UnexpectedValues

I'm a 2nd year PhD student at Columbia. My academic interests lie in mechanism design and algorithms related to the acquisition of knowledge. I write a blog on stuff I'm interested in (such as math, philosophy, puzzles, statistics, and elections): https://ericneyman.wordpress.com/

Pseudorandomness contest: prizes, results, and analysis

For what it's worth, the top three finishers were three of the four most calibrated contestants! With this many strings, I think being intentionally overconfident as a bad strategy. (I agree it would make sense if there were like 10 or 20 strings.)

Pseudorandomness contest: prizes, results, and analysis

Yours was #104 -- you did well!

Great minds might not think alike

Two things, I'm guessing. First, there's the fact that in baseball you get thousands of data points a year. In presidential politics, you get a data point every four years. If you broaden your scope to national and state legislative elections (which wasn't Shor's focus at the time), in some sense you get thousands per election cycle, but it's more like hundreds because most races are foregone conclusions. (That said, that's enough data to draw some robust conclusions, such as that moderate candidates are more electable. On the other hand, it's not clear how well those conclusions would translate to high-profile races.)

Second, electoral politics is probably a way harder thing to model. There are many more variables at play, things shift rapidly from year to year, etc. Meanwhile, baseball is a game whose rules of play -- allowable actions, etc. -- are simple enough to write down. Strategy shifts over the years, but not nearly as much as in politics. (I say that without having much knowledge of baseball, so I could be... um, off base... here.)

Overall numbers won't show the English strain coming

That's true. My response is that if I recall correctly, people didn't seem to react very strongly to what was happening in Italy a couple weeks before it was happening here. So I'm not sure that a surge in the UK would inform the US response much (even though it should).

Overall numbers won't show the English strain coming

Yeah, you're right. 1.3 was the right constant for Covid in March because of a combination of not being locked down and having more and more tests. This was my attempt to make a direct comparison, but maybe the right way to make that comparison would be to say "if R=1.65 (which I'll trust you that it leads to a constant of 1.08), we will react about X days slower than if we started from scratch."

What is X? The answer is about 60 (two months). On the one hand, that's a lot more than the 2-3 weeks above; on the other hand, it's less scary because R is lower.

Great minds might not think alike

Thanks! I've changed the title to "Great minds might not think alike".

Interestingly, when I asked my Twitter followers, they liked "Alike minds think great". I think LessWrong might be a different population. So I decided to change the title on LessWrong, but not on my blog.

Predictions for 2021

I like logarithmic better in general, but I decided to use Brier for the pseudorandomness contest because I decided I really cared about the difference between a 60% chance (i.e. looks basically random) and a 40% chance (kind of suspect). The log rule is better at rewarding people for being right at the extremes; Brier's rule is better at rewarding people for being right in the middle.

Regarding bets: I'm willing to make bets, but won't have a blanket policy like "I'll take a bet with anyone who disagrees with me by 10% or more", because that opens me up to a ton of adverse selection. (E.g. I wouldn't bet with Zvi on COVID.) So... feel free to message me if you want to bet, but also be aware that the most likely outcome is that it won't result in a bet.

(Also, the better I know you, the more likely I am to be willing to bet with you.)

2021 New Year Optimization Puzzles

Puzzle 3 thoughts: I believe I can do it with

1

coins, as follows.

First, I claim that for any prime q, it is possible to choose one of q + 1 outcomes with just one coin. I do this as follows:

- Let p be a probability such that (Such a p exists by the intermediate value theorem, since p = 0 gives a value that's too large and p = 1/2 gives a value that's too small.)
- Flip a coin that has probability p of coming up heads exactly q times. If all flips are the same, that corresponds to outcome 1. (This has probability 1/(q + 1) by construction.)
- For each k between 1 and q - 1, there are ways of getting exactly k heads out of q flips, all equally likely. Note that this quantity is divisible by q (since none of 1, ..., k are divisible by q; this is where we use that q is prime). Thus, we can subdivide the particular sequences of getting k heads out of q flips into q equally-sized classes, for each k. Each class corresponds to an outcome (2 through q + 1). The probability of each of these outcomes is which is what we wanted.

Now, note that 2021*12 - 1 = 24251 is prime. (I found this by guessing and checking.) So do the above for q = 24251. This lets you flip a coin 24251 times to get 24252 equally likely outcomes. Now, since 24252 = 2021*12, just assign 12 of the outcomes to each person. Then each person will have a 1/2021 chance of being selected.

Conjecture (maybe 50% chance of being true?):

If you're only allowed to use one coin, it is impossible to do this with fewer than 24251 flips in the worst case.

Question:

What if you can only use coins with rational probabilities?

Great minds might not think alike

Thanks for the feedback. Just so I can get an approximate idea if this is the consensus: could people upvote this comment if you *like* the title as is (and upvote mingyuan's comment if you think it should be changed)? Thanks!

Also, if anyone has a good title suggestion, I'd love to hear it!

I'm not conditioning on any configuration of points. I agree it's false for a given configuration of points, but that's not relevant here. Instead, I'm saying: number the intervals clockwise from 1 to n + 2, starting with the interval clockwise of Z. Since the n + 2 points were chosen uniformly at random, the interval numbered k1 is just as likely to have the new point as the interval numbered k2, for any k1 and k2. This is a probability over the entire space of outcomes, not for any fixed configuration.

(Or, as you put it, the average probability over all configurations of the probability of landing in a given interval is the same for all intervals. But that's needlessly complicated.)