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

Well, anyways!  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)
log2(3)  ≈-  1.5
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; 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, uh, Keltham doesn't actually have a calculator to compute 19/12 but it'll presumably be closer to 1.58496 than the previous estimate of 1.5, like it'll be 1.57 or 1.6 or something like that.

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 the score is about -0.17 'bits', to borrow the Baseline term.  Does anybody see how that figure gets derived?

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)
log2(3)  ≈-  1.5
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, uh, Keltham doesn't actually have a calculator to compute 19/12 but it'll presumably be closer to 1.58496 than the previous estimate of 1.5, like it'll be 1.57 or 1.6 or something like that.

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?

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, uh, Keltham doesn't actually have a calculator to compute 19/12 but it'll presumably be closer to 1.58496 than the previous estimate of 1.5, like it'll be 1.57 or 1.6 or something like that.

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?

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, uh, Keltham doesn't actually have a calculator to compute 19/12 but it'll presumably be closer to 1.58496 than the previous estimate of 1.5, like it'll be 1.57 or something like that.

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?

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?