I am a Dragon. I emerge from my Egg. While I still have a possible bet in the pipeline, Brian Schmidt now has a real bet lined up for a total of $6-$9k. Personally I think that 0.15 oC/decade is more likely than not, but that less is a realistic possiblity given natural variablity. So while its good from an expected return view it could potentially be bad from a PR view. I wonder how important the latter is... probably not very, in 10 years time.
Brian has a spread of bets across 10, 15 and 20 years which is an interesting way of doing things. I'm still looking forward to making big money (a-la James :-) when a real market for these things developes and reselling good-odds bets for a sure fire profit becomes possible.
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Posted by: copyworld | April 25, 2007 6:40 AM
The other question is, what will $9,000 be worth in twenty years' time?
Posted by: Joseph j7uy5 | April 25, 2007 8:11 AM
Legitimate point on the PR risk. Hopefully I'd only lose on the even-odds bet but win on the 2:1, so the PR effect would be limited. And I only lose on even-odds if the increase is less than 0.13C, as opposed to voiding, so the risk is lower.
Posted by: Brian Schmidt | April 25, 2007 12:46 PM
Adjusted for inflation, I hope.
Posted by: uBeR | April 25, 2007 1:19 PM
Not inflation-adjusted. I'd guesstimate the 20 year bets will be worth half their present value, but YMMV.
Posted by: Brian Schmidt | April 25, 2007 4:03 PM
The first comment from "copyworld" is some kind of spam link.
Posted by: Benjamin Franz | April 25, 2007 9:11 PM
I too am looking forward to making big bucks out of the alarmists.
If +0.15 is 1:1 and +0.1 is 2:1, then assuming the second derivative of odds is reasonably small, a 0.0 rise in 10 years should be worth 8:1 and -0.05 16:1.
I'll take the 0.05 fall in 10 years at 16:1. US$10,000.
Posted by: mugwump | April 27, 2007 2:57 PM
Mugwump - you're showing the danger of extrapolating. The 1:1 and 2:1 bets were rough guesses of what both my opponent and myself perceived as a relatively good bets - trying to draw mathematical equations from that extending to the outer edges of a graph line is ridiculous.
But, if you think the 16:1 fall is a good bet for your side, then by your logic (not mine) you should also think the 2:1 bet I negotiated is good. Contact me if you're truly interested.
Posted by: Brian Schmidt | April 30, 2007 1:12 PM
Brian, I did say "assuming the second derivative of odds is reasonably small" (more accurate would have been "assuming the second derivative of log-odds is reasonably small").
Your bet is interesting because it gives us two points on your log-odds curve, from which we can get a first-order estimate of your log-odds function, which is all I am doing (fitting a straight line to the log-odds).
Assuming your log-odds function is twice-differentiable (definitely not an unreasonable assumption), a straight-line fit is locally accurate. So the question comes down to what is "local". Local here is measured in terms of the independent variable, temperature. I would have thought -0.05C/decade wasn't so far from +0.15C/decade and +0.1C/decade as to make my estimate "ridiculous" (as you put it). Eg, if you do a Taylor-series expansion about +0.1C/decade the second-order term at -0.05C/decade is 0.0225 (= 0.15 squared). At 0.0C/decade it's 0.01 (=0.1 squared). So the second derivative has to be pretty big to make the second-order terms significant.
I might take you up on your 2:1 bet after I model my own subjective probability function in more detail. I know enough already to definitely take you up on the implicit 16:1, 8:1 and perhaps even 4:1 bets.
Posted by: mugwump | April 30, 2007 5:19 PM