Dath ilani kids, before they run into logarithms, have prior experience with seeing numbers as bags of prime factors.  Maybe running over that for a few minutes will help with priming this pump?

15 is a bag of a 3 and a 5.
4 is a bag of two 2s.
15*4 = 60, so 60 is a bag of two 2s, a 3, and a 5.
If you multiply 2 and 3, you get 6.
So if you divide 60 by 6, you should get a bag of one 2 and a 5.
2 times 5 is 10.  Checks out, right?

Now make up your own bags with numbers and play with that to see if your reasoning by bags-of-factors gets you the right answer.

Well, sure, you can use 4s as factors and see what happens?  But if you want to turn numbers into unique bags of numbers, each number in the bag has to not be made up of any numbers smaller than itself.

This the students can follow along with. - the application to creating a scoring rule is not clear.

Does it help if he mentions that a bag of 1.58496 2s is almost exactly 3?

....how would you possibly figure that out if you didn't know it, though.

Did anybody happen to learn calculus since Keltham mentioned they should do that?

Yep!! It was one of the major things they did while Keltham was at the Imperial Palace, along with triage on the library.

....they didn't get very far, since that was only one day and they didn't have textbooks or anything.

That's weird, he'd expect significant progress on calculus if you were spending a significant part of a day on it.

How were they studying calculus at all without textbooks?  Tutoring from somebody?

Some of Security knew some and were willing to trade favors once they were back alive. 

Those favors need to be charged to the project budget somehow.

Onward then!  They're going to need calculus anyways, to get all the way through proving that the logarithmic scoring rule works correctly, and the calculus you need for that exact thing shouldn't be hard to teach in a few minutes even if Keltham has to do it from scratch.  But let's keep the focus on logarithms for now.

So first of all, remember that Asmodia had already worked out that since 9 is a bit more than 8, there should be slightly over three 2s inside a bag of two 3s.  So 1.58496 2s inside a bag of one 3 shouldn't be surprising.

And is that one fact Asmodia found, going to be the only fact like that which exists?  Three 3s is 27, and two 5s is 25, so there should be slightly less 2s in a bag of two 5s than in a bag of three 3s.  Say there's a thrice-bit-more than 4.5 2s in a bag of three 3s, then a little fewer 2s in a bag of two 5s, so there ought to maybe be 4.5 2s in a bag of two 5s and 2.25 2s in a bag of one 5.  The actual number is 2.32193 or so, which is, as one would expect, a tad more 2s than are in a 4.

You could also notice that a bag of seven 2s is 128, and a bag of three 5s is 125, so you'd expect a tad less than 7/3 2s in one 5, which would give you an estimate of 2.333... 2s per 5.  Not far off at all, right?

Yes, Keltham is writing this down on the whiteboard:

3*3*3 = 27   <=>   log3(27) = 3
5*5 = 25      <=>   log5(25) = 2

2*log2(5) = log2(25)  ≈+  log2(27) = 3*log2(3)
3*log2(2) = 3  ≈+  log2(9) = 2*log3(3)
log2(3)  +≈  1.5
actually log2(3) ≈ 1.58496
2*log2(5)  ≈  3*1.5 = 4.5
log2(5)  ≈  2.25
log2(125) = 3*log2(5)  ≈+  log2(128) = 7
log2(5) ≈ 7/3 = 2.333...
actually log2(5) ≈ 2.32193

Now there's cleverer ways to compute this once you actually get calculus.  But it so happens that 3^12 = 531,441, and that 2^19 = 524,288.  There's slightly more than nineteen 2s in a bag of twelve 3s.  So you'd expect log2(3) to even more precisely be a tad more than 19/12, which will be 1/12 more than 1.5, so 1.58333, which is nicely closer to the true 1.58496 than the previous estimate of 1.5.

Problem time!  If you happened to have memorized the figure of 1.58496 2s per 3, you could derive that log2(8/9) ≈ -0.08496*2, for purposes of scoring a prediction of 8/9 on something that actually happened.  So score(8/9) is about -0.17 'bits', to borrow the Baseline term.  Does anybody see how that figure gets derived?

There's a lot of silent scribbling. 

Well, says Gregoria after a bit, log2(8/9) is log2(8) + log2(1/9) - that's the entire desirable scoring property that got them on this horrible tangent in the first place.

And log2(8) is 3.

And log2(1/9) is going to be negative, fractions always are. log2(1/2) was -1. log2(1/4) was -2. log2(1/8) is going to be -3, and log2(1/9) is going to be - log2(9).

She doesn't actually know why this works but she can see that 3 - (1.58496)*2 is about -.17. 

Sure.  It's just saying that you have to take around 0.17 2s out of a 9 in order to get an 8.  9 × 8/9 = 8.  3.17 2s minus 0.17 2s equals three 2s so an 8.  8/9 just literally means the number you multiply 9 by in order to get 8, so it's the number you multiply by to take 0.17 2s out of the bag.

If it's a probability of something happening 8 times out of 9 it's the same number and will score the same way, according to the scoring rule that counts 2s in things.  Which is the scoring rule that gives you the same cumulative score whether you assign 1/4 to two events, or 1/16 to their product event.

- nods.

Message: If you don't actually understand it don't act like you do.

"I don't understand why taking .17 of a 2 out of a bag of twos is a thing you're allowed to do," says Pela.

"Well, look at it this way.  A 16 is a bag of two 4s.  What happens if you take half a 4 out of the bag?"

"....you take a two out, and now you've got a bag that multiplies up to eight."

"A million is a bag of two thousands.  What happens if you take a third of a thousand out of the bag?"

"I don't know what a third of a thousand is. ...I mean I know what it is when it's 333. But it's not, here."

"A thousand is a bag of three tens.  What happens if you take a third of three tens out of a bag of twice three tens?"

"You have five tens left in the bag. So - a hundred thousand."

"Yep."  His smile goes away after a moment; it's impossible to have any sense of how well this is going when everybody is supposed to learn this at age five or six and they're adults.

"Well, if you can take half of a four out of a bag of fours, and a third of a thousand out of a bag of thousands, why not take 17 100ths of a 2 out of a bag of twos?"

"....I guess."

"I mean, there's the problem of figuring out that taking out 0.17 twos from a bag works out to multiplying the contents by roughly 8/9, but you can get that fairly precisely off nineteen twos being a bit less than twelve threes.  Possible self-study problem: rederive that yourself, convince yourself of it, prove it, without looking back at the whiteboard."

"Now, or after class?"

"How many people in this group think that's now so obvious that there's no point in proving it themselves?  Because if the answer is no, then yeah, maybe everybody pauses and tries to rederive the logic."