Greg D

Karma: 1

Greg D 20 May 2024 1:42 UTC
1 point
0
on: D&D.Sci (Easy Mode): On The Construction Of Impossible Structures
I’m not a data scientist, but I love these. I’ve got a four-hour flight ahead of me and a copy of Microsoft Excel; maybe now is the right time to give one a try!

!It seems like the combination of materials determines the cost of the structure.

!Architects who apprenticed with Johnson or Stamatin always produce impossible buildings; architects who apprenticed with Geisel, Penrose, or Escher NEVER do. Self-taught architects sometimes produce impossible buildings, and sometimes they do not.

!This lets us select 5 designs from our proposals which will certainly produce impossible buildings. To do better, we need to understand how to tell when a proposal by a self-taught architect is likely to produce an impossible building.

!~44% of designs by self-taught architects are impossible. This more-or-less matches the ²⁄₅ of masters whose apprentices reliably produce impossible buildings. So I hypothesize that self-taught students pick a favorite master at random and crib their style, acting (illegibly) like a typical apprentice thereafter. So now I need to see if there are particular materials, structure types, or blueprint types which are favored by students of any of the known master architects. By choosing designs by self-taught architects which have those properties, maybe I can tease out whose style they’re probably using.

!A structure can contain either dreams or nightmares, but not both.

!I’m too smooth-brained to tease out complex correlations on this flight while just using Excel: if there’s something weird going on (like, buildings made with either Dreams -or- Glass are likely to be impossible, but if you use both at once they cancel one another out somehow), I don’t know how to find it. So I’ll just assume everything is independent of everything else and do a Bayes to it.

!We can down-select our variables to match those which appear in the Self-Taught proposals; it does us no good to learn whether the “good” architects make use of Nightmares or not, if none the proposals before us make use of Nightmares.

!Good properties: Towers; buildings of Dreams and / or Glass; Hastily-Sketched blueprints. Bad properties: Mansions, Mechanisms; buildings of wood and / or Steel; Obsessively Detailed blueprints.

!So I choose proposals D, E, G, H, and K (probability 1); and also proposal A (probability ~62%) if we’ve got room.

!Ok, I just got off the plane and checked the puzzle description. Turns out we only get to choose 4 buildings, and there was no reason to try and tease out what Self-Taught architects are doing. In that case, I need to rank proposals D, E, G, H, and K by likely price.

!Structure price looks vaguely exponential, so I’ll take do a linear fit to minimize RMS(log10(error)). If I minimize RMSE directly then it always screws up the low-price structures to get marginally better fits on high-priced ones.

!It really looks like for each structure, you pick two materials; each material contributes a random amount to the price, with every material having its own distribution of price contributions. I can’t figure out what dice or whatever are being rolled for each material, but the fit gives me the average contribution for each one.

!So I choose proposals K, E, D, and H, with expected prices 30k, 73k, 78k, and 78k. Proposal G should be impossible too, but it’ll probably cost about 572k.

Greg D 2 Feb 2024 0:25 UTC
2 points
0
on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
I was an inveterate thirder until I read a series of articles on repeated betting, which pointed out that in many cases, maximizing expected utility leads to a “heavy tailed” situation in which a few realizations of you have enormous utility, but most realizations of you have gone bankrupt. The mean utility across all realizations is large, but that’s useless in the vast majority of cases because there’s no way to transfer utility from one realization to another. This got me thinking about SB again, and the extent to which Beauties can or can not share or transfer utility between them. I eventually convinced myself of the halfer position.

Here’s the line of reasoning I used. If the coin comes up H, we have one awakening (experience A). If the coin comes up T, we have two awakenings—either in series or in parallel depending on the variant, but in any case indistinguishable. By Bayes, Pr(H|A) = Pr(A|H)Pr(H)/Pr(A). The core insight is that Pr(A|H) = Pr(A|T) = Pr(A) = 1, since you have experience A no matter what the coin flip says. SB is akin to drawing a ball from one of two jars, one of which contains one red ball, and the other of which contains two red balls. Having drawn a red ball, you learn nothing about which jar you drew from.

What about making bets, though? Say that SB is offered a chance to buy a ticket worth $1 if the coin was T, and $0 if it was H. To maintain indistinguishability between the “three Beauties,” each time she is awakened, she must be offered the same ticket. In this case, SB should be willing to pay up to $2/3 for such a ticket. But this is not because the probability of T is really ²⁄₃ - it is because the payoff for T is larger since the bet is made twice in sequence. In the “clones” variant, SB’s valuation of the ticket depends on how she values the welfare of her clone-sister: if she is perfectly selfish she values it at $1/2, whereas if she is perfectly altruistic she values it at $2/3. Again, this is because of variations in the payout—obviously SB’s estimate of the probability of a coin flip cannot depend on whether she is selfish or not!

A lot of anthropic arguments depend on simply “counting up the observers” and using that as a proxy for probability. This is illegitimate, because conditional probabilities must always be normalized to sum to one. Pr(Monday|T) + Pr(Tuesday|T) = ¹⁄₂ + ¹⁄₂. Any time you use conditional probability you have to be very careful: Pr(Monday|T) != Pr(Monday and T).