Talk:Chest loot/MethodAttempt

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  • 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 * 16.51.755