209

All this does is linearly adjust the normally-flat 5% probability for each number to occur. What results is a increased or decreased probability of any number above or below average to occur, positively for advantage and negatively for disadvantage. See this AnyDice function set, which yields the following: Black is d20, orange is highest of 2d20, blue is ...


200

It doesn't do anything. As a chemist, I had to give this a try. First, I started with a die that had a clear bias toward ~18, according to the saltwater test. The first picture is the initial drop into the water, and the second is after poking the die. I put the die on a paper plate, and microwaved it for 4 minutes. I made sure to put it on its "side," ...


87

(SKILL+CONSTANT) dX, keep highest CONSTANT I don't know exactly the behavior you're going for, so there'll be a few arbitrary numbers in my example: skills can be ranked from 0 to 5 dice rolled are d12 ('cause I think they don't get enough love) we're going to keep highest 3 dice. In this case we're looking at rolling (3+SKILL) d12, keep highest 3. It ...


83

Let's not overcomplicate it: a simple Nd6 + X formula will work just fine. Specifically, the effects of Greater (4d4 + 4) and Superior (8d4 + 8) healing potions are quite closely replicated by rolling 2d6 + 7 and 4d6 + 14 respectively. These formulas yield exactly the same average number of hit points healed as the originals (14 for Greater and 28 for ...


82

From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. The averages are different, which you already know. They obviously have different minimums (10 vs 5), and so ...


78

There's a 59.5125% chance of survival. Naively, we might have thought there'd be a 55% chance of survival as 55% of the roll results are good. But the 20 is a slightly better result than the 1 is a bad one, so that pushes up the probability a bit. Let's see how. The approach The simplest way to tackle this is to look at the probabilities of surviving in ...


76

Yes it does. Your instinct is right. The more dice, the more likely you are to roll some 1s. If I'm reading you right you're just interested in whether any 1s appear in your rolled pool. It may not seem obvious but the easiest way to think of this is to model the probability of rolling all 2s-through-10s. That is $$P(\text{not }1)=\frac 9 {10} = 0.9$$ ...


74

For "at least one" probability problems, it's usually easier to start by calculating the chance that none of the dice crit, as that saves you the hassle of combining the probabilities of getting 1/2/3 crits. Best of three rolls with 18-20 crit range: ~39% chance to crit Chance that a single die will not crit: 17/20 = 0.85 Chance that all three dice will ...


70

Yes, a d100 is the same as 2d10 with one as the percentile. A d100 goes 1–100, a d10 goes 0-9. Neither allows you to roll a 0, because of the way you count a percentile dice. (00 on the percentile and a 6 on the other dice forms 6, 00 on one and 0 on the other is 100, no option will result in 0.) Do remember to use different colors of dice, else you will ...


68

Given the example of (2d6)*2 (henceforth referred to as 'Doubled Damage') vs (4d6) (referred to as 'Doubled Dice'): When you double the damage rolled instead of doubling the dice rolled, you create a more evenly distributed curve. Using either method, you have the best odds of rolling the average damage for the dice you are using but in the doubled damage ...


67

I've forgotten the formal proof for this, but hopefully this is correct: Consider a D6 (for the sake of concrete language). When you roll a 1, you reroll the die and keep the result. This produces an average value of 3.5, and happens 1/6 of the time. When you roll a 2, you reroll the die and keep the result (even if it's lower). This produces an average ...


65

Overall, this will make area/multi-target spells less reliable, but more potent. Numbers The chance of a single individual to save is not affected by this change. The number of individuals affected over multiple castings is also not (or just slightly) changed. What changes is the number of individuals in a given group that make the save. Let's assume a ...


65

This is a classic clustering fallacy Wikipedia defines the clustering fallacy as: ...the tendency to erroneously consider the inevitable "streaks" or "clusters" arising in small samples from random distributions to be non-random. Basically, a perfectly random die will have long (potentially very long) steaks of seemingly non-random behavior, either very ...


62

TL;DR A GM shouldn't roll all-or-none saves. If reduced rolling is necessary, instead they should figure the expected number of saves, then add a d4 and subtract a d4. Below are pictured the results of this method for various numbers of enemies and probabilities of saving. Read on to see how these are derived. All-or-none is a bad idea. But the GM need ...


62

The other answers do a good job of answering the question, but I'll point out how you can answer questions like this in the future: https://anydice.com/ is a very powerful (if slightly complicated) calculator for these sorts of questions. In your case, you'd enter the query: output [highest 1 of 3d20] And then select "At Least" from options below to get ...


61

Rolling 12d20 and taking the 6 highest is not equivalent Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 ...


53

It's not just in extreme cases that this doesn't add up, though. Let's say that you hit a target on a 16+. If you get a +1, you now hit them on a 17+, which is a 6.25% increase in your chance of hitting, and a 25% reduction in your chance of missing. People are using "increased by a percent" sloppily — or, I guess, if you sigh and admit that language ...


52

Dice work just fine for the most common numbers of players I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of ...


51

About 27%, or a little over 1/4 This is a pretty easy calculation to run in anydice: As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner. How do you calculate that without a simulation? The probability of getting ...


50

The d3 is a rare die, and the d2 is a coin. Substituting a D3 can be done with any die whose total number of faces can be divided by 3. These include in your case the d6 and the d12. The easiest way to do this is to use a d6 and say the following: 1 → 1, 2 2 → 3, 4 3 → 5, 6 Or in other words, divide by 2, rounded up. You can also work in cycles, with one ...


49

Possible exceptions: The player already has advantage; granting them advantage again does nothing, but imposing disadvantage on the monster does. The 'Lucky' feat allows the underdog with disadvantage to go from "roll two, choose the worst" to effectively "roll three, choose the best". Asymmetry: Stopping the bad guy from doing something might be more ...


47

There sure is! Pick a size of a pool of d20 dice. The bigger the pool, the stronger the results. Next grab a d6, d10, a different colored d20, or even a coin. Roll the die pool and roll the extra die. If you got an even number on the die, pick the smallest roll from of the pool of die and use this as a result. Odd? You pick the largest die value from the ...


47

The math is straightforward With an advantage you are looking for best of two results. To figure out your odds you need to multiply the chance of FAILURE together to find out the new chance of failure. For example if you need 11+ to hit rolling two dice and taking the best means instead of a 50% of failing you have only a 25% chance of failing (.5 times .5)....


46

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 ...


46

This would be a great change if you were in a fight where you were almost certainly going to lose. Imagine a spell that defeats 1/2 of the creatures it targets. You use it on 2000 creatures. If more than 10 survive after the spell is cast, you lose. With one roll, you have a 50% chance of winning the fight. With one roll per creature, you certainly lose ...


44

D100 and d%+d10 have exactly the same probabilities. If all 3 dice involved are fair, then they should come up with very similar distributions when rolled repeatedly. Obviously this isn't always the case as dice aren't consistent and there is a lot of randomness unless you roll a lot of times. It seems there might be some confusion as to why d% doesn't have ...


40

It's inherently unbalanced You're using ways of determining success/failure using methods yielding different probabilities. For this reason it's not fair/balanced. That being said, let's do a more detailed analysis of the probabilities of achieving a certain result using both methods. The probability of rolling at least n is 1-(probability of rolling ...


Only top voted, non community-wiki answers of a minimum length are eligible