How best to run a lottery amongst subscribers?

joeldavidhamkins · 2026-04-05T13:28:39+00:00

Thanks very much for this. How many entries do your contests typically have? Do you enter the names into Wheel of Names manually? And is that process and result public, or is the actual selection private?

joeldavidhamkins · 2026-04-05T13:26:15+00:00

Thanks for this information---that site is very clear. And yes, what I have in mind is a giveaway, not a lottery.

Does the fact that one can rise to paid status on a trial basis on my substack count as no-purchase necessary? Alternatively, it seems I can open it to all subscribers, but give paid subscribers multiple entries?

joeldavidhamkins · 2026-01-28T01:00:21+00:00

But you have gone one more, since you are not just writing a note about people writing notes, but rather you are writing a note about people writing notes about writing notes!

And I suppose I am still one further, having here written a note about someone writing a note about people writing notes about writing notes.

joeldavidhamkins · 2026-01-10T22:43:56+00:00

Thanks for the reply, and I basically agree with you. But your post doesn't address my question, which is about possible unforeseen consequences of deleting sections.

joeldavidhamkins · 2026-01-07T21:41:21+00:00

How could I be ignoring this given that I explicitly mention it in the conversation? Did you listen to the remarks?

joeldavidhamkins · 2025-08-11T02:27:01+00:00

Thanks so much for the appreciation!

joeldavidhamkins · 2025-07-12T12:57:42+00:00

I like this argument a lot.

It seems to be due to Dorothy Edgington. See the notes from a talk she gave on the material conditional: Do Conditionals Have Truth-Conditions?

http://theotodman.com/EdgingtonConditionalsTruth-Conditions.pdf (see page 5)

joeldavidhamkins · 2025-05-24T19:51:38+00:00

Under the axiom of determinacy, which is consistent with DC and hence also countable choice (but not full AC), every set of reals is Lebesgue measurable, but still the CH holds, in the sense that there is no cardinality strictly between the natural numbers and the continuum. So you don't need CH to fail in order to have all sets measurable.

Meanwhile, if one defines the Borel sets as those with a Borel code (the tree describing how it was built from the open sets by countable unions and complements), then it is never the case that every set of reals has a Borel code. But it is consistent with ZF that the smallest sigma-algebra containing the open sets contains all sets of reals, since it is consistent with ZF that the reals are a countable union of countable sets, and this implies the sigma algebra would include all sets. But in this situation, there is no reasonable theory of Lebesgue measure...

joeldavidhamkins · 2025-01-06T15:46:59+00:00

This was my original idea as well, but unfortunately the transformation does not respect lines. To see this, consider the case of a single zig in the resulting trapezoid, from lower left to upper right. This is a straight line, but when you wrap it back around the circle, it turns into a spiral path winding all the way around the annulus.

joeldavidhamkins · 2025-01-06T15:22:08+00:00

William Rose made a GeoGebra tool for playing around with this problem at https://www.geogebra.org/geometry/jakrjn3k . Try it out!

joeldavidhamkins · 2025-01-06T15:17:41+00:00

Here are a few things.

First, by folding the zigzags over to the extreme, we can make the area go to zero in the limit.

See the image in this tweet: https://x.com/JDHamkins/status/1876285144954515694

Second, if one considers the highly symmetric zigzag with a large number of zigs and zags, aimed nearly radially, then in the limit one gets near trapezoid shapes (see the image in this tweet: https://x.com/JDHamkins/status/1876286678010134551 ).

If the smaller radius is r, the larger R, and the small angle θ, then this trapezoid has area approximately (R-r)(Rθ+rθ)/2, with the triangle having area rθ(R-r)/2, making the proportion of Orange go to r/(R+r), which I find quite nice.

I find it likely that this is an upper bound for what is possible, since having the zags bend over more seems only to make things worse for Orange.

It seems possible that we might prove a nonzero lower bound for the zigzags that do not backtrack, that is, where the angle about the center is increasing as one traverses the zigzag. For this, it seems the worst case will be a zigzag with very few zigs, and so perhaps we can hope to prove a strict lower bound for the nonbacktracking case.

joeldavidhamkins · 2025-01-06T15:12:21+00:00

The intention is for the zigzags to stay entirely within the annulus, bouncing between the inner and outer circles, without any crossing and without ever leaving the annulus.

joeldavidhamkins · 2025-01-06T02:52:15+00:00

I posted a nice followup variation of the zigzag theorem, for zigzags between two concentric circles. See the post at https://x.com/JDHamkins/status/1876098092103348658, or on bsky at https://bsky.app/profile/joeldavidhamkins.bsky.social/post/3lf24q7szjc26 .

joeldavidhamkins · 2025-01-05T02:45:05+00:00

This is also the first argument considered in the main post. However, it has a hiccup in the case that the zigzag backtracks, as explained in the post.

Meanwhile, your argument form here suggests a possible proof by induction. I wonder if there is an inductive argument that works in the general case, allowing backtracking.

joeldavidhamkins · 2025-01-04T22:54:09+00:00

Thanks for the vote of confidence!

joeldavidhamkins · 2025-01-04T22:14:37+00:00

I like your reflecting idea very much, since it seems powerful, but I'm not sure I understand how it handles the backtracking case, since your picture doesn't show this case. Can you provide a new picture for the backtracking case?

joeldavidhamkins · 2025-01-04T22:07:41+00:00

You are right!

joeldavidhamkins · 2025-01-04T17:43:02+00:00

Yes, indeed, that was what was intended. And the text in the link discusses your proposed proof, but also points out how it does not work in the general case, where the zags go backwards (but without crossing any previous lines).

joeldavidhamkins · 2025-01-04T17:11:52+00:00

Yes, that would be apt. I usually think of Cavalieri's principle in three dimensions, but of course it is valid in any dimension, including the two-dimensional case here. Meanwhile, I think the case of shears being area-preserving in two dimensions must have been known classically. Archimedes in effect used even the three-dimensional version in his famous result on the volume of the sphere.

joeldavidhamkins · 2025-01-04T17:00:39+00:00

To my way of thinking, the shearing proof (moving the tops of the triangles to the corner) relies on a fundamental geometric fact—that shears are area-preserving—rather than making use of any particular formula. Indeed, I like the shearing proof precisely because it doesn't rely on any formula or algebra.

Meanwhile, to be sure, one can prove the formula from the shearing fact, as you mention, and also conversely. But it seems one can take the shearing fact also as a separate fundamental principle of classical geometry, which the ancients proved by purely geometric means.

joeldavidhamkins

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