Do you have any thoughts on aggregating probabilities using the MLE of the mean of a beta distribution? In this case, 1 %, 1 %, and 60 % would be aggregated into 21.8 %. The geometric mean of odds is 0.0535 (= ((0.01/​0.99)^2*0.6/​0.4)^(1/​3)), so it would result in 5.08 % (= 0.0535/​(0.0535 + 1)). So the 2 methods can differ quite a bit.
Hi Jaime,
Do you have any thoughts on aggregating probabilities using the MLE of the mean of a beta distribution? In this case, 1 %, 1 %, and 60 % would be aggregated into 21.8 %. The geometric mean of odds is 0.0535 (= ((0.01/​0.99)^2*0.6/​0.4)^(1/​3)), so it would result in 5.08 % (= 0.0535/​(0.0535 + 1)). So the 2 methods can differ quite a bit.
Here are some relevant experiments
https://​​forum.effectivealtruism.org/​​posts/​​acREnv2Z5h4Fr5NWz/​​my-current-best-guess-on-how-to-aggregate-forecasts?commentId=odJfbbZLyrPjNtzC3