Talk:3015: D&D Combinatorics

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). JBYoshi (talk) 00:54, 23 November 2024 (UTC)

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. --162.158.154.160 01:41, 23 November 2024 (UTC)

Yup, if you look at 3015's JSON you see that title and safe_title differ, and if you look at the HTML page source you'll see 3 different things: <title>xkcd: D Combinatorics</title>, <meta property="og:title" content="D&amp;D Combinatorics">, and <div id="ctitle">D&D Combinatorics</div>! So probably what happened is Randall entered D&D but was supposed to enter D&D, and the openGraph tags adder code, having to be HTML-aware, decoded & normalized D&D as HTML would, but the other parts of the pipeline just ate it for some reason.

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. Kaisheng21 (talk) 01:33, 23 November 2024 (UTC)

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? Darkmatterisntsquirrels (talk) 01:29, 23 November 2024 (UTC)

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. 172.68.54.139 01:41, 23 November 2024 (UTC)

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.) 172.68.54.139 01:41, 23 November 2024 (UTC)

Clever, but dice rolls in D&D involving summing all the dice, applying modifiers, if any, and then comparing to one or more threshold values. Your method makes it very difficult to apply modifiers. 162.158.41.8 02:49, 23 November 2024 (UTC)

Minor quibble, arrows aren't fired (unless they're flaming or self-propelled, perhaps), they are shot. (Shotguns are fired of course.) 162.158.41.73 02:52, 23 November 2024 (UTC)

Rolling 22 or lower on percentile dice (or, equivalently, 79 or higher) is close enough, and easier to come up with. (Give or take whether 00 is treated as 100 or zero.) Or directly represent the action: roll a d10. If it's 1-5, you lose. If it's 6-10, roll again; if it's 1-5 you lose, 6-9 you win, 10 roll again. (Modify slightly if you want to distinguish the case of grabbing *two* cursed arrows.) Jordan Brown (talk) 03:26, 23 November 2024 (UTC)

Alternative exact solution for getting this probability using dice

Roll: 1d8, 2d6, 1d4 succeed on 19 or higher.

Alternative way to calculate the probability of drawing two non-cursed arrows

I couldn’t remember the formula for binomial coefficients (“n choose k”), but there’s an easy way to calculate that the probability of drawing no cursed arrows is 2/9 without that formula. You just need to multiply the probabilities that each of the arrows drawn is not cursed. Since only two arrows are drawn, you only have to multiply two numbers.

The probability that the first arrow is not cursed is 5/10 – there are 5 non-cursed arrows and 5 cursed arrows out of 10 total. After taking out one non-cursed arrow, there are 4 non-cursed arrows and 5 cursed arrows out of 9 total, so the probability that the second arrow is not cursed is 4/9. Multiplying the two probabilities, the probability of drawing two non-cursed arrows is (4*5)/(10*9) = 20/90 = 2/9.

I was considering writing this observation in the Explanation section of the page, but I’m not if it belongs there. This solution avoids using formulas from combinatorics, so it might not be connected enough to the comic.—Roryokane (talk) 06:02, 23 November 2024 (UTC)

My simple-minded approach

• Roll d10 once for your first arrow: if 1 to 5, the arrow is cursed, otherwise not;

• Roll d10 again for your second arrow: same rules, but repeat until you have a different number from the first one (so d10 is in fact only a d9 this time)

• I won't calculate probabilities – these are your arrows, live with it ;-) 172.69.109.51 07:33, 23 November 2024 (UTC)

That has the benefit (over 3d6+1d4) of telling you which arrow(s) (if either) was cursed. RegularSizedGuy (talk) 07:52, 23 November 2024 (UTC)