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I play a lot of D20 games. If you're familiar with the various systems you probably know that rolling a 20 on the die is a way to guarantee a hit in an attack roll. It doesn't matter how big and powerful the enemy, how weak and useless the character (or vice-versa), a natural 20 hits… somehow.
In mathematical terms rolling a specific number on a d20 is a 5% chance. It doesn't come up all that often, and it feels special when it happens at a important moment. Natural 20s are instigators of wonderful memories in RPGs.
But I digress. This post is not about fantastic memories of epic monsters slain. This post is about a character I'm playing in one of the side campaigns run with my gaming group. My character, Ukru, is a scoundrel of a halfling who wears a bandolier of daggers and puts them to use pestering his enemies with a thousand cuts. Through the gathering of feats and magical items this little troublemaker has also picked up a number of other ways to attack his enemies: a bite, a gore, and a sting. These are called "Natural Attacks" and are all available to him as additional attacks on his turn. He has a high enough level to allow him multiple attacks per round normally, and he also has taken feats to minimize his penalties from fighting with two weapons, giving him even more attacks. He's a whirlwind of stabbing!
(D20 gamers out there are probably wondering about his class features about now and yes, he has backstab.)
Tonight as I was getting ready for our adventure I counted up his attacks. He has 6 per round normally, but also wears boots granting him haste 10 rounds a day. When using those he has 7 attacks in a round.
And so here's where my aged math-brain woke up. It whispered to me in the dark, "You know probabilities and statistics. Why don't you do the math!" With 6 or seven attacks in a round, how likely is it for me to score at least one natural 20 on a roll?
I dusted off a long-dormant cluster of brain cells which reminded me of the easy way to think about the problem. The question is easier to think about in the negative. What are the chances that after N rolls I won't have ever rolled a 20? The probability of failing to roll a 20 is 95% or (1 - 0.05), which means our formula to answer my first question is:
1 - (1 - P)^n 1 - (1 - 0.05)^6 1 - (0.95)^6 1 - 0.735 0.265 or 26.5%
Pretty simple, right? What about 7 tries? 30%
Nice! Those are better odds. And in my head I immediately turned into an idiot and thought to myself, "30%? Well then in 3 or 4 rounds I'll surely get a 20!"
It took me a few seconds for the fallacy to hit me, then DUHHHH. 30% turn after turn doesn't add together. It's the same formula as above! 3 rounds of 30% chances is:
1 - (1 - P)^n 1 - (1 - 0.3)^3 1 - (0.7)^3 1 - 0.343 0.657 or 66%
Do the math over 4 rounds and you get 76%. Still awesome, but we're not just adding percentages.
Anyway, it's pretty cool to know that I have a 76% chance of scoring a natural 20 inside 4 rounds if I square up with a baddie and unload all I've got. Of course, my little dude is pretty squishy, so surviving 4 rounds up in a monster's face like that is questionable. Figuring out that math, though, is a bit trickier.
Originally Published 2022-05-20 at:
gemini://tilde.team/~tomasino/journal/20220520-some-rpg-math.gmi
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