Unlike most other roleplaying games, Dungeons and Dragons and recently spun off Pathfinder RPG allow for multiple ways to generate your characters starting characteristics. A point-buy system has become the most popular way by most players because it will result in each player having a starting character that is roughly of the same power level. This is a concern for a lot of gaming groups because the games they play are dominated by combat (which chews up quite a bit of game session time per encounter) and players tend to feel resentful if one player is significantly outshining the other players in terms of body count. In more social games, I think disparity in character power levels doesn’t mean as much. If we are roleplaying out Bilbo Baggins’s birthday party does the difference in power level between Gandolph and Samwise Gamgee really matter all that much?

Personally I’ve discovered that I tend to enjoy game sessions where characters tend to be subpar and the power levels are asymmetrically dispersed. First off, it’s a bit more real because not everyone brings equivalent skills to the table in life and why should games that imitate life be all that different. Secondly, something about rolling three six-sided dice (which is abbreviated 3d6 in gamer notation) and having those dictate what your characters scores are in the exact order in which you role them disconnects you from your character. If you’re using a point buy system, or even a method of assigning the stats in the order you like, you become invested in your grand design for the character. Having luck determine what your character’s stats are reinforces that you are not designing this character but instead being asked to play a character who is very different from you and, for that matter, very different than you might have preferred.

Some of the most beloved characters have come from subpar stats. The wizard Raistlin from the Dragonlance Series of books rolled a very low stat for his Constitution, which is all but a death sentence for a first level wizard in the first edition of Dungeons and Dragons. However, as I understand it, the character survived and became one of the most beloved characters in all of D&D fiction in large part because of his limitations. Growing up, I never liked Superman in comparison to other super heros because Superman was simply too perfect and too powerful. To present Superman with any kind of challenge was a challenge itself, and that removed the character too far from the everyday struggles of the reader.

On the Paizo board, it was recently asked what the exact statistics are for the character generation method of 4d6 drop the lowest. That is, if you roll four six sided dice and then disregard the lowest drop the lowest score (because the D&D convention is that you want starting stat numbers that are between 3 and 18). Of the dice conventions used to generate stats, it is the most common, but most people seem at a loss for exactly what the probabilities are for a given score. Heck, even figuring the average seems daunting.

Being a gambler who has done many calculations regarding card distributions, it was second nature for me to generate these numbers. In fact, I did it in my spare time between hands while at work with a pen and paper and only had to resort to the built in calculator of my cellphone for the hardcore number crunching. I am reproducing the exact way to do it here in case anyone wants to do their own dice number crunching.

Firstly you need to recognize, that you are dealing with a distribution of four dice. The end result may only seem like the handiwork of three dice because you only use three out of the four rolled, but you’ll be hopelessly lost in attempting to understand the nature of the problem. The distribution of values is given by the four six sided dice and then a convention is applied to convert the results of these four dice to a number between 3 and 18. Since we are dealing with four dice, not three, the total number of possible results are given by 6x6x6x6. In other words, there are 1296 different ways that four dice can fall.

Now if you’re familiar with other discrete probability distributions, then this should come as no surprise to you. There are 36 different ways that two six-sided dice can fall, but they will only generate a number between 2 and 12 when added together. In order to understand the probability of a given score, all one needs to do is compare the number of different ways the dice can give you that score and then compare it to the total number of distributions. Also, the total number of distributions is a critical way to check your work because all of the total combinations that render a given score should add up to the total number of possible combinations when you add the combinatorial score of every possible result.

With all that out of the way, let’s get started with the easiest result to understand, the lowly three. There’s only one possible combination of four dice ignoring the lowest that will result in a score of three, and that’s if every die yields a one. Therefore, if someone wants to know the probability of getting a three with this method, it’s simply the one divided by 1296, or .00077. Obviously, you’re not going to be getting many threes with this method. So, I’m going to abbreviate this score as (1,1,1), which means that, of the dice that you are scoring, each will have a value of one.

Now let’s take a look at the result of (2,2,2), which is one of the possible combinations that yield a six. You might be tempted to guess that there is only one combination of the dice that will yield (2,2,2) but you’d be wrong. Since we’re dropping the lowest score, the dice results of all twos will yield exactly the same score as three twos and a one. Furthermore, since there are four different dice on which this one could be, there are four combinations of three twos and a one. Therefore, there are five combinations of (2,2,2). Be extension, there are nine combinations of (3,3,3) because we have one combination of all threes, four combinations of three threes and a two, and four combinations of three threes and a 1. It’s a bit formulaic at this point to understand that there are 13 combinations of (4,4,4), 17 combinations of (5,5,5) and 21 combinations of (6,6,6).

Now that we’ve looked at those combinations, another should be fairly easy to understand, and that’s the combinations that yield four. In order to generate a four, you need a single two and three ones or (2,1,1). Not that order is not important in my notation, so understand that (2,1,1) includes all combinations of a single two and two ones. Consequentially, you’ll never see a (1,2,1) because those combinations are already included in (2,1,1). Since the 2 can be on any of the four dice, there are four combinations of (2,1,1). Therefore, you now know the exact probability of getting a four with the 4d6 drop the lowest and it’s 4 divided by 1296.

Now let’s look at all the results that give us a 5. There are two separate combinations that yield a five and that’s (3,1,1) and (2,2,1). We know from our last example that there are four combinations of (3,1,1), but what about (2,2,1). To figure that out, I just wrote out all the different combinations of (2,2,1) which are: 2211 2112 1122 1221 2121 1212. So there are six combinations of (2,2,1). Added to the four combinations of (3,1,1) we have a total of ten combinations that give us 5, and our probability is 10/1296.

Now let’s look at the combinations which result in six which are (4,1,1) (3,2,1) and (2,2,2). We already know that (2,2,2) has five combinations and that (4,1,1) has four, but how many does (3,2,1) have? Well let me write them all out:

3211 3121 3112 2311 2131 2113 1123 1132 1231 1213 1321 1312. That’d be 12 combinations of (3,2,1). I don’t really know any shortcuts for this except just writing them out. It’s a bit of a pain, but the great thing is that we can start applying this method to all combinations that have three different scores between the brackets. In the rest of this article, I’ll call this a three value distribution and you just need to know that it has 12 combinations.

We now know the odds of rolling a six on 4d6 drop the lowest. Since the combinations are (3,2,1) (2,2,2) and (4,1,1) the total number of combinations is 12 + 5 + 4 which is 21. The odds of rolling a six on 4d6 drop the lowest is 21/1296.

Now let’s look at the combinations of seven. We’ve got (5,1,1) (4,2,1) (3,3,1) and (3,2,2). Well the first three combinations we’ve seen before and know that there are four combinations of (5,1,1) 12 of (4,2,1) because it’s the three value distribution we saw above, and six of (3,3,1) because it’s exactly the same as the (2,2,1) I solved earlier. But what about (3,2,2)? Well since we’re dropping the lowest score, the dice results that can generate (3,2,2) are four separate combinations when the extraneous die equals two plus however many combinations there are when the extraneous die equals one. How many combinations of the four dice 3221 are there? Well, it’s just another three value distribution like we did earlier, so there are 12. Adding the 12 from 3221 plus the four from 3222 we get a total of 16 results. So now we can solve for the result of seven.

As you can start to see, the math is going to get a bit tedious, so I’m going to adopt a convention. To see the number of combinations that result from when the extraneous die is equal to a given score, I’ll use the abbreviation L=that score and C will equal the total number of combinations. As an example of what I wrote earlier, here it is in the new notation:

(3,2,2) when L=2, C=4 because there’s only one varying value to distribute over four dice [3222 2322 2232 2223]

when L=1 C=12 because it’s a three value distribution.

Now let’s express seven in a more abbreviated notation so I don’t have to write these explanatory paragraphs each time. For the result of seven, we have the following combinations: (5,1,1) (4,2,1) (3,2,2) (3,3,1) which when expressed in total combinations is 4 + 12 + 16 + 6 = 38.

If you’ve been able to follow me so far, we’re almost to the finish line. Let’s look at the combinations that result in a score of eight:

(6,1,1) (5,2,1) (4,3,1) (4,2,2) (3,3,2). The first is a single value distribution which will result in four combinations, the next two are three value distributions for 12 each, (4,2,2) is the same as (3,2,2) above for 16 combinations. Now let’s look at (3,3,2). When L=2 C=6 because it’s a two value distribution as we saw earlier, and when L=1 C=12 because it’s a three value distribution. So the total number of combinations for (3,3,2) is 18.

Therefore, (6,1,1) (5,2,1) (4,3,1) (4,2,2) and (3,3,2) yields 4 + 12+ 12 + 16 + 18 = 62.

Now let’s look at the results for nine. Nine’s combinations are (3,3,3) (6,2,1) (5,3,1) (4,4,1) (5,2,2) and (4,3,2). (4,3,2) is the only thing that’s really new to us, so let’s look at it. When L=2, C=12 because it’s a three value distribution. When L=1, there are four separate values to distribute over four dice, which is something we haven’t seen before. Fortunately there IS a formula for how to chose four when order is important and it’s 4x3x2x1 which gives us 24. So when L=1, C=24. I’ll call this a four value distribution for future reference. Therefore, there are 36 combinations of (4,3,2), so nine’s combinations total 9+12+12+6+16+36 for a total of 91.

At this point the rest of this should become a exercise for the reader, but I’ll go ahead and work out the hardest one which is (6,5,4). When L=4, C=12 because it’s a three value distribution. When L=3, C=24 because it’s a four value distribution. We also get 24 combinations for L=2 and L=1, so the total number of combinations for (6,5,4) are 24×3 + 12 which results in 84.

Here’s a table with the result complete with the total number of dice results that will give it in parentheses and the number of combinations of each die result given in brackets.

Result | Combinations | Total Combinations | Percentage Chance |
---|---|---|---|

3 | (1,1,1) [1] | 1 | .0772% |

4 | (2,1,1) [4] | 4 | .3086% |

5 | (3,1,1) [4] (2,2,1)[6] | 10 | .7716% |

6 | (4,1,1) [4] (3,2,1) [12] (2,2,2)[5] | 21 | 1.62% |

7 | (5,1,1) [4] (4,2,1) [12] (3,2,2) [16] (3,3,1) [6] | 38 | 2.932% |

8 | (6,1,1) [4] (5,2,1) [12] (4,3,1) [12] (4,2,2) [16] (3,3,2) [18] | 62 | 4.784% |

9 | (3,3,3) [9] (6,2,1) [12] (5,3,1) [12] (4,4,1) [6] (5,2,2) [16] (4,3,2) [36] | 91 | 7.022% |

10 | (5,3,2) [36] (6,2,2) [16] (6,3,1) [12] (5,4,1) [12] (4,3,3) [28] (4,4,2) [18] | 122 | 9.414% |

11 | (6,4,1) [12] (6,3,2) [36] (5,5,1) [6] (5,4,2) [36] (5,3,3) [28] (4,4,3) [30] | 148 | 11.42% |

12 | (4,4,4) [13] (6,5,1) [12] (6,4,2) [36] (6,3,3) [28] (5,5,2) [18] (5,4,3) [60] | 167 | 13.18% |

13 | (6,6,1) [6] (6,5,2) [36] (6,4,3) [60] (5,5,3) [30] (5,4,4) [40] | 172 | 13.27% |

14 | (6,6,2) [18] (5,5,4) [42] (6,5,3) [60] (6,4,4,) [40] | 160 | 12.35% |

15 | (5,5,5) [17] (6,5,4) [84] (6,6,3) [30] | 131 | 10.11% |

16 | (6,5,5) [52] (6,6,4) [42] | 94 | 7.253% |

17 | (6,6,5) [54] | 54 | 4.167% |

18 | (6,6,6) [21] | 21 | 1.620% |

So what are the chances you will write up some “math-hammer” on Warhammer 40K stuff?

Rather remote considering I don’t play that game. Is there a specific problem you wanted me to solve.

yes, which tyranid units do I purchase to beat Space Marines, Eldar and Orks. ðŸ˜‰

Unfortunately the person best qualified to answer that question died recently. Sorry.

Preston, can you change this 4d6 table so that it has two significant digits? I’d like to see the precise numbers. Thanks!

There ya go.

A bit of other number crunching reveals that the Mode (most common number) is 13. the medium (the middle number in the distribution) is 12 and the mean (average) is 12.2446. This means even using the Standard fantasy 15 point system in Pathfinder will yield close to (you could have 3 12’s and 3 13’s averaging 12.5) the same results as 4d6 drop the lowest. Though I suspect most players will not decide to even out all their attributes.

True. The comparative advantage of dice 4d6 dice rolling is that you will have some big numbers and the rest mostly medium numbers. So you don’t have to pay for a big number by making another number smaller.

thanks a load! I was not looking forward to doing all the probability shenanigans, so finding this was a godsend!

Interesting observations. Thanks for the probability calculations Preston! I’d be curious to know whats the probability for 3D6 REROLL 1s ahev you ever done it? If so could you link me up or email me? Thanks again

I presume that you can invert the table for ‘4d6 drop the highest’ score i.e. Only 1 chance to get 18 and 21 chances to get 1 etc?

Yes

This is an excellent post and exactly what I was looking for. Thanks!

I’m working on a system where checks are made rolling 3d6 and for each ‘advantage’ a character has (including relevant proficiencies), they add an extra d6 to the pool and keep the highest 3 dice. Conversely, any ‘disadvantage’ they have to a skill check subtracts an ‘advantage’ die or (if there are more ‘disadvantages’ to the check than ‘advantages’) then they roll that many d6s and keep the lowest 3.

This way, critical successes (17s and 18s) and critical failures (3s and 4s) become appropriately more/less probable given the conditions and expertise of the character: a skilled character with a couple of relevant proficiencies is much less likely to ‘botch’ a check than an unskilled one, and disadvantageous situations make a ‘botch roll’ (rather than a merely low one) more likely even if they have a high ability score. Additionally, this makes multiple ‘advantages’/’disadvantages’ have diminishing effects on checks (ie, 4d6(drop lowest) is on average a +2.24, yet 5d6(drop 2 lowest) is only a +3.43 on average, and so forth).

Here’s the math for up to 7d6 for both ‘advantaged’ and ‘disadvantaged’ rolls:

https://anydice.com/program/16aed

That’s quite similar to how GURPs does it.

Technically, you mean permutation every time you said combination. (And there are 42 permutations of 2d6, if I did my work correctly.)

Still, this helped me find my problem with the math, so thank you.

You’re welcome.

What am I doing wrong in my calculations?

I only ask because shouldnâ€™t it be 24 possible combinations for 18?

4d6 – 6,6,6,x where X can be any value of 1 through 6?

Shouldnâ€™t that be 6*4 possible combos that end up as an 18?

Die 1 = 6

Die 2 = 6

Die 3 = 6

Die 4 = 1-6

Then repeat with each other die falling in the 1-6 potential?

Also, wouldnâ€™t that lead to 17 being harder to roll than than 18 specifically, as both need a precise combination (6,6,5 and 6,6,6 respectively) but the fourth die can be a 1-6 and be an 18 but it has to be a 1-5 to be a 17 (as any time the 4th die is a 6 the score is now an 18)?