ejacob comments on Prediction and Calibration—Part 1

[ejacob]

There are two reasons why this number is so large: 1) Scott made a lot of predictions and 2) Scott is a very good predictor.

To address #1, you can take Nth root of this number where N is the total number of predictions made. This gives you scott’s edge vs randomness per each prediction (on average)

[Jan Christian Refsgaard]

you mean the N’th root of 2 right?, which is what I called the null predictor and divided Scott predictions by in the code:

random_predictor = 0.5 ** len(y)

which is equivalent to ${0.5}^N$ where $N$ is the total number of predictions