but the only really arbitrary seeming port is the choice about how to order the limits.
Deciding that the probabaility of overtaking needs to be >0 might also count here. The likely alternative would be ≥0.5. I've picked 0 because I expect that normally this limit will be 0 or 1 or 0.5, and if we get other values then ≥0.5 might lead to intransitive comparisons - but I might rethink this if I discover a case where the limit isn't one of those three.
The tit-for-tat strategy unravels all the way back to the beginning, and we're back at total defection.
Is even that true? Yes, defecting the last n rounds dominates defecting the last n-1 rounds, but defecting the last 5 vs tit-for-tat all the way the equilibrium can be either depending on the initial population. So for all I know there might be a stable equilibrium involving only strategies that cooperate until close to the end.
I'm also curious whether this can be applied to other problems, like the St. Petersburg Lottery.
I would think so. We can model the St. Petersburg game as the limit of a sequence of n-Petersburg games just like the original but ending after n turns no matter the cointoss then. I still can't fully solve this analytically, but I can lower-bound it: Consider a price p for the petersburg game that is higher than the minimum 2$ you get . Then a sample of n people need to earn more than p*n for the game to be worth it. The probability that one bettor earns that much or more is (1/2)log2(pn), or equivalently 1/pn. The probability that at least one of the n participants manage to do this is 1−(1−1/pn)n, and as n goes to infinity that comes out as 1−e−1/p, a finite value larger than 0. So that sort of is a case where a probability other than 0, 0.5, or 1 occurs - sort of, because one bettor getting very lucky is only one way the betting team can win. But this tells us that according to my definition above, the petersburg bet is at least (again, because thats only one way) as good an any finite fixed price, which if it isn't paradoxical means that its more valuable than any fixed price.
We can also modify the petersburg game to triple the payoff after every cointoss instead of double. In that case the probabilities above are (1/2)log3(pn)=(1/pn)log3(2) for the single, and for team 1−(1−1/(pn)log3(2))n=1−(((pn)log3(2)−1)/((pn)log3(2))n, which comes out to 1 (note: link missing the initial 1). So that version is definitely more valuable than any fixed price. I think the original is a bit strange here because 1/x is also different from the other powers.
I have also by now done some monte carlo tests (the distributions here have infinite mean and variance, but I should be allowed to sample with this method - for essentially the same reasons that pulling the comparison into the limit works). These indicate that the original petersburg game is more valuable than 6$ and likely more valuable than 100$ (it definitely wins more than the bound I gave above). I've also tested the Kelly game with decay, but with a 10% chance of ending each round so that its easier to calculate. I unfortunately don't have enough compute to be very detailed about this but I can say that betting 60% is better than betting 90%, which supports my thesis about an increase in recommended bet to less than 100%.
You might also have meant some kind of repeated St. Petersburg bet with fractional betting, but the normal petersburg bet doesn't have a wagered amount. Perhaps we could try assigning it some price with the option of buying multiples, and then ask what fraction of your money you would want to spend on them each round? Maybe I'll look into that next.
When we don't know how to solve a problem even given infinite computing power, the very work we are trying to do is in some sense murky to us.
I wonder where this goes with questions about infinite domains. It seems to me that I understand what it means to argmax a generic bounded function on a generic domain, but I don't know an algorithm for it and as far as I know there can't be one. So it seems taking this very seriously would lead us to some form of constructionism.
But you're saying we should change that game from f:Ω->[0,inf] to g:Ω,?R?->[0,inf]
No. The key change I'm making is from assigning every strategy an expected value (normally real, but including infinity as you do should be possible) to having the essential math thing be a comparison between two strategies. With your version, all we can say is that all-in has EV 0, don't bet has EV 1, and everything else has EV infinity - but by doing the comparison inside the limits, we get some more differentiation there.
R isn't distinct from Ω. The EV function "binds" the randomness inside what its applied to, so when I roll it out I need to have it occur explicitly inside the limit. I think its fine to say the Rs are normal random variables. Lets say that each r(i) is uniformly distributed in [0;1) iid. game() then uses that for its randomness. For the game at hand, we could say that the binary digit expansion becomes the sequence of heads and tails thrown.
As you might have noticed I wrote the post in a bit of a hurry, so sorry if not everything is hammered out.
If I understand that context correctly, thats not what I'm doing. The unconventional writing doesn't pull a lim outside an EV, it replaces an EV with a lim construction. In fact, that comment seems somewhat in support of my point: he's saying that limt→∞E(U(game(s,r,t))) doesn't properly represent an infinite game. And if the replacing of E with the n-lim that I'm doing works out, then thats saying that the order of limits that results in Kelly is the right one. Its similar to what I said (less formally) about expected utility maximization not recommending all-in for infinitely many rounds.
I've worked out our earlier discussion of the order of limits into a response to this here.
I now think you were right about it not solving anthropics. I interpreted afoundationalism insufficiently ambitiously; now that I have a proof-of-concept for normative semantics I can indeed not find it to presuppose an anthropics.
It seems like there's some mystery process that connects observations to hypotheses about what some mysterious other party "really means"
The hypothesis do that. I said
We start out with a prior over hypothesis about meaning. Such a hypothesis generates a propability distribution over all propositions of the form "[Observation] means [propositon]." for each observation (including the possibility that the observation means nothing).
Why do you think this doesn't answer your question?
but if this process always (sic) connects the observations to propositions that are always true, it seems like that gets most favored by the update rule, and so "it's raining" (spoken aloud) meaning 2+2=4 (in internal representation) seems like an attractor.
The update rule doesn't necessarily favor interpretation that make the speaker right. It favours interpretations that make the speakers meta-statements about meaning right - in the example case the speaker claims to mean true things, so these fall together. Still, does the problem not recur on a higher level? For example, a hypothesis that never interpreted the speaker to be making such meta-statements would have him never be wrong about that. Wouldn't it dominate all hypothesis with meta-statements in them? No, because hypothesis aren't rated individually. If I just took one hypothesis, got its interpretations for all the observations, and saw how likely that total interpretation was to make the speaker wrong about meta-statements, and updated based on that, then your problem would occur. But actually, the process for updating a hypothesis also depends on how likely you consider other hypothesis:
To update on an observation history, first we compute for each observation in it our summed prior distribution over what it means. Then, for each hypothesis in the prior, for each observation, take the hypothesis-distribution over its meaning, combine it with the prior-distribution over all the other observations, and calculate the propability that the speakers statements about what he meant were right. After you've done that for all observations, multiply them to get the score of that hypothesis. Multiply each hypothesis's score with its prior and renormalize.
So if your prior gives most of its weight to hypothesis that interpret you mostly correctly, then the hypothesis that you never make meta-statements will also be judged by its consistency with those mostly correct interpretations.
I'm not sure what you don't understand, so I'll explain a few things in that area and hope I hit the right one:
I give sentences their english name in the example to make it understandable. Here are two ways you could give more detail on the example scenario, each of which is consistent:
Sentences in interpreter language are connected to the epistemology engine simply by supposition. The interpreter language is how the interpreter internally expresses its beliefs, otherwise it's not the interpreter language.
"It's raining" as a sentece of the interpreter language can't be taken to mean "2+2=4" because the interpreter language doesn't need to be interpreted, the interpreter already understands it. "It's raining" as a string sent by the speaker can be taken to mean "2+2=4". It really depends on the prior - if you start out with a prior thats too wrong, you'll end up with nonesense interpretations.
I've thought about applying normativity to language learning some more, its written up here.
∀x∈Monsters:(P((boojum(x)∧snark(x))−1(true))/P(snark(ω)(x)−1)=0.4)This is simply saying that, given we've randomly selected a truth table, the probability that every snark is a boojum is 0.4.
This is simply saying that, given we've randomly selected a truth table, the probability that every snark is a boojum is 0.4.
Maybe I misunderstand your quantifiers, but I don't think it says that. It says that for every monster, if we randomly pick a truth table on which it's a snark, the propability that it's also a boojum on that table is 0.4. I think the formalism is right here and your description of it wrong, because thats just what I would expect P(boojum|snark) to mean.
I think the source of the problem here is that Chapman isn't thinking about subjective propability. So he sees something like "Cats are black with 40% propability" and wonders how this could apply to one concrete cat. and he similarly thinks about the dummy variable of the quantifier that way, and so how could all these different cats have the same propability? But the statement actually means something like "If all I know about something is that it's a cat, I give it a 40% propability of being black".