Still gonna guess that somehow this is from a distribution more easily generated than the triangular. But I may be overfitting on the last time I said something was triangular and it turned out to be exponential. Still, it looks to me like “start at age=0; roll a d6; if you rolled 1-5, add 1 to age and repeat, otherwise on 6 stop; if you reach age 12, discard and start again from age=0” is a decent match for this distribution, and it involves a d6, so that’s my guess I think.
Looks like the likelihood for triangular is over a million times better (to log-nearest order of magnitude ~10^-1672 v. ~10^-1679) than the 1⁄6 drop per turn exponential.
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.
The age 0 amount is higher than expected from the model distribution, but it’s nowhere near 2.5x the age 1 amount. I have:
Overall 289 age 0, 233 age 1 (expected 258, 236.5)
Snow Serpents 103 age 0, 92 age 1
Winter Wolves 141 age 0, 108 age 1
Yeti 45 age 0, 33 age 1
Thank you, my bin size was bad. :(
Still gonna guess that somehow this is from a distribution more easily generated than the triangular. But I may be overfitting on the last time I said something was triangular and it turned out to be exponential. Still, it looks to me like “start at age=0; roll a d6; if you rolled 1-5, add 1 to age and repeat, otherwise on 6 stop; if you reach age 12, discard and start again from age=0” is a decent match for this distribution, and it involves a d6, so that’s my guess I think.
Looks like the likelihood for triangular is over a million times better (to log-nearest order of magnitude ~10^-1672 v. ~10^-1679) than the 1⁄6 drop per turn exponential.
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.