# Kindly comments on Typicality and Asymmetrical Similarity - Less Wrong

25 06 February 2008 09:20PM

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Comment author: 15 June 2012 11:31:49AM 0 points [-]

But the relation you're describing is itself percentage-based! If you go from the rates to, say, the time it takes to cover a distance of 100 miles, then you get (roughly) 102 and 100 hours in the first case, and 10.2 and 10 hours in the second case. These only have the same relation if we use percentage differences or ratios to think about how close two times are.

Comment author: 15 June 2012 01:47:39PM 0 points [-]

It looks to me like that was iemnitable's goal.

Comment author: 15 June 2012 05:47:01PM 1 point [-]

I thought that iemnitable was trying to justify the use of ratios when comparing speeds (as an example), and I pointed out that this requires us first to justify the use of ratios when comparing times.

Comment author: 15 June 2012 05:51:34PM 0 points [-]

Ah; I got a different impression from the great-grandparent. I agree with your point in the parent.

Comment author: 15 June 2012 09:20:49PM 1 point [-]

I was thinking of it more like: if there's a certain place I can get to in (roughly) 102 hours going 98 mph, and I want to get there in 100 hours, I need to speed up to 100 mph. Similarly, if there's a another place that I can get to in roughly 102 hours going 980 mph, and I want to get to that place in 100 hours, I need to speed up to 1000 mph.

I kind of wanted to clarify that in the original post but I hadn't really thought of a good way to express it at the time.

Furthermore, I think that your interpretation of the example even makes it more clear that it makes sense to think of it in terms of a ratio. In the first case, you've sped up by 2 mph and gotten a gain of about 2 hours, straightforward enough. But in the second case, you've sped up by 20 mph, and only gotten a gain of about 0.2 hours. Here's where I think most people's intuition is probably screaming "whaaaaaaat!?"

But if we think of it in terms of the ratios, then everything fits together nicely again and the screaming intuition voice shuts up. Plus the math we have to do to get to the right answer is a lot easier.

Comment author: 15 June 2012 10:40:13PM 0 points [-]

Oh, I see what you mean now.

(Incidentally, Eliezer's original objection can be resolved by taking logs. Suddenly although the ratios 102/100 and 100/102 are not symmetrical, log(102/100) and log(100/102) are.)