From: Lee Daniel Crocker (lee@piclab.com)
Date: Mon Aug 24 1998 - 16:12:04 MDT
One more try at analyzing the Carter-Leslie argument. Knock
it down if you can.
First, let us clarify what is meant by "mathematical probability",
which is that quantity to which the calculus of probability (including
Bayes' Theorem) applies. "Probability" is a property of propositions.
Further, the property exists for any proposition only with respect to
a set of propositions to which it belongs, exactly one of which can be
true at a given sampling time. The probability of that proposition,
then, is a measure of the relative frequency at which it is true at
sampling time compared to the frequency at which the others in the set
are true (the sum of all of which must be 1). For probability
calculations to be meaningful, then, we must be clear about what our
universe of propositions is, and keep it consistent throughout our
calculations.
Let's start with the usual example from the DA: two urns, one with
10 numbered balls and one with 100. A ball is drawn from one of the
urns and we find it numbered 7 (any value from 1 to 10 will do). We
are then asked to estimate what is the probability that it was drawn
from the urn with 10 balls rather than the urn with 100? What is our
set of propositions? There are three sets here: one set is {the ball
was drawn from urn1, the ball was drawn from urn2}. From each urn,
there is the set {ball 1 was drawn, ball 2 was drawn ...}. We must
assign to each of these propositions an initial probability by some
empirical means or by fiat before we can use the calculus to further
study their implications. The traditional method of doing this is
the principle of indifference: if we have no reason to suspect that
any proposition is more probable than another, then we assign them
equal values. In this case, then, we will assume that a fair coin was
flipped to choose which urn to draw from, giving an initial probability
of 1/2 to each urn selection. Likewise, we assume a ball is chosen
fairly from each urn. Given those ideal assumptions, Bayes' Theorem
gives us a probability of 10/11 our ball was chosen from urn 1.
Bayes' Theorem, to review quickly, is the method for calculating the
probability of an hypothesis given a piece of evidence [P(H|E)]
from our knowledge of the probability of the hypothesis itself [P(H)]
and the probability of the evidence given the hypothesis [P(E|H)].
The formula is P(H|E} = (P(H)P(E|H))/(P(H)P(E|H)+P(~H)P(E|~H)).
It is important to note that those initial probabilities we assigned
to P(H) are critical: what if the urn was not chosen fairly? What if
we are only offered this bet in those cases when a 10 or less is drawn?
Let's change the method of draw, for example. Let's say that all 110
balls are first placed together in a bag: one person reaches into the
bag and randomly places a beeping device on one of them. They are then
sorted into the urns, and the beeping device is activated. The beeping
ball is retrieved, and discovered to be numbered 5. What, now, is the
probability that it was drawn from urn 1? In this case, 1/2. By this
method of selection, the probabilities of {ball was drawn from urn 1,
ball was drawn from urn 2} are not 1/2 - 1/2, but 1/11 - 10/11. Bayes'
Theorem then gives the posterior probability as (1/11) / (1/11 +
10/11 * 1/10), or 1/2 (which we could also have determined without
Bayes in this case, but let's be consistent).
In propositions about ourselves, such as the DA, it is similarly of
critical importance to determine _how we came to be selected to make
the bet in question_. From which urn were we drawn, and by what
method of selection? In the example above, if we only offer the bet
in those cases where a 10 or less is drawn, the correct odds to take
are 1/2 - 1/2, not 10/11 - 1/11. Think of the old stock scam: you
receive a letter in the mail predicting a rise in a certain stock,
and it does indeed rise. The next week you receive another letter
predicting a fall, and it falls. For 5 weeks in a row, you receive a
letter with a prediction about some stock, and 5 times it's correct.
The sixth week the letter asks you to pay for its prediction. How
much is the prediction worth? A naive Bayesian would calculate the
chance of another correct prediction at 32/33. A wiser person would
realize that the scammer started out by sending 320 letters with a
"rise" prediction and 320 with "fall", then to those who got the
correct one, he sent 160 "rise" and 160 "fall". To those who were
right on round 2, he sent 80 "rise" and 80 "fall". You were simply
among those (un)lucky 10 who got 5 correct predictions in a row.
Your correct odds are 1/2 which is worth nothing, not 32/33.
These, then, are the questions to be answered if we are to attribute
meaning to the DA: (1) what is the set of propositions from which
our hypothesis "the world is ending soon" is drawn, and what are
their initial probabilities? and (2) how did we come to be selected
to make this bet? The original DA seems to want us to take as our set
of propositions {the world will end soon, there will be billions more
people before the world ends}. This, on its face, is absurd. There
is no reason at all to suspect that those two choices represent the
only possibilities--and of equal probability--in reality. Bayes'
Theorem requires that P(H) + P(~H) = 1, i.e., that there are no other
options. In that extremely contrived case, and further given the
appropriate self-sampling assumption (that I'll deal with below), the
DA would indeed be spectacularly true, but this is not the least
bit remarkable, because we have assigned by fiat absurd initial
probabilities contrived to produce just such a result. A realistic
set of propositions would be something more like {there will be fewer
than 100 billion persons, there will be 100-200 billion persons, there
will be 200-300 billion persons, ...} up to a suitable upper bound
chosen empirically based on the size of the planet or universe.
Without changing the results, this is equivalent to a set of urns with
{1 ball, 2 balls, 3 balls, ... N balls}, where N is the multiple of
100 billion persons we choose as our upper limit. If we then assign
equal initial probabilities to each of those urns (i.e., possible
civilizations), select an urn randomly, draw a ball (a birth order)
and find it to be "1" (less than 100 billion), then our Bayesian
probability that we were drawn from urn 1 (i.e., that there will only
ever be 100 billion persons) depends on our choice of N. For N=2,
those odds ar 2/3. For N=4, they are 1/2. As N approaches infinity,
the odds approach 0, albeit very slowly. For N=1 billion, the odds
that the world will end soon are much better than 1 in a billion, but
hardly anywhere near 1.
Finally, we must deal with the issue of our self-selection. Have we
really chosen an urn randomly, or have we fallen for a sucker bet?
It might be a more satisfying argument if we had a bigger sample of
possible civilizations around--aliens perhaps, or Atlanteans, or a
peek at earth's long-distant future, to give us an idea of what sizes
we can really expect a civilization sufficiently similar to ours to
reach before collapse. Let's say we had such evidence--that advanced
civilizations really seemed to grow to a random size from 100 billion
to 100 billion billion, and then collapse. Would the fact that we
find ourselves numbering less than 100 billion truly give us reason
to suspect we are among the unfortunate slated to perish soon? That
depends upon whether or not the question is being asked by someone
who _fairly chose our civilization to pick on_ by offering this bet.
In other words, is our civilization a truly random sample from all
those we postulate, and is this philosopher a fair sample of persons
from this civilization? Frankly, I don't see how either can be. We
chose to deal with our civilization in the question because it is the
one we are most interested in, not because we chose fairly. We have
no more reason to put that P(H) into the equation than does any other
alien civilization.
Let's change the bet this way, to make it correctly mirror our balls
and urns: assume there are many alien civilizations, and we know
precisely when each of them will end, at somewhere between 100 billion
and 100 billion billion. We make a big wheel with the names of those
civilizations on it and give it a spin. Now, knowing how long that
fairly-chosen civilization will last, we choose a ball (birth order)
from that total number, and offer to someone this bet: "I chose a
random civilization, and then a random birth order from that, and it
is X. What are the odds that this civilization will end soon?" If
all of these conditions are perfectly fair, i.e., we really chose a
random civilization, and we really chose a random birth order from it
knowing how long it would last, and we were shown that it is less than
100 billion, then we would be correct to infer that the odds of this
being one of the shorter-lived civilizations is much better than 1 in
a billion by the Bayesian calculation suggested by the DA. But I don't
think it is possible for us to make the bet fairly in this case. For
one thing, if we chose fairly, odds are good that we would choose a
birth order representing someone who has not been born yet, but we are
selecting ourselves from among the subset of those who have been born,
and can thus contemplate this question. And we really don't know what
our sample set is. We do not know what all the propositions are from
which we are to calculate this probability, and there is great reason
to suspect that we are biased. In short, I think this is a sucker bet
perpetrated on ourselves.
-- Lee Daniel Crocker <lee@piclab.com> <http://www.piclab.com> "All inventions or works of authorship original to me, herein and past, are placed irrevocably in the public domain, and may be used for any purpose without permission, attribution, or notification."
This archive was generated by hypermail 2.1.5 : Fri Nov 01 2002 - 14:49:30 MST