0.088 * 328e6 / 150 = 192k 🤔
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Fatality% respondents report knowing someone in their closest 150 who've died. At current poll result, our collective pool of all our 150 closests is averaging about 9%*1.1/150 dead w/ C19. Which is 0.066%. Current reported US # is 207k/328M = 0.063%. Extremely close!
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Poll's now dropped to 8.8%, which (0.0645%) puts our own pool's estimate at only 2% off of the US official reported total rate.
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Better estimate: (1-x)^150=(1-fatality%). 150*ln(1-x) = ln(1-.088). x=0.0614%. So, running with all the spherical-horses assumptions, we're "actually" running about 7.5% off of the current reported USA-overall totals. Still strangely confirmatory!
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It's combinatorics: Let's say you & 1000 friends each receive 10 random marbles from a giant bag of marbles. Only 8.8% got at least one red marble in your 10 marbles. To estimate: What % of the marbles in the bag are red? Solution:
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Each side of this equation (1-x)^150=(1-0.088) is the probability that an observer sees zero fatalities.
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The big weakness here is multi-counting due to such opt-in polls being within social graphs. In the limit, if you happen to administer it to a single Dunbar village where 1 person died, it would show 100%, way off from 0.064. But that’s a feature if you treat it as a local sample
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Hmm sampling with replacement from the same bag would account for that. There are x% red balls. Everybody takes out 150 balls and then puts them back in. Model graphs with partial replacement. I hand you my 150 same, you randomly replace 50 with a different 50 if our overlap ~100


