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Continuing from: Day 24 (20) / Morning Lecture
did you expect a different branch of humanity to have familiar math customs?
lawful chaotic
The Rule of Succession! This is sort of picking up the topic from last time about Science!, how to do statistical inference in cases slightly more complicated than the trivial ones discussed previously. Asmodia has now supposedly had the time to teach the new candidates some additional Law of Probability, aided by Meritxell. People have cast a few Fox's Cunnings on themselves and others. They will now try again to see if that sufficed for old hands to teach new hands.
Suppose Keltham takes a sphere, puts it on a table with bouncy boundaries, shoves it rolling quickly, and lets it bounce back and forth many times until it comes to a halt, somewhere random. Object!Keltham should theoretically be considered as doing this in private, where the object!others can't see him, though for illustration purposes, meta!Keltham will write down everything plainly for people's meta!selves in class.
[--------0---------------------]
Object!Keltham marks down where Ball 0 landed, picks up the ball, and again rolls it hard.
[--------0------------1--------]
Object!Keltham then asks the object!classroom for their odds that Ball 1 landed to the left or the right of Ball 0. Remember, their meta!selves can see where Keltham made the marks on the whiteboard representing where Ball 0 and Ball 1 landed, but their object!selves can't.
Before being told the answer, what probability should their object!selves assign that Ball 1 landed to the left or right of Ball 0, after both balls bounced back and forth on the table many times? Their object!selves know the experimental setup, they just don't know the detailed results until object!Keltham tells them.
.....seems like they should assign a 50% chance? Because the situation is - symmetrical, from the information they have, there's no reason to expect Ball 1 to be more left or more right than Ball 0.
lawful chaotic
Object!Keltham tells the object!class that Ball 1 was indeed to the RIGHT of wherever Ball 0 landed. They know nothing else about where Ball 1 or Ball 0 landed, as yet, just that Ball 1 was to the RIGHT of ball 0.
[--------0------------1--------]Object!class only knows:
1: RIGHT
Ball 2 is now rolled, same setup.
What probability do they assign that Ball 2 falls to the LEFT or RIGHT of Ball 0?
" - okay, I see why it's not just 50% now. Because our guess about where the first ball lands shifts when we learn the second ball landed to the right of it. It's - the same as spinning a coin, where the Ball 0 is like - deciding if the coin is weighted - and then the subsequent spins give you information about how weighted and in which direction."
lawful chaotic
Indeed. The object!class can now be quite confident that Ball 0 did not stop exactly up against the RIGHT wall, in which case the chance of Ball 1 landing further to the RIGHT would have been zero.
Surely they have no more information than this, however.
......nnnnnno they have more information than that, like, it's also less likely the ball is very near the RIGHT wall, it's just not immediately obvious how to plug it into the formulas from last class.
#math-investor-chaos
"If there were just three positions it could be in, you've ruled out it being at the right end, and if it was in the middle, Ball 1 result was 50/50, if it was on the left end, Ball 1 result was certain, so in that example we should guess it's at the left end twice as often, so 2/3 vs 1/3. So then 2/3 * 2/3 Ball 2 is to the right, plus 1/3 *1/3 Ball 2 is to the right, so 5/9."
this-is-fine
"I think I see what you were trying to do there but I don't think the numbers work out? Ball 0 could be all the way on the left, in the middle, or on the right. The new chance that it's all the way on the right is zero, and if Ball 0 was in the middle, we're half as likely to hear 'RIGHT' than if Ball 0 was all the way on the left. So the new chances are 2/3 that Ball 0 was all the way on the Left, times a probability of 1 of hearing 'RIGHT' then, plus a 1/3 chance that Ball 0 was in the middle, times a 1/2 probability of hearing 'RIGHT' then. 2/3 * 1 + 1/3 * 1/2 = 5/6 chance of hearing 'RIGHT' next time."
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."
know-it-all
"Pretty sure we should expect to hear 'RIGHT' more than we expect to hear 'LEFT'."
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 }