https://www.explainxkcd.com/wiki/api.php?action=feedcontributions&feedformat=atom&user=162.158.154.160explain xkcd - User contributions [en]2026-06-24T06:29:19ZUser contributionsMediaWiki 1.30.0https://www.explainxkcd.com/wiki/index.php?title=Talk:3015:_D%26D_Combinatorics&diff=357652Talk:3015: D&D Combinatorics2024-11-23T01:41:27Z<p>162.158.154.160: </p>
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The bot originally created this page as “D Combinatorics”. I renamed it to the correct title and tried to get as many of the references as possible (including a few redirects). [[User:JBYoshi|JBYoshi]] ([[User talk:JBYoshi|talk]]) 00:54, 23 November 2024 (UTC)<br />
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:The title in the Atom feed (which I'm assuming the bot consumes) is "D Combinatorics". I'm guessing something in Randall's pipeline didn't like the ampersand. --[[Special:Contributions/162.158.154.160|162.158.154.160]] 01:41, 23 November 2024 (UTC)<br />
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What are the odds of rolling 16 or higher on 3D6+D4? 3D6 average 10.5, D4 average is 2.5, total average should be 13. I do not know how to proceed from here.<br />
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:By raw combinatorics: 71 + 52 + 34 + 20 + 10 + 4 + 1 ways to get each of 16 - 22 respectively, for a total of 192, out of 4(6^3) = 864 total. 192/864 simplifies to exactly 2/9. I have no idea how Randall found this; if anyone has an idea, please let me know. [[User:Kaisheng21|Kaisheng21]] ([[User talk:Kaisheng21|talk]]) 01:33, 23 November 2024 (UTC)<br />
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It seems like we edited the transcript at the same time. The odds of rolling 16 or higher in this situation seem to be 2/9? [[User:Darkmatterisntsquirrels|Darkmatterisntsquirrels]] ([[User talk:Darkmatterisntsquirrels|talk]]) 01:29, 23 November 2024 (UTC)<br />
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: There are 864 possible rolls (6 * 6 * 6 * 4). If you enumerate all of the rolls you will find that 192 are 16 or higher. 192/864 = 2/9, the value from the explanation. [[Special:Contributions/172.68.54.139|172.68.54.139]] 01:41, 23 November 2024 (UTC)<br />
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A much simpler approach: Roll two six sided dice and sum the result. You are successful if the result is 5 or 9. That happens 8 times out of 36. 8/36 = 2/9. (Or successful if the sum is 4 or 6, or 2 or 7, or 2,3,4 or 11, or several other combinations.) [[Special:Contributions/172.68.54.139|172.68.54.139]] 01:41, 23 November 2024 (UTC)</div>162.158.154.160