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
 * = { [ 15 * 36 ] - [ 0 ] - [ 15 * 3 ] + [ 0 ] } / 30
 * = { [ 540 ] - [ 0 ] - [ 45 ] + [ 0 ] } / 30
 * = 16.5
 * 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 94,
 * so, avg number of iron ingots per village chest = 10/94 * 16.5 ≈ 1.755

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)


 * , would you perhaps be able to comment? –Goandgoo ᐸ Talk Contribs 09:20, 27 May 2015 (UTC)
 * The data on pages like Dungeon and Stronghold was taken from the code of the game. I'm too tired right now to do the math myself to see whether the table on this page is correct or not or whether Sealbudsman is on the right track, or even to try to figure out whether it's actually useful to say you'll get an average of 0.652 bread per dungeon chest over the long term versus just stating the stacks per chest and the weights and quantities of the possible items. Anomie x (talk) 12:14, 27 May 2015 (UTC)