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Math Help with 2d12 (Patreon)

Published:
2023-08-23 11:41:28
Imported:
2026-06

Content

Hi all. I'm working on the Lost City but Daggerheart is in the back of my mind. The main mechanics require players to roll 2d12. Doubles means a critical success or failure.

Question: what is the % chance or rolling doubles on 2d12?

I know some math people will be able to figure that out. Thanks!

--PDM


Comments

Kristen Solindas

Doubles on 2d12 are a 1 in 12 chance. About 8.33%.

xoxoclvbs

Came here to give the same answer as Kristen!

Bob Freeman

Is that right? I thought it was .083. Dang it. Math is hard.

Ron F.

This is interesting. So, while a chance of rolling doubles is 1/12, if your standard 2d12 difficulty is 13 on the dice (which I'm guessing it is), the chance of critical failure is 1/24 and crit success is 1/24: 4.167%. Pretty close to a nat 1 or nat 20. Just an observation.

Graham Kane

Yeah a significantly higher chance than on a D20. My guess is it applies crits in a more narrative way, rather than straight damage.

Brian Leet

FWIW, I’ve also run numbers on the extra d6 for advantage rather than roll twice and take the higher. For middle range target DCs, it’s essentially identical change in odds of success (about 20-25% more likely).

Ron F.
>>115851100

It is. .083 is 8.3%. Unless you were joking, in which case I missed it 😆

Vincent Pelletier

The dice will roll doubles 1/12 times. So about 8.3%.

John
>>115851334

Life is biology, which is chemistry, which is physics, which is math, or maybe music but music is math too, so there will ALWAYS be math! :)

Donald Koenig

Probability of rolling a 12 on one die = 1/12 Probability of rolling a 12 on the second die = 1/12 Probability of rolling 12 on both dice = (1/12) * (1/12) = 1/144 So, the odds of rolling 2 12s on 2 12-sided dice are 1 in 144, or approximately 0.69%.

Richard Jensen

Probability = (Number of ways to roll doubles) / (Total number of possible outcomes) = 12 / 144 = 1/12 ≈ 0.0833 or about 8.33%. To calculate the chance of rolling doubles on 2d12, where each die has 12 sides, you need to consider the number of ways you can roll doubles (i.e., the same number on both dice) and then divide it by the total number of possible outcomes. There are 12 ways to roll doubles: (1,1), (2,2), ..., (12,12). The total number of possible outcomes when rolling 2d12 is 12 * 12 = 144. So, the probability of rolling doubles on 2d12 is: Probability = (Number of ways to roll doubles) / (Total number of possible outcomes) = 12 / 144 = 1/12 ≈ 0.0833 or about 8.33%.

Ron F.
>>115850975

I believe he was asking the chance for rolling any doubles, not just double 12's.

JP Trudel
>>115850975

No the probability of rolling a single unique result is 1/144. There are 12 doubles out of the 144 unique results 12/144 = 1/12. Kristen is correct, the reply is to explain to David.

Richard White
>>115852105

I don’t think this is correct. Although I am far from a maths expert, so I’m not the best person. The problem I have with your equation is that you’re assuming the first roll only counts if it’s a 12. That is incorrect. The first roll is correct 100% of the time. Doesn’t matter if it’s a 1, a 5, or a 12. Let’s say the first die rolls a 7. The second number only has to match the first outcome. In this scenario, a 7. The player will roll a 7 to get doubles 1/12 times. = 8.33% If both rolls had to be a single target number (eg. 12) then I would agree with your formula.

Baron de Ropp

According to a simulation I just ran on my computer with 1 million dice rolls, it was 8.3265%. Doing it with 100 billion dice rolls got a result of 8.3311% and my CPU consumed an entire dinosaur's worth of energy.

JP Trudel
>>115851191

What’s also interesting is that your critical chance go up when you get better.

Cee-Jay
>>115850975

Ron F. I think you are correct... at first I interpreted his question the way David did (probability of rolling a 12 on both).

JP Trudel
>>115852634

According to a study by Dr. Timothy L. Clarey and Jeffrey P. Tomkins in 2015 the average biomass of a single dinosaur is about 630 kg. Another study done by Jeff Dukes, back in 1997, calculated that the “efficiency” rate of making oil from carbon biomass is about 10,000 to one. So 0,063kg or 49mL of crude oil, about 509KCal… Could be worse. (I had to know)

Christian Bynum

Indeed. If you keep it simple and think about it with common sense, the first die roll, regardless of type of die, has a uniform 1/n chance of any result, and there are n possible results (1/n)(n)=1. Then, for the second die, there is a 1/n chance for the second result to match the first one. When you aggregate, you get 1/n. So 2D4 roll doubles 25%, 2D6 roll doubles, 16.7%, 2D8 roll doubles 12.5%, 2D10 roll doubles 10%, and 2D12 roll doubles 8.3%.

Cee-Jay
>>115851191

In DCC, fighters get a crit with a 19 or a 20. And on the flip side, when magic users fumble, their critical fumble escalates to a 1 or a 2, all the way up to 4 after multiple failures. It's fun.

Pyram King

Each die has 12 sides, so when you roll two of them, there are a total of 12×12= 144 possible outcomes, because each die can land on any of its 12 faces, and the outcome of one die is independent of the outcome of the other. There are 12 possible doubles: (1, 1), (2, 2), (3, 3), ..., (12, 12). To determine the probability of rolling doubles, divide the number of successful outcomes (doubles) by the total number of possible outcomes: 12/144 This simplifies to 1/12 or approximately 8.33%. Yet, there's a more intriguing aspect to consider. Given that a 2d12 system forms a basic bell curve, it's fascinating to examine the probabilities of rolling specific sets of doubles. For instance, if double ones represent a catastrophic failure or double 12s signify an exceptional success, assigning unique outcomes to particular double combinations can lead to a richer diversity of results.

Bark Madly

How does Death Bringer do Maths? Does he consider it another form of sorcery and beneath him? Or does he love fractions because he likes chopping things up? 🤔 🤣

Kristen Solindas
>>115852634

However, the carbon-oil we obtain from drilling is actually not from dinosaurs at all, but from the plant and microscopic life that came before and during their reign - plants, algae, and bacteria, mostly. The hydrocarbons were sequestered over the eons until we dug them up and started using them as fuel. So the efficiency is correct, but the actual source is not. :)

JP Trudel
>>115852634

I went with closest relatives ;-). But you are correct that would actually be 0 energy! Sorry for being off a few million years. On a positive side note Baron solved free energy!

Stefan

What does it mean when you post a math question and you get more responses than when you post a D&D question???

Kristen Solindas
>>115856303

I'd posit it's because math is broad, and applies to all kinds of games, as well as just about everything. I don't know, but that feels right, anyway. 😄

Kristen Solindas
>>115851191

I really like it when this sort of thing is baked into the system, though, not just an exception.

Dave Kuzyk

anydice.com is my favourite tool for figuring these things out. See probabilities of any dice combo you can think of; even add +1s to the roll and everything. Cheers! (sorry if anyone's already mentioned this)

Kaz Faren

So, basically you go from a 5% chance of a Crit with the traditional D20 to an 8.3% chance. That is a 66% increase on the average frequency of rolling a crit.

Kaz Faren

Whether that is a good thing or not I guess depends on whether you feel it should happen more often than it already does.

Kristen Solindas
>>115857424

I like this, as it gives the mean and standard deviation. However, I cannot quite figure out if there's a way to determine odds for number of successes from a pool, especially if the dice "explode". I can do it on my own, but a website or app to do it like this one would be quite useful. Otherwise, I'd have to send out my Excel spreadsheet to those who ask. Maybe I'll just include it with my game when I finally get it done.

Benjamin Marra

Interesting to see that the chances of a crit on a 2d12 is higher than a 1d20. I like 2dX systems for the chances of rolling doubles and how that can be used mechanically.

Dave Kuzyk
>>115857424

There is a mechanism in the code to add exploding dice (output [explode d6]), advantage/disadvantage (lowest/highest), and many other things. Just check the Function Library. Cheers!

Maatlock

There are 144 combinations and 12 instances of doubles. The odds are 1/12 I believe… but my math degree is 12 yrs old and combinatorics is not my strong suit

Shane Gallant

This is fascinating to me! (I am also not a math expert, so take what I say with a grain of salt). I just did a quick calculator thing, and 12 divided by 1 is 0.083 repeated, or the general consensus of 8.3%. But it would seem to me that rolling doubles on TWO d12's would have to be LESS than that! I would be tempted to to half it to 4.15-ish%, but I think that might also be wrong, and I don't know why. I suspect god like math would somehow come into play here... (Calculus or something like that. I haven't done calculus in 30 years. Just thinking about it makes my brain bubble)

Maatlock
>>115859974

I am more confident in my response now. I literally just wrote a Python script that rolled 10 million times, and 8.33% doubles was the experimental result, which is the expected 1/12.

Dinko

@Shane the odds of rolling any number on the first d12 are 100%. (This is because we don't care which number comes up). The odds of the second d12 rolling the same number are 1/12.

Old Man Umby

Pyram King is correct with the odds overall, plus he mentioned the bell curve, which will change the odds for particular pairs. I've plugged in the dice; here is the bell for anyone interested in the exact odds of rolling a particular pair ... https://anydice.com/program/6

MrGunn

@PDM Math is hard, l’m pretty? 🤣🤣🤣

Steven Bogart
>>115854439

Nice explanation, Pyram. Another way to think about it -- the first d12 can be anything, but whatever it is, the second must match it. Let's say the first d12 is a 7. The probability of doubles = probability the second d12 is a 7 = 1/12. No matter what the first d12 shows, the probability the second d12 matches it is 1/12. Hence, the probability of doubles is 1/12, about 8.33%.

Old Man Zed

OK mathies, compare and contrast the impact of a +1 in d20 versus 2d12. Is that +1 sword about the same in both systems or wildly different? Thoughts?

Kristen Solindas
>>115866183

This assumes they are added together (I don't know, as I haven't looked into the game, yet). However, the odds of rolling any SPECIFIC pair on 2d12 (two 4s, two 7s, etc.) is 1 in 144, or 0.6944%.

Christophe Jégou
>>115859974

yes it's 1/12 (8,3%) well done! First draw any score between 1 and 12, second draw 1/12 to get the same result

Kristen Solindas
>>115857756

Yes, but it is a critical success OR failure. The probability of each of those on a d20 are 5%, but for either one is 10%. So the odds are actually lower. 😊

Kevin Beck

Unless my math is way off it would be 12/144. So you would get doubles 12 times out of 144 rolls. How did I come up with this? The chance of rolling doubles on two six sided dice is 6/36.

Dustin Hitzing

It is an 8.333% chance of critical success only. According to the designer interview I just watched there does not seem to be critical failure.

Jay

Ah crap others have already beaten me to it but here is my answer anyway. Please check my math busy day today. Gonna be embarrassed if I messed this one up. When rolling 2d12 there are 12 ways that both dice can land on the same number and there are 144 total possible ways for both dice to land. The probability that both dice land on the same number can be calculated as: Probability = 12 / 144 Probability = 1 / 12 (Reduced to Lowest Common Denominator) The probability that both dice land on the same number is 1/12 or 0.083 Rounding to a Max of 3 decimal places or 8.33%.

Jay
>>115880003

Yes you are correct yet if I'm right you forgot to reduce to the LCD. So 12 goes into 12 one time and into 144 twelve times to it reduces to 1/12 and 6 goes into 6 one time and into 36 six time so that reduces to 1/6. The rest is spot on.

Christian Lindke

Just for fun, I did a simulation in R to answer the question. The exact answer is the mathematical one given above 12/144 = 8.33%, but simulating 1,000,000 rolls I got 8.3%. I'll post the code and illustrations later on my Substack, but I'll give you the illustration on Twitter.

Kaz Faren
>>115857811

Good points, and you may be right in your interpretation. I was only thinking about a critical success. I don't recall if the doubles for the 2D12 system were for failures and successes, or just successes.

John Delphia III

One way to check probabilities is using the anydice.com website. Looking at 2d12 the probability of getting a '2' or '24' (doubles) are both 0.69%. Multiply 0.69 by the 12 doubles you could get makes 8.28% which is basically 1:12.

Graham Kane
>>115854905

He lops off heads and creates his own makeshift abacus. He also doesn’t care if he gets it right or wrong. You know, show your work!

Kristen Solindas
>>115857424

I hope it meets expectations in the modern era of TTRPGs. But I'm making the game I want, and that's what matters, right?

JP Trudel
>>115857424

Indeed, make the the game you want to play. General public expectations have a tendency to brew mediocre games. Intricacies and imperfections are what makes a game shine not streamlined mechanics.

Jeff Boggs
>>115866183

This. I don't think the Normal distribution is relevant in this instance because we only need both dice to match. The Normal distribution will just gives us the mean value of a pair and the expected variation around that mean. To extend @Kristen Solindas' observation, we can add these twelve probabilities together to get the probability of rolling a matching pair, regardless of which pair. One solution is (1/144)+(1/144)+do that ten more times = 12/144. This reduces to 1/12. ca. 0.083 or 8.3%. Which is the same as if we summed 0.00694 or 0.6944% twelve times. Apologies if it appears I'm mansplaining because I suspect (correction after reading more comments, SHE KNOWS but I can't figure out strike through) Kristen already knows this, but others may not.

Ian Butterworth

As others have said, the chance of rolling a double on 2D12 is 8.3%. The first D12 is irrelevant, it rolls whatever it rolls. The second D12 has a 1 in 12 chance of matching the first die. The calculation is 1 divided by 12 and multiplied by 100 = 8.33 recurring.

Christopher Gauci

It's much simpler. To get any number on a d12 it's 1 in 12 chance. For the 2nd die to match that number it's 12 to the power of 12

Christopher Gauci

Expressed as a percentage it's 100 ÷ 144 = 1.44%

Christopher Gauci

My working assumes you try and guess what the double will be before the roll. If you roll one die and then just hope the second die matches the first then it's 1 in 12 once the first number reveals itself. I'm not sure how to work out what the probability is if you throw simultaneously and hope for any double...

Christopher Gauci

I got it....There are 12*12 = 144 possible outcomes when you throw the dies and 12 of those outcomes are doubles so the probability is 12/144 = 1/12

Christopher Gauci

So it's a bit of a redundant mechanic. It's just like guess a number on a 1d12

Kage Holman

Actually, you’re mostly right. Depending on the rules, but if it is like how Warhammer RPG works, its 12^2(144) divided by n-12: where n is the number you must roll under.

Kage Holman

So the chances go up as the needed number increases

Charles Fout

1 in 12. The second die just has to match whatever the first die shows.

Jay
>>115860692

I think you may have missed something. Although you are correct in the 8.33%, the objective is to roll doubles AKA two of a kind not just 12's.

Todd Sharp

Charles is correct. Here's the long version.. You have 144 possible combinations that two 12-sided dice can fall in. Of those, only 12 combinations have both dice showing the same number. 12:144 = 1:12 As a percentage, there's an 8.33% chance that you get doubles on any given casting. By contrast, there's a 16.66% chance you get doubles with 2d6. Seems like they want to increase the chance of getting a critical hit or miss in Daggerheart, without just declaring that a 1/2 or 19/20 constitutes a crit. (I consider the 2% difference negligible) I know this is late. But, I needed a break from work and this was interesting.