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>
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