bzax · HackerTrans

29 giorni fadiscuss

You could observe future random numbers by taking combat actions, and then reset to the start of the fight and play a line which consumes fewer random numbers in order to manipulate your card rewards. Maybe you could generate the card reward at the start of the fight, but what if they play a card which impacts the card reward, e.g. by creating an extra card reward.

bzax

3 mesi fadiscuss

Well, once you've derived unary exp and ln you can get subtraction, which then gets you unary negation and you have addition.

bzax

7 mesi fadiscuss

It doesn't mean anything. The point is that the language of lean, and its proof derivation system, are able to express (and prove) statements that do not correspond to any meaningful mathematics.

bzax

2 anni fadiscuss

I feel obliged to mention that this does feature prominently in Kim Stanley Robinson's Red Mars trilogy. The single most important piece of infrastructure on Mars is a space elevator, but not everyone on the planet is happy with how the owners of the space elevator are running things.

bzax

2 anni fadiscuss

https://www.osha.gov/hexavalent-chromium

bzax

2 anni fadiscuss

It sure would be great if they just owned the supplier outright and could align incentives that way.

bzax

3 anni fadiscuss

Not only does Southwest exclusively fly Boeing aircraft, they exclusively fly 737s, which enables their unusual routing style. Essentially every pilot and crew at Southwest can fly any aircraft the company has for them. Presumably this gives Boeing a strong incentive to keep making new 737s that push the engineering envelope, instead of making a new narrowbody aircraft.

bzax

3 anni fadiscuss

Not just any two couples, one couple was her ex-husband's parents...

bzax

3 anni fadiscuss

I should admit I'm being very generous to Peters here - I came to the conclusion that this is what he means only because the math of ergodicity (https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorem...) talks a lot about "except on a set of measure zero". He provides no explanation of how he moves from "the time average of values in a particular run of the process" (which is ergodicity) to "what does a typical process round do, with probability 1" (which is perhaps what someone computing a utility function cares about).

I asked a friend who is an econ professor "Why does this Peters guy explain this so poorly" and his response was more or less, yes, all of economics has been wondering that too since he first published his Nature Physics paper on this a decade ago.

bzax

3 anni fadiscuss

The expected value of this distribution goes up with every iteration, there is no such Kelly point. You could try this with

heads: double your money tails: lose all your money

in which case the expected value is always $1, as you have a 1/2^n chance of having $2^n dollars after n rounds, and 0 otherwise.

The point of discussing ergodicity here, however, is whether you can describe the behavior of the iterated distribution deterministically if you exclude a portion of that distribution which has measure zero.

bzax

3 anni fadiscuss

I believe the entire point of the ergodicity question here is "If you apply this process n times, with n approaching infinity, obviously the result may depend on what point in the n-times iterated distribution you sample, but if you choose a volume of vanishingly small measure to exclude, can you make a single concrete statement about what the process is doing without taking an expected value over the different outcomes"

And the answer is yes - with probability approaching 1 as n increases (ie excluding a portion of the distribution whose measure decreases to 0), the random process matches a deterministic process which is described by "you lose 5% each round".

bzax

3 anni fadiscuss

The "average" of the distribution goes up as you increase the number of rounds, but the probability that you get an average or above value when you sample that distribution once goes to zero as the number of rounds increases.

bzax

3 anni fadiscuss

If you repeat this game n times (as n goes to infinity), you will have Θ(n) pairs of (heads, tails) and O(sqrt(n)) unpaired wins or losses, except for a vanishingly small fraction of the time when the results fall outside of any fixed number of standard deviations.

The point is that you as an individual playing a repeated game don't get to meaningfully sample the expected value of the distribution. You only get to sample once, and you will almost surely (i.e. with probability approaching 1 as n goes to infinity) sample a point in the distribution where you lose nearly all of your money.

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