Quick question on log odds: from what I've read, you calculate the log odds in decibels by treating the odds as a fraction (i.e. dividing one "odd" from the other), taking the base 10 log of that, and multiplying by ten. But at http://www.overcomingbias.c..., it seems that they are calculated using the natural log:

But when you transform to odds ratios, 0.502 and .503 go to 1.008 and 1.012, and 0.9999 and 0.99999 go to 9,999 and 99,999. And when you transform to log odds, 0.502 and 0.503 go to 0.08 decibels and 0.12 decibels, but 0.9999 and 0.99999 go to 92 decibels and 115 decibels.

Earlier in the same post, base 10 seems to be used:

For example, let's say that the prior probability of a proposition is 0.0001 - this corresponds to a log odds of -40 decibels.

So my question is, am I mistaken as to the definition of log odds, or was this just a mistake in the post?

Also, why multiply by 10? I do not see how that helps out as far as further calculations go.

Does Bayes' theorem make racial stereotyping okay?

Quick question on log odds: from what I've read, you calculate the log odds in decibels by treating the odds as a fraction (i.e. dividing one "odd" from the other), taking the base 10 log of that, and multiplying by ten. But at http://www.overcomingbias.c..., it seems that they are calculated using the natural log:

But when you transform to odds ratios, 0.502 and .503 go to 1.008 and 1.012, and 0.9999 and 0.99999 go to 9,999 and 99,999. And when you transform to log odds, 0.502 and 0.503 go to 0.08 decibels and 0.12 decibels, but 0.9999 and 0.99999 go to 92 decibels and 115 decibels.

Earlier in the same post, base 10 seems to be used:

For example, let's say that the prior probability of a proposition is 0.0001 - this corresponds to a log odds of -40 decibels.

So my question is, am I mistaken as to the definition of log odds, or was this just a mistake in the post?

Also, why multiply by 10? I do not see how that helps out as far as further calculations go.