[RPG] Dice rolling mechanic where modifiers have a predictable and consistent effect on difficulty

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I am looking for a dice rolling mechanic that makes it such that increasing or decreasing a modifier on the roll has a constant multiplier effect on the probability of the outcome.

Say you have to make a roll for STAT. Such a roll has a probability of success of 50%. Now say you roll with a mod of -1, this roll has a probability of success of 25%. -2 has a probability of 12.5%. -3 is 6.25% and so on, always halving. The other way around it should be the same but for the probability of failure, always being divided by the same factor.

It doesn't have to be a multiplier of 0.5, in fact I'd much rather it was a multiplier of 0.66-0.75, not such an extreme change. The default unmodded value doesn't need to be 50% chance of success either, it can be something else.

Is there any kind of dice rolling mechanic I can use to simulate something like this?

Best Answer

Roll a number of extra dice along with the base roll

I will start with an example for 0.75 and then explain how this can be more generalized. First, you have your base STAT roll with no modifiers which can be whatever you want. For this example, let's assume it has a 60% chance of success (such as a d20 that needs to roll at most a 12, it doesn't really matter here).

Now if you want to roll STAT-1, you still do the base STAT roll, but you also roll an extra 1d4. In order for STAT-1 to succeed, the base STAT roll must succeed and the 1d4 can't have a result of 1. This gives STAT-1 a chance of success equal to 0.6*0.75.

For STAT-x, you simply roll and extra xd4 and if any of these d4 rolls a 1 it is a failure, giving you the progression you want.

For STAT+1, you roll an extra 1d4 but this time, it is a success if the base STAT roll succeeds, or if the 1d4 rolls a 1 (or if both things happen). This causes the chance of failure to go from 0.4 to 0.4*0.75. Similarly for STAT+x, you roll x additional 1d4 and you get a success if any of them is a 1.

Using an extra d4 like this gives you a progression of 0.75 because that's the probability of not rolling a 1 on a d4. You can adjust this to other numbers by using different dice and different ranges of numbers. For example using a 1d3 would allow a 2/3 (0.66...) progression.

The Standard Check

This is pretty straight-forward (written as %success):

\begin{array}{l|r|r|r} \text{Difficulty} & \text{Fumbles} & \text{Joe} & \text{Rambo} \\ \hline \text{ 5 Easy} & 55\% & 80\% & 100\% \\ \text{10 Normal} & 30\% & 55\% & 95\% \\ \text{15 Tough} & 5\% & 30\% & 70\% \\ \text{20 Really Tough} & 0\% & 5\% & 45\% \\ \text{27 WTF} & 0\% & 0\% & 10\% \end{array}

No surprises here, Fumbles is almost as likely to fail as to succeed in trivial stuff, and can barely pass tough challenges - anything above that is beyond him. Joe is more likely to pass a tough test than to fail in an easy one, and Rambo generally kicks butt, passing really tough challenges about half the times, and nailing a "WTF" challenge one time out of ten.

The Dungeon World Variant

This works differently, now the odds for a clean success are (success)×(success) of the standard check, and the odds for "success with consequences" are 2×(success)×(failure) (written as %success/%success with consequences, rounded to whole percentage):

\begin{array}{l|r|r|r} \text{Difficulty} & \text{Fumbles} & \text{Joe} & \text{Rambo} \\ \hline \text{ 5 Easy} & 30\%/50\% & 64\%/32\% & 100\%/ 0\% \\ \text{10 Normal} & 9\%/42\% & 30\%/50\% & 90\%/10\% \\ \text{15 Tough} & ~0\%/10\% & 9\%/42\% & 49\%/42\% \\ \text{20 Really Tough} & 0\%/ 0\% & ~0\%/10\% & 20\%/50\% \\ \text{27 WTF} & 0\%/ 0\% & 0\%/ 0\% & 1\%/18\% \end{array}

In your variant, the results are less extreme, as the two dice tend to cancel out and reduce the odds of unlikely results (since now you need to get them twice for them to count...).

So, Fumbles now succeeds at easy tasks 80% of the times, and is more likely to pass normal tests than fail them - but he is very likely to have consequences and not a clean success. In fact, the only way he can pass a tough challenge now is with consequences, but the odds for that are doubled (10% now instead of the clean 5% from before).

Joe now passes easy tests 96% of the times (instead of 80% in the standard method), and tough tests 51% of the times. but the harder the difficulty, the more likely he is to face some consequences.

Finally, Things got really better for Rambo - he always succeeds in normal or lower challenges and he passes tough and really tough challenges 91% and 70% of the time respectively (it was 70% and 45% in the standard method), and he nearly doubled his chances for nailing a "WTF" (19% now, was 10% before). He'll rarely face any consequences for normal or lower DCs, but for tough ones, it'll happen about half the times, and almost always for "WTF" challenges...

Bottom line: Your variant makes static bonuses more powerful, as the randomness is now averaged - extreme results are more rare, the weak become weaker (but less prone to complete fiascoes), and the strong can almost always count on winning. I'm not criticizing that, just making an observation - whatever works better for you and your group is your best choice.

The D&D Next Twist

This is somewhere in between the standard version and your variant. The odds for an advantaged success are (success)×(success)+2×(success)×(failure) of the standard check, and the odds for a disadvantaged success are (success)×(success) (written as %adv. success/%disadv. success, rounded to whole percentage)

Note that since advantage/disadvantage are not part of the same scenario, you can't add their percentages for a combined success rate as in your variant:

\begin{array}{l|r|r|r} \text{Difficulty} & \text{Fumbles} & \text{Joe} & \text{Rambo} \\ \hline \text{ 5 Easy} & 80\%/30\% & 96\%/64\% & 100\%/100\% \\ \text{10 Normal} & 51\%/ 9\% & 80\%/30\% & 100\%/ 90\% \\ \text{15 Tough} & 10\%/ 0\% & 51\%/ 9\% & 91\%/ 49\% \\ \text{20 Really Tough} & 0\%/ 0\% & 10\%/ 0\% & 70\%/ 20\% \\ \text{27 WTF} & 0\%/ 0\% & 0\%/ 0\% & 19\%/ 1\% \end{array}

It may not be immediately obvious, but what you have here is that the odds for an "Advantaged Success" are exactly the odds for success (with or without consequences) in your variant. And the odds for a "Disadvantaged Success" are exactly the odds for a "clean success" in your variant.

So, here Fumbles succeeds at easy and normal tasks 80% and 51% of the times respectively when advantaged, but only 30% or 9% with a disadvantage (compared to a flat 55% and 30% in the standard method). Joe now passes a tough challenge at either 51% or merely 9% (compared to 30% in the standard method) - so an advantage means succeeding half of the times instead of less than once out of ten.

Rambo is having the time of his life here, as his odds for success when he has the advantage are the same as in your variant - without the risk of any consequences. Seriously, 70% chance for success at DC 20?! A bit more and he'll qualify as Chuck Norris's stand-in!

When he doesn't have the advantage, though, his odds are worse compared to the basic method (49% vs. 70% for tough, 20% vs. 45% for really tough and 1% vs. 10% for the WTF DC).

Bottom Line: In this variant, having the advantage raises your chances of success to the roof, and being disadvantaged severely pushes them down. Compared to your variant, here the players (who tend to become closer to "Rambo" than to "Joe") are rewarded for making the effort to gain an advantage, instead of always being able to count on their superior static bonuses.

The Ultimate Cocktail

Calculating this isn't that straight forward, and we need to come up with 4 different figures for each cell in our benchmark table (both advantaged / disadvantaged and success / success with consequences for each). Again, based on the results of the standard method, the odds for an advantaged clean success are (success)×(success)×(success)+3×(success)×(success)×(failure).

The odds for an advantaged success with consequences are 3×(success)×(failure)×(failure)
The odds for a disadvantaged clean success are (success)×(success)×(success).
And the odds for a disadvantaged success with consequences are 3×(success)×(success)×(failure)

I'll split the results to 2 tables to keep it readable, (results are written as %clean success/%success w. cons. (%Total Success) - same as in your variant, with total success added in parentheses):

With an advantage

\begin{array}{l|r|r|r} \text{Difficulty} & \text{Fumbles} & \text{Joe} & \text{Rambo} \\ \hline \text{ 5 Easy} & 57\%/33\% \; (91\%) & 90\%/10\% \; (99\%) & 100\%/ 0\% \; (100\%) \\ \text{10 Normal} & 22\%/44\% \; (66\%) & 57\%/33\% \; (91\%) & 99\%/ 1\% \; (100\%) \\ \text{15 Tough} & 1\%/14\% \; (14\%) & 22\%/44\% \; (66\%) & 78\%/19\% \; ( 97\%) \\ \text{20 Really Tough} & 0\%/ 0\% \; ( 0\%) & 1\%/14\% \; (14\%) & 43\%/41\% \; ( 83\%) \\ \text{27 WTF} & 0\%/ 0\% \; ( 0\%) & 0\%/ 0\% \; ( 0\%) & 3\%/24\% \; ( 27\%) \end{array}

With a disadvantage

\begin{array}{l|r|r|r} \text{Difficulty} & \text{Fumbles} & \text{Joe} & \text{Rambo} \\ \hline \text{ 5 Easy} & 17\%/41\% \; (57\%) & 51\%/38\% \; (90\%) & 100\%/ 0\% \; (100\%) \\ \text{10 Normal} & 3\%/19\% \; (22\%) & 17\%/41\% \; (57\%) & 86\%/14\% \; ( 99\%) \\ \text{15 Tough} & 0\%/ 1\% \; ( 1\%) & 3\%/19\% \; (22\%) & 34\%/44\% \; ( 78\%) \\ \text{20 Really Tough} & 0\%/ 0\% \; ( 0\%) & 0\%/ 1\% \; ( 1\%) & 9\%/33\% \; ( 43\%) \\ \text{27 WTF} & 0\%/ 0\% \; ( 0\%) & 0\%/ 0\% \; ( 0\%) & 0\%/ 3\% \; ( 3\%) \end{array}

Right, so, what have we here? With an advantage, even Fumbles will almost always pass easy challenges, and can count on passing normal DCs 2 out of 3 times. He'll even beat a tough challenge 14% of the time. Without an advantage, his odds are much lower, but are still way better than what he'd get in the D&D Next Twist (57% vs. 30%, 22% vs. 9% and 1% vs. 0%) - yes you read that right, this cocktail means that even a disadvantaged Fumbles is more likely to succeed than fail an easy challenge.

Joe seems to have turned from average into J.I.Joe - passing tough challenges 66% of the time with an advantage, and with negligible odds of failure at easier tasks. Disadvantaged, he is still almost guaranteed to beat the easy DC, and his odds of beating harder DCs is comparable to the standard method (90% vs. 80%, 57% vs. 55%, 22% vs. 30% and 1% vs. 5%) - so he feels a slight disadvantage in the higher DCs, but easy to normal are still easier.

And finally, Rambo. While he is still not quite Chuck Norris, this cocktail turned him into Captain America. When disadvantaged, he will still beat DC 20 more than 2 out of 5, and pass tough DCs about 4 out of 5! Normal and easy are now a none-issue, he can't fail them. period. With an advantage, he needn't even roll for tough or lower tests - except to find out if there are any consequences, and the almighty DC 20 is now easier for him than an easy task was for Joe in the standard method. And he'll nail a WTF more than 1 out of 4...

Bottom Line: The ultimate cocktail makes everything very easy - for everyone. This arguably is still a matter of taste, but to me it seems like it totally breaks the system. You may be able to compensate this to some extent by either nerfing the bonuses (don't let Rambo get more than a +3, hit Fumbles with a -10 penalty so he doubts his ability to open doors again) or by raising all DCs - or both.

This idea ends giving you a "success with consequences" result, by all but eliminating the "failure" result - seems to me like too much of a sacrifice, but you may of course disagree.

I suppose there are other ways to gain this "success with consequences" without losing so much (probably also without using this somewhat cumbersome process of rolling 3d20, removing one, and checking the rest...). Maybe use two different colored d20 and decide that if one is higher the result has additional consequences (for both failures and successes), maybe treat a "double" result as special, or perhaps use some way to measure margin of success by comparing die result and target DC. There are many options, but this feels like a topic deserving its own question.

Hope this helps (and hope I didn't make any more miscalculations...), please feel free to use the comments for any clarifications or questions still missing from this answer.

P.S Thanks to DiabloMonkey's help in fixing a calculation error - and the inaccurate conclusion derived from it.

[RPG] How to approach the probability of a d100, using 2 d100’s

Average The Skills

If he has to use two skills, average the two skills together and then make one roll. In this case, that'd be a single roll to get 50 or below, since he has 50 in both skills (so the average is 50).

If he was better at one skill than another, it'd look slightly different. Say he has a 50 in Stonecarving and 25 in Artistry. That makes the average of them 37.5, so he'd have to get a 37 or below (or a 38, depending on how you want to round).

That basically treats it like he's using both skills and has to succeed on using them in combination, rather than having to succeed on separate rolls for both. It also keeps it to a single roll with similar odds, and is relatively simple to implement for players.

Alternative - Geometric Mean

The downside to averages is that if you're really good at one skill (say 100 in Stonecarving) and really bad at the other (0 in Artistry), you still have a 50 in the combined skill. That might not be what you had in mind, as someone with no artistic talent doesn't suddenly gain it just because they are working with stone.

In this case, an alternative method is to take the Geometric Mean. For two skills, that is this formula:

$$\sqrt{skill_1 \cdot skill_2}$$

So, if you have 100 in Stonecarving and 0 in Artistry, you do \$100 \cdot 0\$, which is 0. The square root of that is 0. As a result, you now need to at least have 1 skill point in Artistry in order to attempt the combined result. If you did have Artistry 1, you'd get \$100 \cdot 1 = 100\$, the square root of which is 10. As you add points in Artistry, your chances will quickly increase.

For my previous example of 50 and 25, you'd get \$50 \cdot 25 = 1250\$, the square root of which is 35.3.

The main downside to this method is that in a tabletop game, it's extremely hard to calculate without a calculator. Even with one, it requires a more complicated understanding of math and is more time consuming. If you put this in a rule book, there will be people who won't understand what you want them to do. For something like a video game where it's calculated by the software, that isn't a problem.

(Thanks to Peteris and Vatine for the suggestion!)

Alternative - Minimum/Maximum

A very simple method for combining skills is to use either the minimum skill in the two of them, or the maximum skill in the two of them. The maximum means you're just using the skill you're better at, while the minimum means you're using the skill you're worse at.

In the case of the minimum, it simulates the idea that you have to succeed on what you're weaker at in order to accomplish the goal. This lets you do it in a single roll, and is very easy to understand. It also has some issues, in that if you're extremely good at Stonecarving and so so at Artistry, your Stonecarving gets ignored in this system as you only roll on your lower one (Artistry).

Because of that, I don't think it really accomplishes what you intend very well, but it's ease of use is a significant upside over the other suggestions.

(Thanks to Neil Slater and Ellesedil for suggesting.)