Anyway, in the spirit of tumbling platonic solids:
One possible distribution for the age numbers would be the distribution generated by min(d12,d12)-1. This is not the same as the 1,2,3,4...12 triangular distribution, but rather a 1,3,5,7,...23 triangular distribution. (The 1,2,3...12 distribution would be generated by min (d12, d13)-1).
And checking the likelihood—this one is actually better.
LOG10 likelihood
-1672.05 for 1,2,3,...12
-1671.43 for 1,3,5,...23
P.S. I was terse in the previous comment because of time constraints. About the difficulty of the triangular distribution, I was thinking it wasn’t that unlikely anyway because in the previous problem abstractapplic generated a weighted average by taking a random entry from a list that contained duplicates, and a suitable list could be generated easily enough using a for loop.
To acquire karma: respond to my posts with actual data.
Anyway, in the spirit of tumbling platonic solids:
One possible distribution for the age numbers would be the distribution generated by min(d12,d12)-1. This is not the same as the 1,2,3,4...12 triangular distribution, but rather a 1,3,5,7,...23 triangular distribution. (The 1,2,3...12 distribution would be generated by min (d12, d13)-1).
And checking the likelihood—this one is actually better.
LOG10 likelihood
-1672.05 for 1,2,3,...12
-1671.43 for 1,3,5,...23
P.S. I was terse in the previous comment because of time constraints. About the difficulty of the triangular distribution, I was thinking it wasn’t that unlikely anyway because in the previous problem abstractapplic generated a weighted average by taking a random entry from a list that contained duplicates, and a suitable list could be generated easily enough using a for loop.