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Continuing from: Day 24 (20) / Morning Lecture
did you expect a different branch of humanity to have familiar math customs?
someone-somewhere
"Let us, possibly, try this with five positions before going straight to ten."
"0/4, 1/4, 2/4, 3/4, 4/4 likelihood of yielding RIGHT, all positions equally probable before then."
"We see one 'RIGHT'. The likelihoods are 0/4, 1/4, 2/4, 3/4, 4/4. The posteriors are - I see the other two things Pilar did wrong."
"We start with five hypotheses 0/4, 1/4/, 2/4, 3/4, 4/4, each of which has prior probability 1/5."
"After seeing 'RIGHT' one time, we've got 'priors-times-likelihoods' of 0/20, 1/20, 2/20, 3/20, 4/20."
"The 'relative-odds' of the posterior will be proportional to the 'priors-times-likelihoods', but they have to sum to 1. There's 1 + 2 + 3 + 4 = 10 total shares, of twentieths but it doesn't matter what exactly, so the 'posterior-distribution' is 0, 1/10, 2/10, 3/10, 4/10, summing to 1."
"Then if you take the likelihoods off each of those, weighted by the new posterior-probability, you get 1 + 4 + 9 + 16 divided by 40 equals 3/4."
lawful chaotic
Keltham will helpfully write down the calculation Asmodia actually did, there:
0/10 * 0 + 1/10 * 1/4 + 2/10 * 2/4 + 3/10 * 3/4 + 4/10 * 1= 1/40 + 4/40 + 9/40 + 16/40= 30/40
curse-of-laughter
"With 7 positions it's 0 + 1 + 4 + 9 + 16 + 25 + 36, divided by (0 + 1 + 2 + 3 + 4 + 5 + 6) * 6, so 91/126."
curse-of-laughter
...all... right then...
"I get 1^2 + 2^2 + 3^2 out to infinity, divided by... 1 + 2 + 3 out to infinity, times infinity."
...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.
Parenthetical retroactive class chatter
Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?
......Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.
Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2]. Proof by generalizing examples:
sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly hasty intuitive reasoning that will create a good sense for what's really going on. So instead he'll just quickly write on the board that:
let f(N) = N(N+1)(2N+1)/6
in f(0) = 0
f(N+1) = f(N) + (N+1)^2
proof:
0(0+1)(2(0)+1)
= 0(1)(3)
= 0
f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6 - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1) * [(N+2)(2N + 3) - (N)(2N+1)] / 6
= (N+1) * [2N^2 + 7N + 6 - 2N^2 - N] / 6
= (N+1) * [6N + 6] / 6
= (N+1)^2If they're not happy with this logical proof by certain reasoning - because, like, why - they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty much looks like it should take up about a third of a cube of size N.
Now, back to the problem of evaluating:
[1^2 + 2^2 + 3^2 + 4^2 + ...] / { [1 + 2 + 3 + 4 + ...] times infinity }
curse-of-laughter
Then Pilar will obtain the sum N(N+1)(2N+1)/6, divided by N*[N(N+1)/2].
Letting N = infinity, this yields INF*(INF+1)*(2INF+1)/6, divided by INF*INF*(INF + 1)/2.
Simplifying produces (2INF+1)/(INF*3). This then cannot be simplified further.
lawful chaotic
Why, of course it can be!
Keltham writes on the wall:
2*INF + 1 2 * INF 1
--------- = ------- + -------
INF * 3 INF * 3 INF * 3
1 divided by three infinities simplifies to zero, obviously. Think of how little cake each person gets, if you divide one slice of cake among all the billion people in Golarion; one-third divided by infinity is much less than that.
So this whole expression simplifies to 2/3.
lawful chaotic
It's Lawful when he does it, in a way that carefully doesn't take any mathematical reasoning steps that would actually be invalid. Keltham would not necessarily advise Pilar trying this on her own for a little while. Or rather, she should try it, but be prepared to fail.
Pilar may, for her own reassurance, note that her previous answer of 91/126 is not too far from 90/135, also known as 2/3.
#math-investor-chaos
"There are 3 permutations which have 0 before 1: 012, 021, 201."
"Of these, 1/3 have 2 left of 0 and 2/3 have 2 right of 0."
lawful chaotic
"So what you're saying here is that any of the possibilities..."
[---2----0------------1--------]
or [--------0------2-----1--------]
or [--------0------------1-----2--]
"...are equally probable? I'm not sure how reasonable that is, it looks to me like the middle stretch is wider than the two ends."
lawful chaotic
"So what you're saying here is that if I roll three balls back and forth, and they bounce back and forth a lot, they're equally likely to land in any of the six possible orders, so far as you know?"
"Can that really be true? After all, there could be bumps in the table, the bouncy walls might not be perfect, I might have rolled the balls harder on some occasions than others..."
know-it-all
"If we don't know any of that on the 'object-level', then, so far as we know, all six permutations are equally likely."
"You could say that sort of thing about any problem with permutations in it."
lawful chaotic
"Oh, I suppose. Just don't forget, in real life, always cheat the shit out of anything if you can get any extra information like that."
"After all, if you think otherwise, why, that's the sort of thinking that might lead to you always insisting that all of the chemical elements were equally likely to be in any order on the Periodic Table, any time you don't know for sure what order they have."
"Soooo how about if I then tell you that Ball 2 landed to the RIGHT of Ball 0, again? What's the probability of RIGHT next time, with the third ball?"
"3/4," says the same person who asked about permutations last time. (They've unfortunately written their nametag in letters too small to see from where Keltham is standing.)