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Continuing from: Day 24 (20) / Morning Lecture
did you expect a different branch of humanity to have familiar math customs?
know-it-all
"If, however, we actually look at what - 'meta-Keltham' - drew on the wall, it seems like a pretty reasonable way things could be, and in that way things could be, we're not supposed to say 5/6."
"I mean, think about it, if this actually happened, and Keltham said 'RIGHT' one time, would you immediately afterwards go, 'Well, I guess the chance of hearing "RIGHT" next time is 5/6?' Hearing 'RIGHT' one time shouldn't actually make you that confident about it happening again."
ruin-of-souls
"It's still information, though. In all of the situations where ball 0 is left of center, ball 1 was more than 50% likely to be right of it, and in all of the situations where ball 0 is right of center, ball 1 is less than 50% likely to be right of it. So the fact that it's to the right is some update towards ball zero being left of center, which suggests that ball two is also more likely to be to the right of ball zero, but - I don't know how you'd tell what the numbers are."
curse-of-laughter
"Let's say we divide it into eleven possible positions. If Ball 0 arrives all the way on the right, chances of hearing RIGHT are 0/10. If Ball 0 is 90% of the way to the right, chances of hearing RIGHT are 1/10."
"So it's 1/10 squared plus 2/10 squared plus 3/10 squared which is... 1 plus 4 plus 9 plus 16..."
"I get 285/100."
"What'd I do wrong."
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."