You should fly anyone who wins over 5k to meet with you in person. They have 1 hour to shift your credences by the same amount the already did (in bayesian terms, not % difference[1]). If they do, you’ll give them the amount of money you already did.
I imagine some people arguing in person will be able to convince you better, both because there will be much greater bandwidth and because it allows for facial expressions and understanding the emotions behind an intellectual position, which are really important.
That footnote is an important point. People need to learn to use odds ratios. Though I think that with odds ratios, the equivalent increase is to 1 - ((1/99) x ((1/99) / (10/90))) = 99.908%, not the intuitive-looking 99.9%.
Also, the interpretation of odds ratios is often counter-intuitive when comparing test groups of different sizes. If P(X) >> P(~X) or P(X) << P(~X), the probability ratio P(W|X) / P(W|~X) can be very different from the odds ratio [P(W,X) / P(W,~X)] / [P(~W,X) / P(~W,~X)]. (Hope I’ve done that math right. The odds ratio would normally just use counts, but I used probabilities for both to make them more visually comparable.)
You should fly anyone who wins over 5k to meet with you in person. They have 1 hour to shift your credences by the same amount the already did (in bayesian terms, not % difference[1]). If they do, you’ll give them the amount of money you already did.
I imagine some people arguing in person will be able to convince you better, both because there will be much greater bandwidth and because it allows for facial expressions and understanding the emotions behind an intellectual position, which are really important.
If you move someone from 90% to 99%, the equivalent increase is to 99.9% not to.. 108%??
That footnote is an important point. People need to learn to use odds ratios. Though I think that with odds ratios, the equivalent increase is to 1 - ((1/99) x ((1/99) / (10/90))) = 99.908%, not the intuitive-looking 99.9%.
Also, the interpretation of odds ratios is often counter-intuitive when comparing test groups of different sizes. If P(X) >> P(~X) or P(X) << P(~X), the probability ratio P(W|X) / P(W|~X) can be very different from the odds ratio [P(W,X) / P(W,~X)] / [P(~W,X) / P(~W,~X)]. (Hope I’ve done that math right. The odds ratio would normally just use counts, but I used probabilities for both to make them more visually comparable.)