Reply to thread

Message: <blockquote data-quote="EzekielRaiden" data-source="post: 8783263" data-attributes="member: 6790260">The problem is, this ensures that the first array always gets the best result of a given block of 4, and the fourth array always gets the worst result in that block. The ox-plow-like alternation is an effort to ensure fairness.If you really wanted to ensure fairness you'd look for a way to distribute these things via the Thue-Morse sequence. That would give you a table like this:<table style='width: 100%'><tr><td>---</td><td>Roll #N</td><td>Roll #N</td><td>Roll #N</td><td>Roll #N</td><td>Roll #N</td><td>Roll #N</td></tr><tr><td>Array A</td><td>1</td><td>8</td><td>11</td><td>14</td><td>17</td><td>24</td></tr><tr><td>Array B</td><td>2</td><td>5</td><td>12</td><td>15</td><td>18</td><td>21</td></tr><tr><td>Array C</td><td>3</td><td>6</td><td>9</td><td>16</td><td>19</td><td>22</td></tr><tr><td>Array D</td><td>4</td><td>7</td><td>10</td><td>13</td><td>20</td><td>23</td></tr></table>It's imperfect, because we don't have a clean power of 4 number of elements. But it's the best that can be done with four arrays. You'd need to do a total of six arrays if you wanted it to work out neatly (because then we'd use the first nontrivial instance.) That would look like this (skipping the labels above to make it fit in a smaller space):<table style='width: 100%'><tr><td>1</td><td>12</td><td>17</td><td>22</td><td>27</td><td>32</td></tr><tr><td>2</td><td>7</td><td>18</td><td>23</td><td>28</td><td>33</td></tr><tr><td>3</td><td>8</td><td>13</td><td>24</td><td>29</td><td>34</td></tr><tr><td>4</td><td>9</td><td>14</td><td>19</td><td>30</td><td>35</td></tr><tr><td>5</td><td>10</td><td>15</td><td>20</td><td>25</td><td>36</td></tr><tr><td>6</td><td>11</td><td>16</td><td>21</td><td>26</td><td>31</td></tr></table>The bottom row kinda gets shafted here 'cause it takes so long for it to be the "first" choice, but it's not the worst thing ever. The top row will likely end up with only one amazing stat (plausibly 16-18) but several middle-of-the-road stats, while I would expect the second or third array to deliver the best overall performance.Edit: On reflection, probably not worth it. You'd need many more 'rounds' of sharing to make it actually shift toward fairness, and with stats being a highly uneven spread, the 5th and 6th arrays are almost guaranteed to be shafted, while the first three arrays are almost guaranteed to be excellent. Alas. I find the Thue-Morse sequence fascinating but it really only works for binary splits, anything bigger starts to get unwieldy.</blockquote>

Verification