My Faults My Own

Any human’s death diminishes me,

because I am involved in humankind.

IN  WHICH Ross Rheingans-Yoo—a sometime economist, artist, trader, expat, poet, EA, and programmer—writes on things of int­erest.

Reading Feed (last update: July 5)

A collection of things that I was glad I read. Views expressed by linked authors are chosen because I think they’re interesting, not because I think they’re correct, unless indicated otherwise.


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Blog: Don't Worry About the Vase | Spoiler-Free Review: Witcher 3: Wild Hunt (plus a Spoilerific section)

Blog: Popehat | The Fourth of July [rerun]

Blog: Tyler Cowen @ Bloomberg View | The NBA’s Reopening Is a Warning Sign for the U.S. Economy — "If so many NBA players are pondering non-participation, how keen do you think those workers — none of whom are millionaire professional athletes — are about returning to the office?"

Comic: SMBC | Saturday Morning Breakfast Cereal - Holism


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Blog: Market Design | Job market technology is diffusing slowly through the armed forces

Blog: Marginal Revolution | Tales from Trinidad barter

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Metaculus has some issues

In Zvi's 2/11 Covid update, he turned to Metaculus for help. He looked at the numbers. Becase the man is an inveterate trader, he saw odds that were Wrong On The Internet and just couldn't stop himself from creating an account to bet against it. And then he saw the payout structure and decided he was done after making a single prediction.

I spent some time with the Metaculus site and figured out how they borked this one up enough to drive away Zvi Mowshowitz. I'll try to explain it here.


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Here's a presumably current description of the scoring function that I found on the FAQ, slightly abridged:

Your score \(S(T;f)\) at any given time \(T\) is the sum of an "absolute" component and a "relative" component: \[S(T;f)=a(N)\times L(p;f)+b(N)\times B(p;f),\] where \(N\) is the number of predictors on the question.

If we define \(f=1\) for a positive resolution of the question and \(f=0\) for a negative resolution, then \(L(p;f)=\log_2(p/0.5)\) for \(f=1\) and \(L(p;f)=\log_2((1−p)/0.5)\) for

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