Given that others seem to be using it to get the right answer, consider that you may rightfully believe SIA is wrong because you have a different interpretation of it, which happens to be wrong.
Huh? I haven't been using the SIA, I have been attacking it by deriving the right answer from general considerations (that is, P(tails) = 1/2 for the 1-shot case in the long-time-after limit) and noting that the SIA is inconsistent with it. The result of the SIA is well known - in this case, 0.01; I don't think anyone disputes that.
P(R|KS) = P(R|K)·P(S|RK)/P(S|K) = 0.01·(0.5)/(0.5) = 0.01
If you still think this is wrong, and you want to be prudent about the truth, try finding which term in the equation (1) is incorrect and which possible-observer count makes it so.
Dead men make no observations. The equation you gave is fine for before the killing (for guessing what color you will be if you survive), not for after (when the set of observers is no longer the same).
So, if you are after the killing, you can only be one of the living observers. This is an anthropic selection effect. If you want to simulate it using an outside 'observer' (who we will have to assume is not in the reference class; perhaps an unconscious computer), the equivalent would be interviewing the survivors.
The computer will interview all of the survivors. So in the 1-shot case, there is a 50% chance it asks the red door survivor, and a 50% chance it talks to the 99 blue door ones. They all get an interview because all survivors make observations and we want to make it an equivalent situation. So if you get interviewed, there is a 50% chance that you are the red door one, and a 50% chance you are one of the blue door ones.
Note that if the computer were to interview just one survivor at random in either case, then being interviewed would be strong evidence of being the red one, because if the 99 blue ones are the survivors you'd just have a 1 in 99 chance of being picked. P(red) > P(blue). This modified case shows the power of selection.
Of course, we can consider intermediate cases in which N of the blue survivors would be interviewed; then P(blue) approaches 50% as N approaches 99.
The analogous modified MWI case would be for it to interview both the red survivor and one of the blue ones; of course, each survivor has half the original measure. In this case, being interviewed would provide no evidence of being the red one, because now you'd have a 1% chance of being the red one and the same chance of being the blue interviewee. The MWI version (or equivalently, many runs of the experiment, which may be anywhere in the multiverse) negates the selection effect.
If you are having trouble following my explanations, maybe you'd prefer to see what Nick Bostrom has to say. This paper talks about the equivalent Sleeping Beauty problem. The main interesting part is near the end where he talks about his own take on it. He correctly deduces that the probability for the 1-shot case is 1/2, and for the many-shot case it approaches 1/3 (for the SB problem). I disagree with his 'hybrid model' but it is pretty easy to ignore that part for now.
Also of interest is this paper which correctly discusses the difference between single-world and MWI interpretations of QM in terms of anthropic selection effects.
I have been attacking it by deriving the right answer from general considerations (that is, P(tails) = 1/2 for the 1-shot case
Let me instead ask a simple question: would you actually bet like you're in a red room?
Suppose you were told the killing had happened (as in the right column of Cupholder's diagram, and required to guess the color of your room, with the following payoffs:
Guess red correctly -> you earn $1.50
Guess blue correctly -> you earn $1.00
Guess incorrectly -> you are terribly beaten.
Would you guess red? Knowing that under...
EDIT: This post has been superceeded by this one.
The doomsday argument, in its simplest form, claims that since 2/3 of all humans will be in the final 2/3 of all humans, we should conclude it is more likely we are in the final two thirds of all humans who’ve ever lived, than in the first third. In our current state of quasi-exponential population growth, this would mean that we are likely very close to the final end of humanity. The argument gets somewhat more sophisticated than that, but that's it in a nutshell.
There are many immediate rebuttals that spring to mind - there is something about the doomsday argument that brings out the certainty in most people that it must be wrong. But nearly all those supposed rebuttals are erroneous (see Nick Bostrom's book Anthropic Bias: Observation Selection Effects in Science and Philosophy). Essentially the only consistent low-level rebuttal to the doomsday argument is to use the self indication assumption (SIA).
The non-intuitive form of SIA simply says that since you exist, it is more likely that your universe contains many observers, rather than few; the more intuitive formulation is that you should consider yourself as a random observer drawn from the space of possible observers (weighted according to the probability of that observer existing).
Even in that form, it may seem counter-intuitive; but I came up with a series of small steps leading from a generally accepted result straight to the SIA. This clinched the argument for me. The starting point is:
A - A hundred people are created in a hundred rooms. Room 1 has a red door (on the outside), the outsides of all other doors are blue. You wake up in a room, fully aware of these facts; what probability should you put on being inside a room with a blue door?
Here, the probability is certainly 99%. But now consider the situation:
B - same as before, but an hour after you wake up, it is announced that a coin will be flipped, and if it comes up heads, the guy behind the red door will be killed, and if it comes up tails, everyone behind a blue door will be killed. A few minutes later, it is announced that whoever was to be killed has been killed. What are your odds of being blue-doored now?
There should be no difference from A; since your odds of dying are exactly fifty-fifty whether you are blue-doored or red-doored, your probability estimate should not change upon being updated. The further modifications are then:
C - same as B, except the coin is flipped before you are created (the killing still happens later).
D - same as C, except that you are only made aware of the rules of the set-up after the people to be killed have already been killed.
E - same as C, except the people to be killed are killed before awakening.
F - same as C, except the people to be killed are simply not created in the first place.
I see no justification for changing your odds as you move from A to F; but 99% odd of being blue-doored at F is precisely the SIA: you are saying that a universe with 99 people in it is 99 times more probable than a universe with a single person in it.
If you can't see any flaw in the chain either, then you can rest easy, knowing the human race is no more likely to vanish than objective factors indicate (ok, maybe you won't rest that easy, in fact...)
(Apologies if this post is preaching to the choir of flogged dead horses along well beaten tracks: I was unable to keep up with Less Wrong these past few months, so may be going over points already dealt with!)
EDIT: Corrected the language in the presentation of the SIA, after SilasBarta's comments.
EDIT2: There are some objections to the transfer from D to C. Thus I suggest sliding in C' and C'' between them; C' is the same as D, execpt those due to die have the situation explained to them before being killed; C'' is the same as C' except those due to die are told "you will be killed" before having the situation explained to them (and then being killed).