I think we can actually do an explicit expected-utility and value-of-information calculation here:
Let one five-star book = one util
Each book’s quality can be modelled as a rate p of producing stars.
The star rating you give a book is the sum of 5 Bernoulli trials with rate p.
The book will produce p utils of value per read in expectation.
To estimate p, sum up the total stars awarded n+ and total possible stars n.
The probability distribution is then Pr(p=P)∝Pn+(1−P)n−n+ (assuming uniform prior for simplicity).
For any pair of books, we can compute the probability that book 1 is more valuable than book 2 as ∫10Pr1(p=P)Pr2(p<P)dx.
Let’s say there’s a prescribed EA reading list.
Let people who encounter the list be probabilistic automata.
These automata start at the top of the list, then iteratively either: 1) read the book they are currently looking at, 2) move down to the next item on the list, 3) quit.
Intuitively, I think this process will result in books being read geometrically less as you move down the list.
For simplicity, let’s say the first book is guaranteed to be read, the next book has a 50% chance of being read, then 25%, …, and then n-th book has 2−n chance of being read (with n starting at zero).
The expected value of the list is then ∑Nbooki=02−iEPri[p]
To calculate the value of information for reading a given book, you enumerate all the possible outcomes (one-star, two-stars, …., five-stars), calculate the probability of each one, look at how the rankings would change, and re-calculate the the expected value of the list. Multiply the expected values by the probabilities et voila.
This is brilliant!
I think we can actually do an explicit expected-utility and value-of-information calculation here:
Let one five-star book = one util
Each book’s quality can be modelled as a rate p of producing stars.
The star rating you give a book is the sum of 5 Bernoulli trials with rate p.
The book will produce p utils of value per read in expectation.
To estimate p, sum up the total stars awarded n+ and total possible stars n.
The probability distribution is then Pr(p=P)∝Pn+(1−P)n−n+ (assuming uniform prior for simplicity).
For any pair of books, we can compute the probability that book 1 is more valuable than book 2 as ∫10Pr1(p=P)Pr2(p<P)dx.
Let’s say there’s a prescribed EA reading list.
Let people who encounter the list be probabilistic automata.
These automata start at the top of the list, then iteratively either: 1) read the book they are currently looking at, 2) move down to the next item on the list, 3) quit.
Intuitively, I think this process will result in books being read geometrically less as you move down the list.
For simplicity, let’s say the first book is guaranteed to be read, the next book has a 50% chance of being read, then 25%, …, and then n-th book has 2−n chance of being read (with n starting at zero).
The expected value of the list is then ∑Nbooki=02−iEPri[p]
To calculate the value of information for reading a given book, you enumerate all the possible outcomes (one-star, two-stars, …., five-stars), calculate the probability of each one, look at how the rankings would change, and re-calculate the the expected value of the list. Multiply the expected values by the probabilities et voila.
Can I get the data please?
Cool idea! Send you a message.