> On 25 Feb 2020, at 12:43, Bruce Kellett <[email protected]> wrote: > > On Tue, Feb 25, 2020 at 10:26 PM Bruno Marchal <[email protected] > <mailto:[email protected]>> wrote: > On 24 Feb 2020, at 23:22, Bruce Kellett <[email protected] > <mailto:[email protected]>> wrote: >> On Tue, Feb 25, 2020 at 12:10 AM Bruno Marchal <[email protected] >> <mailto:[email protected]>> wrote: >> On 23 Feb 2020, at 23:49, Bruce Kellett <[email protected] >> <mailto:[email protected]>> wrote: >>> On Mon, Feb 24, 2020 at 12:21 AM Bruno Marchal <[email protected] >>> <mailto:[email protected]>> wrote: >>> On 23 Feb 2020, at 04:11, Bruce Kellett <[email protected] >>> <mailto:[email protected]>> wrote: >>>> >>>> I don't really understand your comment. I was thinking of Bruno's >>>> WM-duplication. You could impose the idea that each duplication at each >>>> branch point on every branch is an independent Bernoulli trial with p = >>>> 0.5 on this (success being defined arbitrarily as W or M). Then, if these >>>> probabilities carry over from trial to trial, you end up with every binary >>>> sequence, each with weight 1/2^N. Summing sequences with the same number >>>> of 0s and 1s, you get the Pascal Triangle distribution that Bruno wants. >>>> >>>> The trouble is that such a procedure is entirely arbitrary. The only >>>> probability that one could objectively assign to say, W, on each Bernoulli >>>> trial is one, >>> >>> That is certainly wrong. If you are correct, then P(W) = 1 is written in >>> the personal diary, >>> >>> I did say "objectively assign". In other words, this was a 3p comment. You >>> confuse 1p with 3p yet again. >> >> Well, if you “objectively” assign P(W) = 1, the guy in M will subjectively >> refute that prediction, and as the question was about the subjective >> accessible experience, he objectively, and predictably, refute your >> statement. >> >> >> And if you objectively assign p(W) = p(M) = 0.5, then with the W-guy and the >> M-guy will both say that your theory is refuted, since they both see only >> one city: W-guy, W with p = 1.0, and the M-guy, M with p =1.0.. > > That is *very* weird. That works for the coin tossing experience too, even > for the lottery. I predicted that I have 1/10^6 to win the lottery, but I was > wrong, after the gale was played I won, so the probability was one! > > In Helsinki, the guy write P(W) = P(M) = 1/2. That means he does not yet know > what outcome he will feel to live. Once the experience is done, one copy will > see W, and that is coherent with his prediction, same for the others. He > would have written P(W) = 1, that would have been felt as refuted by the M > guy, and vice-versa. > > But if he wrote p(W) = 0.9 and p(M) = 0.1 he would get exactly the same > result. The proposed probabilities are here without effect.
If I toss a perfect coin too. Of course, that would lead directly to some problem with the iterated case scenario. >> If not, tell me what is your prediction in Helsinki again, by keeping in >> mind that it concerns your future subjective experience only. >> >> >> In Helsinki I can offer no value for the probability since, given the >> protocol, I know that all probabilities will be realized on repetitions of >> the duplication. > > In the 3p picture. Indeed, that is, by definition, the protocol. But the > question is not about where you will live after the experience (we know that > it will be in both cities), but what do you expect to live from the first > person perspective, and here P(W & M) is null, as nobody will ever *feel to > live* in both city at once with this protocole. > > And, as I have repeated shown, the first person perspective does not give you > any expectations at all. If I am duplicated like in the 2^(16180 * 10000) * (60 * 90) * 24 “movie” scenario, I do expect seeing white noise, and I certainly don’t expect to see “2001, Space Odyssey” with Tibetan subtitle. I am not sure what you mean by “the first person perspective does not give any expectations”. Do you agree that if you are promised, in Helsinki, that a cup of coffee will be offered to you, both in M and W, you can expect, with probability one, to get a cup of coffee after pushing the button in Helsinki? (Assuming Mechanism, of course). I would expect, in Helsinki, to drink a cup of coffee with probability one (using this protocole and all default hypotheses, like no asteroids hurt the planet in the meantime, etc.). And I would consider myself maximally ignorant if that coffee will be Russian or American coffee. > > The experience is totally symmetrical in the 3p picture, but that symmetry is > broken from the 1p perspective of each copy. One will say “I feel to be in W, > and not in M” and the other will say “I feel to be in M and not in W”. > > Regardless of any prior probability assignment. Exactly. > > >> I cannot infer a probability from just one trial, but the probability I >> infer from N repetitions can be any value in [0,1]. > > But we try to find the probability from the theory. > > And we use the experimental data to test the theory. If you predict p(W) = > p(M) =0.5, after a large number of duplications that prediction will be > refuted by the majority of the copies. In fact, in the limit, only a set of > measure zero will obtain p = 0.5 from their data. Then that is true for the iterated coin tossing too, and there is no probabilities at all. > > As I illustrated with the WMS triplication, unknown to the candidate, we see > that we cannot infer any probabilities, from experiences alone. > > What the WMS example shows is that if you guess the wrong theory, you will > get the wrong answer. Yes, and that shows that the fact the guy in Helsinki knows the true protocol is important, to derive the theoretical first person indeterminacy. If you have a problem with the specific answer P = 1/2, keep in mind that this is not use in the reduction of the mind-body problem the derivation of matter appearance from a statistics on all computation (“all computations” being defined in arithmetic). What is used is only the first person indeterminacy (and some variants) and the fact that the means to calculate the probabilities, or the credibilities, or the plausibilities is invariant for the changes made at each step of the Universal Dovetailer Argument. (P = 1/2 is used just to fix the idea, and also because most people find this to be the natural easiest solution with this “simple” protocol). > > Keep in mind that we *postulate* Mechanism. We work precisely in the frame of > that theory/hypothesis. > > > You might do so. But I do not. I am working with the protocols and data as > they are generated. You seem to use also the assumption that there is a physical reality, even a unique one. That is the point which does no more work when we assume digital mechanism, or you need to say more about that ontological physical universe, and explain how it makes consciousness, and how it deprives the same computation in arithmetic of consciousness. Bruno > > Bruce > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLQh-g%3DmHcxE4Nt5vX8Ro0%2BMJUOckWCZPKCkzPpcAj6GeQ%40mail.gmail.com > > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQh-g%3DmHcxE4Nt5vX8Ro0%2BMJUOckWCZPKCkzPpcAj6GeQ%40mail.gmail.com?utm_medium=email&utm_source=footer>. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/D2E5FB27-263B-4ACE-BE47-0852003C2591%40ulb.ac.be.
