 = {1100, 1010, 1001, 0110, 0101, 0011, 1000, 0100,
0010, 0001, 0000} has P(A_4) = 11/16.
But, I must find the most likely split. Well, if precisely three of
either gender is a split; that is, either three females or three males
satisfies the condition: then P_8 , precisely four puppies each of which
is of either gender, should be a split. P_8 = S and P (S) = 1 by one of
the rules for probability systems. This, as well as many other events
with probability one, is tie for most likely, e.g., P_9 is the event such
that at least one puppy is either male or female. This is a very old
joke. It precedes Charlemagne. It seems to me that indifference with
respect to three puppies is no better than indifference with respect to
one puppy. Marilyn will have trouble defining a split in such a way that
it cannot be generalized, for example, to one female, two males, and one
of either gender.
Alright, what do you think? I shall send this to Marilyn as a mo'
better argument. But, I have a tendency to tinker, which should not be
rigorously suppressed, but rather subjected to a little self-restraint.
Goodnight / The Amateur
P.S. Maybe the server is broken. Could we have some feed back from
sys-op or whatever he or she is called?