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)

I used some simple python code to loop over every dice and confirm and it's 2/9 162.158.158.111 12:11, 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)

I added a table of outcomes to clarify how it works out to 2/9, anyone know how to make it pretty? -- Laurence Cheers

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)

I think you misunderstand the problem here. This is not skill, no modifiers apply, it's purely probability 162.158.158.111 12:11, 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)

Arrows are "loosed", even more accurately. At least to avoid the confusion from how so many things may be shot, or a shot. (Many different nouns, from a physical measure of liquer/coffee/vaccine to a projectile, or an even abstract fundemental of chance; and, as verb, projectiles perhps may be shot, then so may their targets.) 172.68.205.178 14:32, 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)

If you don't like re-rolls, you can make d9 out of 2d3. Nine possibilities, so just assign one of them (perhaps by rolling them one at a time) to be the more significant digit. Don't have a d3 handy? Use d6 and modulo off the extra! (1=1, 2=2, 3=3, 4=1, 5=2, 6=3) 172.68.150.91 05:59, 24 November 2024 (UTC)

The physical difficulties of an M-of-N locking system

There seems to be doubt that a "N locks and M keys to unlock them" system could be easily accomplished. I think it could be trivial, with strategically interlocking locked-restraints. A chain formed of bike-locks can give a larger locked loop that can be unlocked by just unlocking any single one of the constituent locks, leaving the other locked loops to not matter (or you could also try the Borromean rings system, whereby it is again secure against itself, until just one ring is opened up to reveal that the rest now aren't even locked at all...). With almost arbitrary ability to cross-link (or, if you will, repeated/alternating-reflected Borromean triplet connections), you can extend the requirements to more than one unlocking being required (by looping chain elements to mre than just the 'adjacent' loops, sideways onto a parallel meta-loop or up/down the chain, all you might do is allow some slack (could be sufficient to get a thing held directly closed by the taut loop-of-loops, but not enough if the passage of the loop through a hasp/sneck actually prevents the otherwise free movement of the final slide-to-unlock action to occur), but a second (or third, or fourth) unlocking can be required to open-end the whole metaloop of locks. At the top end, M=N solutions are also trivial (e.g. two keys, two locks popularly of safety deposit boxes or other things). Which is not to say that a specific M-of-N puzzle (where 1<M<N) might not need a little bit of thought to actually design and implement, but there's no obvious reason why all such combinations shouldn't be nicely doable. 172.69.79.165 14:56, 23 November 2024 (UTC)

Can we first confirm that the M-of-N Encryption was what Randall was referencing in the first place? 172.71.154.140 03:17, 24 November 2024 (UTC)

No, first confirm that this is what the explanation treats as what Randall was referencing. As it was, "complicated lock mechanics" and/or "magic" were suggested as the only ways of doing this, when this (or what we thought this was) just needs a little thought and N bike-locks suitably entangled. 172.70.58.45 13:17, 24 November 2024 (UTC)

I'm glad someone else chimed in on this, because it is definitely not difficult to require unlocking of multiple discrete locks! I can't even figure out why one might think it would be?

ProphetZarquon (talk) 15:55, 24 November 2024 (UTC)

Polyhedral Dice

"other polyhedral dice, with the number of faces denoted by dX (e.g., d10 is a 10-sided die, with numbers from 1 to 10 on it)." - the d10 may be a poor choice as exemplar here; Back in the last century, when I was playing D&D, d10 were typically (and uniquely) numbered 0-9, not 1-10. This may no longer be the case, and I may be showing my age, but if it is still the norm, the d8 or d20 might be a better choice of example. 172.68.210.6 02:40, 24 November 2024 (UTC)

Typically, I've only seen 0-9 d10s, as part of a "d100" dice pair, with one reading 0-9 & the other reading 0⁰-9⁰... Single d10, mostly seem to come in 1-10? Maybe it depends which reseller one shops at...

ProphetZarquon (talk) 15:49, 24 November 2024 (UTC)

You've all been nerd-sniped.

Caliban (talk) 10:53, 24 November 2024 (UTC)

Combinatorics degree?

Does such a degree really exist?

--162.158.130.37 17:19, 24 November 2024 (UTC)