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----- Unsent message follows ----- Received: from CS.CMU.EDU by FRC2.FRC.RI.CMU.EDU (4.1/5.17) id AA11136; Thu, 5 Aug 99 06:09:05 EDT Received: from CMU1.ACS.CMU.EDU by CS.CMU.EDU id aa13626; 5 Aug 99 6:08:29 EDT Received: by cmu1.acs.cmu.edu (8.9.3/8.9.3) id GAA01606 for [EMAIL PROTECTED]; Thu, 5 Aug 1999 06:08:24 -0400 (EDT) Received: via localmail; Thu, 5 Aug 1999 06:08:24 -0400 (EDT) Received: by cmu1.acs.cmu.edu (8.9.3/8.9.3) id GAA01602 for [EMAIL PROTECTED]; Thu, 5 Aug 1999 06:08:23 -0400 (EDT) Received: via localmail; Thu, 5 Aug 1999 06:08:23 -0400 (EDT) Received: from mx1.eskimo.com (mx1.eskimo.com [204.122.16.48]) by cmu1.acs.cmu.edu (8.9.3/8.9.3) with ESMTP id GAA01598 for <[EMAIL PROTECTED]>; Thu, 5 Aug 1999 06:08:22 -0400 (EDT) Received: (from smartlst@localhost) by mx1.eskimo.com (8.9.1a/8.8.8) id DAA15433; Thu, 5 Aug 1999 03:05:31 -0700 Resent-Date: Thu, 5 Aug 1999 03:05:31 -0700 Message-Id: <[EMAIL PROTECTED]> Date: Thu, 05 Aug 1999 06:00:04 -0400 From: Christopher Maloney <[EMAIL PROTECTED]> X-Mailer: Mozilla 4.61 [en] (WinNT; I) X-Accept-Language: en Mime-Version: 1.0 To: everything-list <[EMAIL PROTECTED]> Subject: Re: predictions Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Resent-Message-Id: <"j10rd2.0._m3.h7Mgt"@mx1> Resent-From: [EMAIL PROTECTED] X-Mailing-List: <[EMAIL PROTECTED]> archive/latest/1094 X-Loop: [EMAIL PROTECTED] Precedence: list Resent-Sender: [EMAIL PROTECTED] Gilles HENRI wrote: > > >Note that one need not bring MWI into this at all. The only big > >assumption is the existence of a copy machine. Instead of MWI, one > >can think of the identical experiment being carried out on an > >ensemble of, say, 100 hapless souls Albert, Bernard, Caroline, etc. > >At time t1, some number close to 50 will have seen heads. At time > >t2, there will be 150 people, 100 of whom remember seeing heads, > >and 50 of whom remember seeing tails. From a bird perspective, > >if you picked any person at random from this group, the chance that > >they'll have seen heads is 2/3. > > ok, let's aside the question of how to duplicate exactly a human being :) > I think the right answer is p=1/2 because this value will be naturally > observed by any Jane who repeats the experiment : if you toss the coin N > times, whatever you do with the Janes, they all will have seen the same > sequence of heads and tails, which entails statistically as much H than T. > Subjectively, suppose you toss your coin without knowing that the > duplicating machine is working (the machine can create another "you" in the > Andromedae Galaxy, say). You will not notice any difference, i.e. you > always will see a sequence of 50% H and 50% T. > With the color cards, each Jane will measure subjectively a probability 1/2 > of yellow, 1/4 of red (1/2 H *1/2 "being chosen as Jane 1") and 1/4 blue , > so again p(H) = p(T)=1/2 with the conditional probability formula. > The probability 2/3 is indeed the chance of finding someone who saw H after > the first experiment from a bird perspective, because duplicating > introduces a bias. However by repeating the experiment a lot of times, you > will find only very few people that have seen more H than T (in fact > exactly the same proportion as if there were no duplication). The bias > disappears statistically because for anybody it is as probable to have a > positive bias (H) than a negative one (T). . I've thought about this response all day, and I think you've shown, Gilles, that if there is only one universe, then the probability of seeing heads must be 1/2. I converted your argument into this thought experiment: Spoze you have 1024 identical booths, each of which has a TV monitor. In the begining of the experiment, Jane0 sits down in booth 0. The TV monitors are viewing a table located elsewhere on which a coin is tossed ten times. If the coin lands tails, then nothing happens to the Janes. If it lands heads, then all of the Janes at that instant (the instant after the last coin toss) are duplicated. After each toss, each monitor flickers to a screen with a solid color for a moment. The color is different in each of the booths, so we are sure that after each toss, all of the Janes are distinct. Now, from a bird perspective, the odds are good that we'll see somewhere close to 5 heads, which would mean that 32 Janes would be produced. They would all remember 5 tails and 5 heads. So for each of them, they could compute the odds of heads at 50%. As the number of tosses increased to infinity, this would become exact. Fine. But what if MWI or AUH is true? Then we have more to consider, because at each toss, the entire universe splits into two distinct histories. So after the first toss, there will be three Janes: Jane0-Tails, Jane0-Heads, and Jane1-Heads. After ten tosses, there will be 3^10 = 59049 Janes. If there were no duplication by copy-booth, the number of Janes who saw a given number m of heads would be given by the binomial theorem: 10! J(m,10) = --------- m!(10-m)! But with copying, the number of Janes who have seen m heads will be 10! J(m,10) = --------- X 2^m m!(10-m)! With this distribution: This many Janes Saw this many heads 1 0 (all tails) 20 1 180 2 960 3 3360 4 8064 5 13440 6 15360 7 11520 8 5120 9 1024 10 (all heads) If we assume that the original Jane is equally probable to end up as any of these final Janes, then Jane can expect to see heads 2/3 of the time. -- Chris Maloney http://www.chrismaloney.com "Donuts are so sweet and tasty." -- Homer Simpson