On 06 Mar 2014, at 01:52, chris peck wrote:
Hi Jason/Gabriel
Thanks for the posts. They were both really clear. I can see that it
was a mistake to hedge my bets on exact figures and also, given
Jason's comments, to think that seemingly regular sequences were
quite common.
I do maintain that proportions of roughly 50/50 splits are a
spurious measure of 'seemingly random' though and that irregularity
of change is a better one.
I agree, and that is why I justify "randomness" by incompressiblity.
It is an exercise to show that in the iterated self-duplication, the 1-
views grows more and more intrinsically non regular, indeed non
algorithmically compressible.
There also seems to me to be a big difference between Tegmark's game
as described in the quote below, and flicking coins. Tegmark's game
is a process guaranteed to generate (over 4 iterations) 16 unique
and exhaustive combinations of 0s and 1s (heads or tails). If 16
people were to flick a coin 4 times and write down the results there
is only a low probability that the resulting set would map on to
that generated by Tegmarks game. There is fair chance there would be
some repetition.
Jason, you say:
>> Even if your pattern were: 0 1 0 1 0 1 0 1 0 1, you still have no
better than a 50% chance of predicting the next bit, so despite the
coincidental pattern the sequence is still random.
I disagree here. In Tegmarks game you know a particular outcome is
not exclusive and that you'll have two successors who get one and
the other. The next outcome is (01010101010 AND 01010101011) not
(01010101010 XOR 01010101011).
That is the 3-1 views. If you predict in that setting your future 1-
view by (01010101010 AND 01010101011), both copies will refute it, and
you loss the bet. If you predict (01010101010 XOR 01010101011), both
copies win the bet. here the bet can be done in the 1p-plural way,
with someone accompanying you in the telebox.
Now this might influence how you bet. If you care about your
successors you might refuse to make a bet because you know one
successor will lose. If we rolled dice rather than flicked coins and
were to bet on getting anything but a 6, in a modified Tegmark game
we might still refuse to bet knowing that one successor would
certainly lose. Its a bet we almost certainly would take if we were
rolling die in a classical world without clones.
More dramatically, if you play Russian roulette in Everettian
Multiverse you always shoot someone in the head. Crossing the road
becomes deeply immoral because vast numbers of successors trip and
get run down by trucks.
A final confusion: Does anything ever seem 'apparently random' in a
Marchalian/Tegmarkian game? Given that you know outcomes are
generated by a mechanical process and given you know exactly what
the following set of outcomes will be, how can they seem random?
Even 1001111010110011 isn't looking very random anymore.
Like John, you keep describing the 3-1 views, which we know already
are deterministic. But the question bears on the 1-views themselves,
and it is easy to see that any specific prediction (without using or
or xor) will fail.
If in helsinki you predict "I will see M and I will see W, when
opening the door", well, both copies, when opeing the doors will have
to assess that they were wrong, as they see only W, xor M.
Bruno
Date: Thu, 6 Mar 2014 10:21:47 +1300
Subject: Re: Tegmark and UDA step 3
From: lizj...@gmail.com
To: everything-list@googlegroups.com
On 6 March 2014 06:45, Gabriel Bodeen <gabebod...@gmail.com> wrote:
Brent was right but the explanation could use some examples to show
you what's happening. The strangeness that you noticed occurs
because you're looking at cases where the proportion is *exactly* 50%.
binopdf(2,4,0.5)=0.375
binopdf(3,6,0.5)=0.3125
binopdf(4,8,0.5)=0.2374
binopdf(8,16,0.5)=0.1964
binopdf(1000,2000,0.5)=0.0178
binopdf(1e6,2e6,0.5)=0.0006
Instead let's look at cases which are in some range close to 50%.
binocdf(5,8,0.5)-binocdf(3,8,0.5)=0.4922
binocdf(10,16,0.5)-binocdf(6,16,0.5)=0.6677
binocdf(520,1000,0.5)-binocdf(480,1000,0.5)=0.7939
binocdf(1001000,2e6,0.5)-binocdf(999000,2e6,0.5)=0.8427
binocdf(1000050000,2e9,0.5)-binocdf(999950000,2e9,0.5)=0.9747
Basically, as you flip a coin more and more times, you get a growing
number of distinct proportions of heads and tails that can come up,
so any exact proportion becomes less likely. But at the same time,
as you flip the coin more and more times, the distribution of
proportions starts to cluster more and more tightly around the
expected value. So for tests when you do two million flips of a
fair coin, only about 0.06% of the tests come up exactly 50% heads
and 50% tails, but 84.27% of the tests come up between 49.95% and
50.05%.
Thank you, that's exactly what I was attempting to say in my cack-
handed way. (And it is almost certainly what Max intended to say.)
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