Talk:Chest loot/Archive 1

Method?
Hello, I was wondering how the values on the page were calculated. I got somewhat different ones.

My method was as follows:
 * Note that each chest has a number of rounds to fill itself, random from x to y.
 * Note that each chest defines a pool of items, each with a weight, and possible stack sizes, random from m to n.
 * So for a given item,
 * the chance of that item being picked to be filled is governed by its relative weight among all the items in the pool.
 * the number in a given chest will range across a matrix from xm to yn: number of rounds multiplied by stack size.
 * simple case: rounds is in the range 1–3, stack sizes is in the range 1–4
 * there are (1+y-x)×(1+n-m) = 3×4 = 12 ways to form products in this example,
 * and keep in mind the sum from 1 to n = ( n^2 - n )/2,
 * and the sum of all products in the product matrix 1–y:1–n = ( n^2 - n )/2 * ( y^2 - y )/2 ,
 * so [( n^2 - n )/2 * ( y^2 - y )/2 ] / [ (1+y-x)×(1+n-m) ] represents the average number of items once rounds are complete.
 * more complicated (real life) case, iron ingots in a village chest: rounds is in the range x–y = 3–8, stack size is in the range m–n = 1–5
 * there are (1+y-x)×(1+n-m) = 6×5 = 30 ways to form products in this example,
 * the sum of products of all numbers x–y and m–n involves (I believe)
 * taking the entire sum of products across 1–y:1–n ,
 * subtracting out sections of the matrix 1–y:1–(m-1) and 1–(x-1):1–n ,
 * then adding back in 1–(x-1):1–(m-1) once, since that quantity was subtracted out twice...
 * giving you: { [ ( n^2 - n )/2 * ( y^2 - y )/2 ] - [ ( (m-1)^2 - (m-1) )/2 * ( y^2 - y )/2 ] - [ ( n^2 - n )/2 * ( (x-1)^2 - (x-1) )/2 ] + [ ( (m-1)^2 - (m-1) )/2 * ( (x-1)^2 - (x-1) )/2 ] } / [ (1+y-x)×(1+n-m) ] represents the average number of items once rounds are complete.
 * so, avg num items in rounds = { [ ( 25 - 5 )/2 * ( 64 - 8 )/2 ] - [ ( 0 - 0 )/2 * ( 64 - 8 )/2 ] - [ ( 25 - 5 )/2 * ( 4 - 2 )/2 ] + [ ( 0 - 0 )/2 * ( 4 - 2 )/2 ] } / 30
 * = { [ 280 ] - [ 0 ] - [ 10 ] + [ 0 ] } / 30
 * = 9
 * avg number of items per chest = item weight * avg num items in rounds
 * for iron ingots in a village chest, that item weight is 10, and the total item weight is 91,
 * so, avg number of iron ingots per village chest = 10/91 * 9 ≈ 0.989

And god bless you if you follow that. I only describe it so that maybe a person can qualitatively say whether I'm down the wrong path or not.

And you know, I wanted to know if there was an easier way, or what your way was,. – Sealbudsman (Aaron) t/c 06:33, 26 May 2015 (UTC)