= 27d567bc2d48d9f40e9ccc20ea370b15)
Coscott comments on Even Odds - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (46)
A less complex way of making bets on probabilities, which I learned at a finance internship, is for one player to "make a market" on the outcome of interest--that is, offer to buy or sell contracts whose value settles to, e.g., $1 if the outcome is true and $0 if the outcome is false. If you make a market with a "spread"--e.g. buy at .40, sell at .60--then you can safely name your probability, as the counterparty in this anecdote asks you to do, without giving up a large advantage. For example, at my internship one common thing was precommitting to trade on any market of a given spread.
This has the advantage of being easier to remember and allowing multiple parties, but the disadvantage of (probably, I think) being subject to gaming and market dynamics which distract from the probabilities (but are interesting in their own right).
I don't think it is too difficult to remember. You put in the square of probability you think you're correct minus the square of probability he thinks you are correct all times 25. He uses the same algorithm.
A: 60% confidence B: 30% confidence
My intent is to demonstrate that, while the above is probably incorrect,
is not an adequate explanation to remember and get the right result out of, because the calculations I specified above are my genuine interpretation of your statements.
(this problem persists for every value of p and q, whether they total to above 1 or not)
Somebody replied with an explanation of how I was basically omitting the relativization of 'you' when considering what values to use.
That is, B should bet according to his confidence that he is correct, which in my case would be 70%..
Since you know it's wrong, how about you try again to make sense of it?
http://lesswrong.com/lw/jgv/even_odds/ad6y
Neither probability should be <50%, you take the probability that your opinion is the right one, not whether the proposition is true or false.
In your example B would be betting against his beliefs, thus the negative result.
The right calculation: A = 0.6 B = 0.7
Edit:
actually, it's sufficient that A and B sum to over 1. Since you can always negate the condition, the right calculation here is:
Also, apparently I can't use the retract button the way I wanted to use it.