did you expect a different branch of humanity to have familiar math customs?
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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.

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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.

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"Is that actually Lawful reasoning?"

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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.

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"...can’t you just think of this as permutations of 3 things?"

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"Hmmm.  Can you?"

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"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."

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"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."

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"We don't know that!  In the - 'object-level' - classroom!"

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"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..."

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"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."

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"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?"

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"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.)

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"This one comes up LEFT, though.  Your prediction FAILED.  All your theories were probably WRONG."

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"No, we just ran into a 1/4 chance, that happens sometimes.  Like with the chance of failure on an Augury."

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"I suppose.  We should just discard the evidence, then, since it's only to be expected that reality gives us the wrong result, sometimes.  You predict 3/4 RIGHT again next time?"

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You can, in fact, get 10-foot range on Prestidigitation, though not when you're a wizard as fresh as Keltham.  One of the students nearest that wall makes their own change to it without leaving their desk.

[----3---0------------1-----2--]

"3/5 right, 2/5 left," they declare.

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What if the 2-ball actually landed to the left of the 1-ball?  They're not told which of those was true!

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...if those two possibilities are equally probable, then they can sum the probabilities, right?

1/2 chance of 3/5 plus 1/2 chance of 3/5 equals 3/5.

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You don't need to label them!

[---|---0---|---|---]

"That's five landing regions, two to the left and three to the right."

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Does anyone want to object to the invalidity of that reasoning -

- or alternatively, state a general rule for what to predict on the next round, if you previously got N balls LEFT and M balls RIGHT of the 0-ball, and that's all you know?

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"If you have N balls LEFT, there are N+1 landing regions LEFT, and M balls RIGHT gives M+1 landing regions RIGHT, so my first thought is (N+1)/(M+1)."

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Ah, yes, so if they have seen one ball go LEFT, and zero balls go RIGHT, the chance of seeing it go LEFT on the next round is (1+1)/(0+1)=2.

So the probability of seeing LEFT next time is 2.  That sounds pretty reasonable to Keltham?  Though clearly anthropics were involved at some point.  Maybe the balls are duplicating people once they start to go LEFT?

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They obviously meant, like, a ratio of (N+1):(M+1), not to calculate the probability that way!

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So, like... after seeing one ball go right, the chance of seeing it go left on the next round, is half the chance of seeing it go right.  That's what the (N+1)/(M+1) quantity is, how many times more likely LEFT is than RIGHT, in this case, (0+1)/(1+1) = 1/2.

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