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This is great!

@prokopetz.bsky.social

IQ of the DM is everything

😄😄😄

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Evil. I love it.

Awesome 🙌

I simplified that part with the 'you only got one arrow' bit (reasonable, I think). The alternative would be to reroll the second die if they tie. The fun bit would be if there were 13 arrows or such. You'd need to roll d20 and reroll anything 14+.

Or reroll the second die until it doesn't match the first. I would make 1 to 5 be the cursed ones, but that's just a preference. Expected number of rerolls is 0.11111, so you're rolling 2 dice most of the time.

Even if there are only 3 arrows, get a couple anyway. They're hella cool.

Roll 2d100, what's the issue?

Easy. Grab the percentile dice.

Roll 4 d10 and use them as digits of course!

"With the sheee number and mixture of your ammo, one is cursed one isn't. If you pull an odd number of arrows coinflip to see which way spare goes."

Roll 2 d100. Like, this is pretty simple to solve, without solving the combinatorics problem. The comic is still funny.

The probability has a factor of 11 in the denominator, so an exact solution can't be based on a standard set. Use coins (If you really gotta, get two decks of playing cards and prune to 50/50 red/black.)

Then you get max 20 of them

I have to know why you decided to sign and date every post

Important caveat: Find out if the player *intends* to grab two arrows, but grabs individually or *does* grab two. If the latter, chances of one of the two arrows being cursed is 40%.

See my earlier reply to alecjohn.

🤣

Alternative title text: “I’m using this trick to get out of all of our ‘Settlers of Catan’ games too.”

Randall, I am , well kinda old. My husband was a mathematician. And seeing this post brought so many memories. soooo remember him studying combinatorics. Never thought would see that word again..and have daughter into D&D. So this I have to follow.

screw! combinatorics! let's! just! override! the! system! choose! everything!

I’ve always counted on you to do better.

honestly I would have the player roll 2 d10s. odds they grab a cursed arrow, evens they don't

I really like the idea but would use color. I think I can feel the diff between d12 and d20

This is amazing. Thank you.

This is legit so funny.

THat's why GURPs is superior. Everything is a bell curve on d6s.

As is often the case, it's the easiest step that I fuck up. Good catch, thanks.

Mine's the much easier approach when you haven't taken probability in 31 years. I did write out the permutations of dice rolls by hand (e.g., 14 - 662 653 644 544 -> 3 + 6 + 3 + 3 = 15). I found it interesting that the combinations for a 18 thru 13 followed the triangular numbers.

there's also whatever this is. basically there's a 1/3 chance of each outcome in increments of 0.1 (because there's 10 arrows total). you can divvy up the numbers however you see fit btw, assuming truly random dice, i just did that bc it's easier to write down.

Look at all you nerds working out how to do this "accurately" (Talking about myself btw)

roll d100, <10 you grabed one. On the 2nd arrow, roll a d100, <11 you grabbed one.

Damn it just realised why. OK, second roll, re-roll if first number comes up again..

Roll for humor. Rolls 5. Silence with a distant cough. Rolls 14. Polite, consistent chuckling. Rolls 1. Rotting fruit is flung at you in rapid succession. Rolls 36, dies from laughter. Ghost reappears and also somehow dies from laughter.

"Good enough" solution: roll a d10 twice. Reroll if you get the same number. 1-5 means you picked up a cursed arrow, and 6-10 means you picked up a regular arrow.

Maybe the artows curse whos hit by them and he just doesnt want to use them yet.

Maybe the arrows are on the temple table under a sign - To gain my favour fire 2 of these identical arrows at the target. If you hit with a cursed arrow, the guardian will awaken. Chose wisely adventurer. Oh, and if you miss with a non cursed arrow they will awaken as well.

My youngest would appreciate this...it's kind of lost on me tho!

Do they play hangman with the losers?

Can't you just roll 2 d10 and assign the values 1-5 to "cursed", reroll 1 die on a double? Am I missing some deeper probability here?

I would have you roll a percentage die and have you pick high or low.

anydice.com/program/1f1f Odds of at least 16 on 3d6 + 1d4 are 22.22% (2/9)

Correct. Because I was curious, I calculated the number of permutations of getting a 12 through 18 on the 3d6. I then reduced the permutations for 12 thru 14 by 25%, 50%, and 75%, respectively, to account for the d4. It add up to 48. There are 216 (6³) possibilities for the 3d6, 48/216 = 2/9

Grab ten playing cards, five red and five black. Shuffle them and draw two.

Would rolling two d10 with 5 and under being cursed and 6 and up being normal not be the same odds?

Love that. I just imagined there could be a puzzle that can only be solved by rolling a prime number, and letting the player roll any dice. Many of them would go straight to d20, not to d4 or d6, without thinking about it.

...but... the alt text. Where's the alt text 😧

So how would you start with the target probability of 2/9 and arrive at the necessary die rolls?

I expect some D+D diehard has done a brute-force search and compiled a lookup table of the simplest dice combinations to achieve any desired result, ie all m/n for m,n in Z > 0 such that m < n < 50. 🤷

Of course, you could use 2d6, but you can't use the top of the bottom range saying you need at least an X. It would be you need "a 5 or 9" or "a lucky 3 or 7," with everything else failing. I don't know how to do an elegant way of finding 2/9 like Randall did without trial-and-error.

Tried to do this by hand. Good to know I got pretty close. 🎉

Interestingly, for 13 through 18 (and 3-8), the combinations follow the triangular numbers: 18 - 1 17 - 3 16 - 6 15 - 10 14 - 15 13 - 21 12 - 25 Odds are out of 216 (6³).

or just 1/2 (p of first arrow) * 4/9 (p of second arrow).

So now the player just needs to roll a single physical D9.....

Yeah, probability is 1/2 × 4/9 = 2/9 That's the easy part. 😄

One for you @anothergreenworld.bsky.social

Easy answer for the DM: "The arrows are CURSED. Of course you grab one."

I would've just told him to roll Arcana and based on the result roll 2 d10's choosing either the lower or higher result.

👍🏼💯🎯🤣🤣🤣🤣

I'm not much of a math or numbers person. Would it also work to say "roll 2d10. If you roll doubles, you only get one arrow. Any odd numbers mean you grabbed a cursed arrow."

THAT’S JUST EVIL RANDALL

Doing this on the fly is fun!

There's a degree for everything.

Roll a d10, numbers 1-5 are cursed arrows. If you roll the same number, roll again until you've got two different numbers. Easy and accurate :)

This is when you have them go onto a digital die roller and have them roll 1d10 and then 1d9. You’ll need the roll of the d10 to interpret the d9 roll.

Welcome to who voted.

for me that was drew, love that guy

Ah yep I missed it in the poster's handle. Love XKCD, always so good.

Cueball engages in nerd sniping.

I studied math: The people taking combinatorics were weird - not as weird as the pure math students, but weird.

Lolololol! Now I want to write .......

Me as someone who loves dice math a little too much.

Rule of thumb when it comes to ammo for me is only if It’s special/unique or if there is like a fumble/sitch where scarcity is essential.

What would the cursed ones do, since they’re in your quiver? Why don’t you just get ‘em out if there harmful to you?

Just Roll a D10 twice. Consider the cursed arrows to be 6,7,8,9,10. If you roll any of them you shot a cursed arrow. Seems obvious to me..... 🤷🏻‍♂️

Followed ✔️

I love this.

Bring dice with exclamation points on them lol

YES!! This is the D&D I grew up with in the 80’s when we had a substitute English teacher with a PhD in Mathematics as a DM. It was a little annoying at first, but peeled my brain like a grape and let so much in. This really takes me back.

Me not realizing that the binomial distribution is necessary and that I can just do 5/10 times 4/9 is why I don't have a degree in combinatorics

for first arrow make a flat DC10 roll, on fail you drew a cursed arrow, could carry this over to roll 2, but to make it more accurate you can make the second roll with DC9 (if you failed first check) or DC11 (if you succeeded), the math's not 1:1 but it's fairly close

This made me snicker

Fun. 3 other ways to resolve: 1. Roll a d10 and a d9. (Love an excuse to use weird dice.) 2. Use dry-erase item cards. Shuffle up and deal. 3. Use a quiver and 10 arrows. Mark the tips of the arrows so the player knows which ones are cursed when they fire.

As if D&D wasn’t already a game of calculated chaos, now we’ve got a math major throwing combinatorics into the mix! 🎲😂 I bet this DM makes every riddle a lesson in probability theory.

Two things: 1. The narrator is the first speaker. The DM is the one ruling on how to roll, as DMs do. SMH that anyone missed that. 2. @xkcd.com, are d2 allowed? I may have a 3-dice solution.

...But to engage the puzzle directly, can we use a d2? This and 2d6 gives 72 combos in a range of 3-14, with success being 11 or better.

Lol

Games helped me Math. ^.^

D&D should be part of the school curriculum. Teaches them collaboration/independent thinking, math skills, consequences... And the most important of all, who has the power!!!! 😜

I took combinatorics in college and, ever since, I only gamble playing blackjack.

Sharing this with my D & D kiddo. He tried to teach me about this stuff but old cats learn new tricks slowly.

7/9 chance of the DM picking the cursed arrow(s)

Thank you for making the DM who loves math a woman (presumably) ❤️

technically you can do a "roll 2d6 and you have to get at least 9, but you cannot get the same number on both dice" if we want minimum number of dices (although, idk if rules like this in DnD are too complex or not)

My favorite D&D line:

The alt text screams pigeonhole principle to me.