My intent is to demonstrate that, while the above is probably incorrect,
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.
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)
A: 60% confidence B: 30% confidence
af = .6 **2 == .36
bf = .3 **2 == .09
A pays (af—bf) * 25 == $6.75
B pays (bf—af) * 25 == -$6.75?!?!
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%..
B bets (.49 - .16) * 25 == $8.25
A bets (.36 - .09) * 25 == $6.75
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.