On Mon, Mar 3, 2014 at 9:18 PM, chris peck <chris_peck...@hotmail.com>wrote:
> Hi Liz > > > > > > > *>> 0000 0001 0010 0011 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 > 1010 1011 1100 1101 1110 1111Of which I'm fairly sure half the digits are 0 > and half 1!What am I missing here?* > > > If you concatenate all those strings together you'll get a bigger string > in which the proportion of 1s to 0s is exactly 50/50. And that will always > be the case no matter how long the individual bit strings are. If they are > 8 bits long then you'll have 256 individual strings. When concatenated > together the proportion will be exactly 50/50. > > But it looks to me like you're misconstruing Tegmark's method here. Its > each individual string that matters. What is the proportion of 1s to 0s in > 0000, or in 1011, or 1100 etc. Because each string represents a sequence of > room - wake ups. In your example, 16 people live through room-wake ups over > 4 nights. Each person's experience represented by an individual string. > > Even over 4 nights you'll see, just by counting, that the number of > occasions where the proportion of 1s to 0s is 50% is 6. Not 8. Not half. > How does that square with his claim that almost all people will experience > a 50/50 distribution of 0s to 1s? not even half will. Now as the individual > strings get longer, as more nights are encountered, that proportion goes > down. Not up. When individual strings are 16 bits long, there are 65,536 > combinations (people). Of whom less than 20% experience a 50/50 split of 1s > and 0s over those 16 nights. > > Now Brent, and Bruno with customary obtuseness, correctly point out that: > > 1) Tegmark talks about 'roughly half', so not an exact 50/50 split. > > 2) if you take that into account, then you can get a figure approaching > 'almost all'. > > in the 16 bit example, if you include strings where there are 7 ones (or > zeros) and you take strings where there are 6 ones (or zeros) then about > 78% of people will experience 'roughly' 50% ones or zeros. Ofcourse now > we're in a situation where personal opinion rears its head. Is 78% 'almost > all'? Is 37% (6/16) 'roughly half'? Right and wrong don't really preside > over these kinds of opinions, but 37% doesn't look like 50% to me. > > In any case both Bruno and Brent miss the bigger picture: > > Consider the following 16 bit strings: > > 1010101010101010 - does that look random? > > how about > > 0101010101010101 > > how about this: > > 1100110011001100 > Those are examples of the comparatively rare exceptions (compressible strings). The overwhelming majority of all possible permutations of binary strings have no discernible patterns and are thus incompressible (and by definition random looking). Think of it like this: the number of ways to randomly shuffle a deck of cards is 52*51*50 ... *3 *2 *1 (or 52 factorial). This is equal to 8 followed by 67 zeros. There is a huge number of ways you might shuffle the deck into some pattern, say the first 26 cards red, and the last 26 cards black. There are (26 factorial)^2 ways to do this, or 1.62 followed by 52 zeros, another huge number. But when you compare those numbers, you see the fraction of possible shufflings that result in such a pattern is 2 quadrillionths<https://www.google.com/search?q=26!+*+26!+%2F+52!>. Even if you add up all the possible patterns that you can describe more succinctly than simply listing out each card, the fraction of those patterns would be miniscule compared to the total number of shufflings. The same is true of bit strings and their representations. For any N-bit long string, almost all of them cannot be described using fewer than N-bits of information, and increasingly so the larger N is. > > Seems to me Tegmark is confusing a roughly equal distribution of 1s and 0s > with apparent unpredictability. > 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. > A better approach considers irregularity of change in 1s and 0s. So where > there is irregular change : 0100111101010011 it looks unpredictable, but > where change is regular : 1111000011110000 it doesn't. The proportion of 1s > and 0s is irrelevant. > True, he used an inaccurate description of randomness, it is more than just an even distribution. > > So has Tegmark convinced me that in his thought experiment I would assign > 50/50 probability of seeing one or the other room each iteration? Not > really. > If you had seen "0 1 0 1 0 1 0 1" would you pay $1000 on the next number being "0" if the payout were only $100? If not, how high would the payout have to be for you to make such a bet? Jason > > I'm sure Tegmark's world won't be shaken too much by any of this, I'm even > more certain that I have something wrong. Though it does seem to have sent > Bruno running for cover behind his little sums. So perhaps I am on to > something.... > > All the best > > Chris. > > ------------------------------ > Date: Tue, 4 Mar 2014 11:59:05 +1300 > > Subject: Re: Tegmark and UDA step 3 > From: lizj...@gmail.com > To: everything-list@googlegroups.com > > I should also mention that in the quote, Max says that you wake up in room > 0 or room 1, so if we WERE omitting leading zeroes, we'd write > "1111111111..." ! > > Shurely shome mishtake! > > > -- > 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 everything-list+unsubscr...@googlegroups.com. > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/groups/opt_out. > > -- > 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 everything-list+unsubscr...@googlegroups.com. > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/groups/opt_out. > -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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