A potential model of the full problem (involves questionable numerology):
There are 13 lots currently, and the number of carcasses in the record is divisible by 13 (129*13). If we include the current auctions, that’s 13*130, or 13*13*10.
So, I’ll assume that all auctions have 13 lots.
The individual monster types aren’t divisible by 13 (except Snow Serpents), nor are they if we include the current auctions (except Winter Wolves). However, the Wolf:Yeti ratio seems very close to 3, and if the overall ratios were 2:5:6 that would fit in with the 13 theme and seems close enough to the Yeti:Serpent:Wolf ratio.
The age distributions look fairly triangularish, with a maximum age of 11. One possible way to express that would be there is on average 1 carcass of age 11 for every 2 aged 10, up to every 12 aged 0. And what’s 1+2+...12? Of course—a multiple of 13. Specifically 13*6.
Now, it would be nice if looking at the data in blocks of 13*6 showed some pattern, but I don’t see one, nor is the data a multiple of 6. No matter, we will press on without such empirical validation.
I also note that, in the current auction, the early lots look newer than the late auctions. Coincidence? Probably.
So, model: Base Lot Generation (low confidence): Auction of 13 lots each day Each lot assigned an age from 0 to 11 by a random distribution weighted by {12-age} Each lot assigned an animal type by a random distribution weighted 2:5:6 for Yeti:Serpent:Wolf
Bids: There are two non-Carver bidders, one of whom bids: 60-20*age for Snow Serpents 50-12*age for Winter Wolves 55-6*age for Yeti
and the other one bids: 9+d8 for Snow Serpents 19+d4 for Winter Wolves 29+d6 for Yeti
whereas Carver bids (not high confidence): 7+2d10 for Snow Serpents 31+2d8-3*age for Winter Wolves 32+2d20-2*age for Yeti
Revenue: (credit to GuySrinivasan) 20+2d6 for Snow Serpents 25+4d6-2*age for Winter Wolves 72+1d6-{age}d6 for Yeti (assuming that the prior for d6′s over d5′s is stronger than the 45 times better fit for a d5 in the base part of this formula)
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.
A potential model of the full problem (involves questionable numerology):
There are 13 lots currently, and the number of carcasses in the record is divisible by 13 (129*13). If we include the current auctions, that’s 13*130, or 13*13*10.
So, I’ll assume that all auctions have 13 lots.
The individual monster types aren’t divisible by 13 (except Snow Serpents), nor are they if we include the current auctions (except Winter Wolves). However, the Wolf:Yeti ratio seems very close to 3, and if the overall ratios were 2:5:6 that would fit in with the 13 theme and seems close enough to the Yeti:Serpent:Wolf ratio.
The age distributions look fairly triangularish, with a maximum age of 11. One possible way to express that would be there is on average 1 carcass of age 11 for every 2 aged 10, up to every 12 aged 0. And what’s 1+2+...12? Of course—a multiple of 13. Specifically 13*6.
Now, it would be nice if looking at the data in blocks of 13*6 showed some pattern, but I don’t see one, nor is the data a multiple of 6. No matter, we will press on without such empirical validation.
I also note that, in the current auction, the early lots look newer than the late auctions. Coincidence? Probably.
So, model:
Base Lot Generation (low confidence):
Auction of 13 lots each day
Each lot assigned an age from 0 to 11 by a random distribution weighted by {12-age}
Each lot assigned an animal type by a random distribution weighted 2:5:6 for Yeti:Serpent:Wolf
Bids: There are two non-Carver bidders, one of whom bids:
60-20*age for Snow Serpents
50-12*age for Winter Wolves
55-6*age for Yeti
and the other one bids:
9+d8 for Snow Serpents
19+d4 for Winter Wolves
29+d6 for Yeti
whereas Carver bids (not high confidence):
7+2d10 for Snow Serpents
31+2d8-3*age for Winter Wolves
32+2d20-2*age for Yeti
Revenue: (credit to GuySrinivasan)
20+2d6 for Snow Serpents
25+4d6-2*age for Winter Wolves
72+1d6-{age}d6 for Yeti (assuming that the prior for d6′s over d5′s is stronger than the 45 times better fit for a d5 in the base part of this formula)
Love it!
The age distribution is very clearly off, though. Age 0 has about 2.5x the count of age 1, for each species. I didn’t see anything come up clearly...
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.