I think it results from the scale independence of log (and as soon as you add external wealth, the scale independence goes away). That makes it so you can treat every question separately (since only its scale is determined by the previous questions, and that doesn’t impact the maximization). It is a pretty result, but as I don’t think people often talk about this sort of problem I don’t know if I would call it well known or not.
It’s also cool to work back from the last question and see how conditional probabilities connect the full knowledge case and the discovery case.
And here is the prettiest part: The whole thing seems to work even if we don’t assume that the answers to the quiz questions are independent random variables. The expected utility is always equal to the entropy of the joint distribution of the variables, and the best strategy is always honesty in every turn. Note the statement about entropy, it’s a new motif. Before generalizing to non-independent variables it did not add too much value, but in the current version, it is a quite powerful statement.
For example, let the first question be a 50%-50% A-B, but let the second question depend on the first correct answer in the following way: if it was A, then the second question is a 50%-50% C-D, but if it was B, then the second question is a 100% C. First it seemed to me that we can win some here by not being honest in the first round, but actually we can’t. The entropy of the joint distribution is 1.5 bits, and the only way to achieve this much expected utility is by betting 50%-50% in the first round.
I think it results from the scale independence of log (and as soon as you add external wealth, the scale independence goes away). That makes it so you can treat every question separately (since only its scale is determined by the previous questions, and that doesn’t impact the maximization). It is a pretty result, but as I don’t think people often talk about this sort of problem I don’t know if I would call it well known or not.
It’s also cool to work back from the last question and see how conditional probabilities connect the full knowledge case and the discovery case.
And here is the prettiest part: The whole thing seems to work even if we don’t assume that the answers to the quiz questions are independent random variables. The expected utility is always equal to the entropy of the joint distribution of the variables, and the best strategy is always honesty in every turn. Note the statement about entropy, it’s a new motif. Before generalizing to non-independent variables it did not add too much value, but in the current version, it is a quite powerful statement.
For example, let the first question be a 50%-50% A-B, but let the second question depend on the first correct answer in the following way: if it was A, then the second question is a 50%-50% C-D, but if it was B, then the second question is a 100% C. First it seemed to me that we can win some here by not being honest in the first round, but actually we can’t. The entropy of the joint distribution is 1.5 bits, and the only way to achieve this much expected utility is by betting 50%-50% in the first round.