But HOW do they square that up with the previous CLAIM as proven by DUBIOUS INFINITARY MATHEMATICS that, after seeing one RIGHT, the chance of the next ball landing RIGHT is two-thirds?

...it's the same claim.  One-third left is half of two-thirds right.

But is it really.

YES.

Somebody would in fact now like to make the general claim that, if N balls go LEFT and M balls go RIGHT, the chances of the next ball landing LEFT and RIGHT respectively are:

  N + 1           M + 1
---------  vs.  ---------
N + M + 2       N + M + 2

Keltham agrees!

Now would anybody like to try to prove that claim using dubious infinitary mathematics?

"I'd love to, but I don't know where to start... well, actually, I guess the obvious place would be with - two RIGHT and one LEFT and seeing if we can make that come out to 2/5 left and 3/5 right and then seeing if we can get it to generalize."

Several of the new students do in fact know calculus, and that seems like the obvious tool to use on this problem?

Ah, yes.  Golarion's notion of 'calculus'.  Keltham has actually looked into it, now.

It looked like one of several boring, safe versions of 'dubious-infinitary-mathematics' where you do all the reasoning steps more slowly and they only give you correct answers.

Dath ilani children eventually get prompted into inventing versions like that, after they've had a few years of fun taking leaps onto shaky ground and learning which steps land well.

Those few years of fun are important!  They teach you an intuitive sense of which quick short reasoning steps people can get away with.  It prevents you from learning bad habits about reasoning slowly and carefully all the time, even when that's less enjoyable, or starting to think that rigor is necessary to get the correct answer in mathematics.

Pilar has not yet been exposed to professional Golarion mathematicians talking about the importance of rigor.  It will probably be a powerful theological experience for her when she does!

Meanwhile she already has a strong sense that Keltham needs to be burned as a heretic.

"Don't I need to reason correctly to arrive at correct answers?"

"I don't think we have several years to spend on freeform mathematical exploration, even if we should make it an educational priority for the next generation."

"Rigor is necessary to know you got the correct answers.  Nonrigorous reasoning still often gets you correct answers, you just don't know that they're correct.  The map is not the territory."

"Often though not literally always, the obvious methodology in mathematics is to first 'rapid-prototype' an answer using nonrigorous reasoning, and then, once you have a strong guess about where you're going, what you're trying to prove, you prove it more rigorously."

"We don't have years to spend doing all of our mathematical reasoning rigorously the first time around, and besides, who does that anyways, and also, I'm here and can potentially check whether your fast crazy reasoning steps were valid or not."

Okay, you know, Asmodia's just going to try this thing with the dubious infinitary mathematics.  If not her, then who?  Asmodia alone is protected from any vengeance the afterlife might visit on her.

Consider the case for one LEFT and two RIGHT first.  They start with an infinite number of points between 0 and 1 probability of yielding LEFT, each with prior probability 1/INF.

For each of those points p on the spectrum from 0 to 1, their likelihood of yielding one LEFT and two RIGHT is:

(p)^1 * (1-p)^2  =  p - 2p^2 + p^3

Then to get the posterior... Asmodia has to...

Well, to be honest, she has in fact tried to get calculus off some Securities and she is having to exert something of a deliberate effort not to use that.  But the calculus also wasn't - the thing that she can feel Keltham is trying to teach?  Asmodia is getting a sense of the way that Keltham teaches intuitions and not just formulas.  She thinks it's meant to teach the thing that Ione does a little better than Asmodia.

A little better.

Asmodia wants to learn it.

The way they got here before, was by Pilar trying it with five exact hypotheses for p, then seven hypotheses for p, then an infinite number of hypotheses for p.

Let's say there were four hypotheses, 0/3, 1/3, 2/3, 3/3 probability of LEFT, and you saw one LEFT and two RIGHT.

Then you'd sum up - 1/4 * 0 + 1/4 * (1/3)^1*(2/3)^2 + 1/4 * (2/3)^1*(1/3)^2 + 1/4 * 0.

That would give you the posterior distribution of - some stuff just cancels out there, actually - 1^1*2^2 + 2^1*1^2 - and then after dividing through by that -

Actually, a lot of this stuff is going to cancel out anyways, isn't it?

But finish the calculation anyways.

After that, we get the 1/3 hypothesis with posterior probability 4/(4+2), and the 2/3 hypothesis with posterior 2/(4+2).

Then to get the prediction of LEFT next time, it's 2/3 probability on the 1/3 hypothesis which contributes 2/9 to LEFT, and 1/3 probability on the 2/3 hypothesis which contributes 2/9 to LEFT, and it comes out as 4/9 total prediction for LEFT.  Which does at least match that they originally saw more RIGHT than LEFT.

Now do the infinitary version.

Okay yeah, if she's not just allowed to throw calculus at it, or take limits in a way that just rederives Golarion-style calculus, Asmodia isn't really seeing how the dath ilan dubious-infinitary-reasoning version works.

Ione already knows the answer is N+1 / (N + M + 2) via a perfectly valid combinatorics argument!  She shouldn't have to rigorously prove things when she already knows the correct answer, or even prove them using dubious infinitary arguments!  This is Nethysian abuse!  Aren't only Asmodeans supposed to treat people like this?  She's sure it's not how Nethysian math classes work.

Keltham starts by considering the case where N balls go LEFT and 0 balls go RIGHT.

Then every point lying a fraction p along the way from the LEFT side to the RIGHT side, has a likelihood p^N of yielding the data we saw.

Retroactive parenthetical class chatter

One student is having some trouble parsing this, despite Keltham's gainful attempts to speak slowly.  The p is a position along a line from 0 to 1, and the p^n is 'if the 0-ball landed at that point, how likely it is that it yielded the data we saw'.  But p=0 at the far left, shouldn't it be infinitely likely it be to the left of the furthest-left point instead of no chance whatsoever -

A moment later, they've worked out that, no, the next ball has probability 0 of being any further left than as far left as possible, and the whole thing flips around in their head.

But by then, of course, Keltham has already continued talking.

The prior probability is the same everywhere, so the posterior function will be proportional to p^N everywhere.  If you wanted the whole thing to sum to 1, you'd divide every point by whatever p^N sums to between 0 and 1.

Retroactive parenthetical Alexandre intertag

        "... Excuse me." Alexandre raises his hand, since showing weakness in this environment appears to be less bad than possessing it. 'Divide every point by whatever p^N sums to between 0 and 1' - can I ask you to expand on that?"

Keltham shall be happy to do so!  Thank you for providing any feedback on where anyone is getting lost.

As a simple example, take the case of N=1, so p^N=p.  If you graph that, you get a straight line from likelihood=0 at p=0, to likelihood 1 at p=1.  That graph looks like a triangle filling half of a square with side 1.  So by visual reasoning, the sum 1/INF * [0/INF + 1/INF + 2/INF + ... + INF/INF] = 1/2 in the end.

Proof by dubious infinitary mathematics, now that they know the correct answer by simpler visual reasoning:

1 + 2 + ... + INF  =  INF*(INF+1)/2
   =>    1/INF * [0/INF + 1/INF + 2/INF + ... + INF/INF]
       = 1/INF * [1/INF * { 1 + 2 + ... + INF }]
       = 1/INF * [1/INF * { INF*(INF + 1)/2 }]
       = INF * (INF + 1) / (INF * INF * 2)
       = 1/2

Then to find a probability distribution proportional to that prior-times-likelihood object, but which sums to 1 to form the posterior, they divide the whole thing by 1/2 or multiply it by 2.

1/INF  *  [0/INF + 1/INF + ... + INF/INF]  /  (1/2)   = 1.

As Keltham is saying this, it occurs to him that it may not be obvious to the candidates why you'd do that, depending on what got covered in previous Probability classes and how well.  He'll expand on that too.

If they were just tracking, say, 1/2 prior probability of propensity 1/3 versus 1/2 prior probability of propensity 2/3, the prior-times-likelihood object would look like {1/2 * 1/3, 1/2 * 2/3} and the sum over the prior-times-likelihood object would look like {1/6 + 1/3} = 1/2.  Which is just what it should be, because if you start with 50-50 on 1/3 vs. 2/3 then your prior probability of seeing the ball go LEFT is indeed 1/2.

Conversely, the prior-time-likelihood object for seeing RIGHT is {1/2 * (1 - 1/3), 1/2 * (1 - 2/3)} = {1/2 * 2/3, 1/2 * 1/3} = {1/3, 1/6} again summing to 1/2.

This should look like it makes lots of sense.  On priors, you think either the ball goes left, or the ball goes right, but not both, nor neither, so your prior probability of LEFT plus your probability of RIGHT should sum to 1.

But after actually seeing the ball go LEFT and updating on that, you throw out all the possible worlds from your probability distribution where the ball went RIGHT, and live only inside the worlds where the ball went LEFT.  After seeing LEFT, then, within this kind of hypothesis where you're not questioning the whole setup from outside, you're certain that the ball went LEFT.  It's now probability 1 that the ball went LEFT.  The probability 1, of what is real, is divided up among the surviving worlds; the surviving worlds should have probability proportional to their prior times their likelihood, but summing to 1.

So you divide through {1/6, 1/3} by its total probability of 1/2 to get {1/3, 2/3} as the posterior probability.

In symbolic terms, there's:

P("propensity 1/3") = 50%
P("propensity 2/3") = 50%
P(TRUE) = P("propensity 1/3") + P("propensity 2/3") = 1
P(LEFT ◁ "prp 1/3") = 1/3
P(LEFT)
  = P(LEFT & "prp 1/3") + P(LEFT & "prp 2/3")
  = P(LEFT ◁ "prp 1/3")*P("prp 1/3") + P(LEFT ◁ "prp 2/3")*P("prp 2/3")
  = 1/3*50% + 2/3*50% = 1/2
P("prp 1/3" ◁ LEFT) = P("prp 1/3" & LEFT) / P(LEFT) = 1/6 / (1/2) = 1/3
P("prp 1/3" ◁ LEFT) + P("prp 2/3" ◁ LEFT) = P(TRUE) = 1

So to get the posterior probability on all the hypotheses between 0 and 1, after seeing N balls go left, they take the prior-times-likelihood object, which is 1/INF * [ (0/INF)^N + (1/INF)^N + ... + (INF/INF)^N ], and divide it through by its sum so that the probability inside the whole thing sums to 1.  If we started out certain that the ball has a propensity between 0 and 1 to go LEFT or RIGHT, and then we observe N LEFTs, we end up still certain that the ball has some propensity between 0 and 1 to go LEFT.

So the question then becomes:  1/INF * [ (0/INF)^N + (1/INF)^N + ... + (INF/INF)^N ] = ?

Now, before they try to prove a correct answer, they should, of course, first figure out what the correct answer is, so they can prove it.

The correct answer in this case obviously has to be that it sums to 1/(N+1).

Why?  Well, you saw N balls go LEFT.  On the first round, that had probability 1/2.  On the second round, you thought there was a 2/3 chance of seeing LEFT, and then you saw it, so your joint probability over all predictions is now 1/2 * 2/3 = 1/3 after seeing two LEFT.  On the third round, you think there's a 3/4 chance of seeing LEFT, so your joint probability goes to 1/2 * 2/3 * 3/4 = 1/4 after seeing three LEFT.

Since your joint probability of seeing N balls go LEFT is 1/(N+1), the sum from 0 to 1 of an infinite number of points each with weight p^N for how much they predicted seeing N balls going LEFT, must clearly end up as 1/(N+1).

Pilar is feeling, on some deep instinctive level, that her soul may be in danger here.  This is not how math works.

It's fine!  They derived that the final answer has to be 1/(N+1) from combinatoric reasoning about the probability, which is simple and reliable, so they now know that's where any dubious infinitary reasoning about the same probability ought to end up.

The problem before them, then, is to sum up:

1/INF * [ (0/INF)^N + (1/INF)^N + (2/INF)^N + ... + (INF/INF)^N ]

To be clear on what this is intended to mean:

The terms 0/INF, 1/INF, 2/INF up to INF/INF are an infinite number of distinct hypotheses covering the spectrum from 0 to 1 of LEFT frequencies;

Raising a hypothesis for a frequency between 0 and 1, to the power of N, gives that hypothesis's likelihood for generating the observed data of N LEFTs;

And the 1/INF term at the start, is being factored out from every hypothesis in the sum having 1/INF prior probability.

Where, again, the answer must be 1/(N+1).

"...am I getting it right that the contribution from (2/INF)^N is - obviously just 0?"

Obviously.  If you pick any finite number, you're starting infinitely close to a 0 probability of generating any LEFTs.

It's only once you start getting a finite fraction of the way towards infinity, that dividing by infinity won't just yield 0 again, intuitively speaking.

Don't worry, only 0% of the numbers between 0 and INF are finite, so it makes sense that they'd contribute 0% of the likelihood.