# Tag Info

126

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 ...

71

A good way to analyze the differences between the two distributions is to imagine a head-to-head contest between characters. First, suppose you have two identical characters, A and B, rolling off against each other with d20. They tie 5% of the time; 47.5% of the time one wins; 47.5% of the time the other wins. In contrast, if you use 3d6, ties occur 9.2% ...

70

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 $$P(\text{not }1)=\frac 9 {10} = 0.9$$ ...

52

Yes, a d100 is the same as 2d10 with one as the percentile. A d100 goes 1–100, a d10 goes 1–10. Neither allows you to roll a 0, because of the way you count a percentile dice. (10 on the percentile and a 6 on the other dice forms 6, 10 on one and 10 on the other is 100, no option will result in 0.) Do remember to use different colors of dice, else you will ...

52

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 ...

48

Answers Below are the expected number of steps/rolls to get to 6 coins: Steps To Go From 2 Coins to 6 Coins = 7.52631 Steps To Go From 3 Coins to 6 Coins = 4.25833 Steps To Go From 4 Coins to 6 Coins = 2.45833 Steps To Go From 5 Coins to 6 Coins = 1.125 Solution The best way to solve this problem in general is not Anydice, but with a tool called Markov ...

46

First off, those little +1s and +2s are going to be much more important. Being flanked is suddenly a matter of, say, a 50% increase in their chance to hit you rather than a 10% increase. You noticed this with Aid Another, but it'll come up other places as well. Any power that forces an enemy to grant combat advantage becomes much, much more powerful. Being ...

43

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 ...

41

I appears that bySwarm is right. Here are the results: along the X axis is the total bonus over the six ability scores. Along the Y axis, the probability, obtained from 1 million runs. Results below a total bonus of +3 have been purged from the count, so the grand total of runs is less than the original 1 million. It appears that the twelve 3d6 ...

41

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 ...

39

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 ...

37

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 ...

35

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 ...

34

As a fellow GM of Earthdawn, and former GM/Player of DnD 4e I have some good news and some bad news: Your player is being somewhat silly if he's actually hardcore about statistics: It's easy enough to perform a numeric analysis on Earthdawn mechanics if you really want to. There's even an article, "The Bare Bones", that RedBrick commissioned for their ...

34

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 ...

31

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)....

30

Just roll a d1000 in anydice. The probabilities for rolling 3d10 as the 3 tens places will be exactly the same as rolling a d1000. These answers shows the math for d100 vs 2d10, it's exactly the same story here just times ten. The point of using d1000 is that probabilities are easy to calculate: the chance of the number or less is equal to the number in ...

29

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 ...

28

Link? No. But... I can do the requirements for them and work the figures out. Rounding to 0.01% increments. 9+ is 160/216 12+ is 81/216 13+ is 56/216 14+ is 35/216 15+ is 20/216 17+ is 4/216 Original D&D Rules Bk1 Fighter: No requirements. 100% Cleric: No requirements. 100% Magic User: No requirements. 100% Original D&D Rules, including ...

27

The mean result goes from 10.5 to 7.175 for disadvantage and to 13.825 for advantage. The odds go from a flat 5% for each of 1 through 20 to (disadvantage results shown; reverse the first column for advantage results): 1 39 9.75% 2 37 9.25% 3 35 8.75% 4 33 8.25% 5 31 7.75% 6 29 7.25% 7 27 6.75% 8 25 6.25% 9 23 5.75% 10 21 5.25% 11 19 4.75%...

27

Existing answers are already very good and this one does not mean to replace them. Wax Eagle and TuggyNE did a great job. I would like to offer an alternative, narrative point of view. In games like D&D mathematics are supposed to express certain things and provide a difference in the mechanics where there is a narrative difference in actions modelled. ...

25

It depends on your players and campaign style The problem, as you've noted, is that players start being able to do specific things really well. However, that's also the solution - force them to do new things. If your campaign is a dungeon crawl, this will be harder than if it's a city-based setting, but you have to remember that the PCs' actions shouldn't ...

23

Here's the link to the program, and here it is in its entirety: DICE:{4,6,8,10,12,20} loop D over {1..(#DICE-1)}{ loop SECOND over {0..5}{ FIRST: 6-SECOND FIRSTD: D@DICE SECONDD: (D+1)@DICE output [count {4..20} in FIRSTdFIRSTD] + [count {4..20} in SECONDdSECONDD] named "[FIRST]d[FIRSTD] and [SECOND]d[SECONDD]" } }

23

You don't. You just roll one d100. As you understand, rolling multiple dice is a useful tool for achieving different result spreads. But rolling multiple dice is a tool with a time and place for when you want various advantages: you can take highest or lowest, you can create a bell curve effect, or do other interesting things. However, you're not ...

21

The result of a roll is a bit more complicated than just 2d6+mod. Some moves, such as Aid, Bolster, and some item effects, grant +N forward or ongoing. Others, such as Interfere or Conditions, provide negative modifiers. That means the fixed range of stats modifiers (-3 to 3) isn't sufficient to show all possibilities in Dungeon World. So, I've expanded the ...

20

This is sort of brute-force, but the computers nowadays...: >>> all = [(i,j,k,l) for i in range(20) for j in range(20) for k in range(20) for l in range(20)] >>> p = lambda X: 1.0*len([t for t in all if max(t[0],t[1]) + X > min(t[2],t[3])])/len(all) >>> print '\n'.join("%3d: %.4f" % (X, p(X)) for X in range(-15,16)) -15: 0.0854 ...

19

Alternatively, you can: Roll an odd number of dice (3d20 is easiest). Add the highest and lowest values, and compare the sum to 21. If it's less (than 21), use the lowest value. If it's more (than 21), use the highest value. If it's equal (to 21), see whether the median (middle) value is high or low (1-10 or 11-20) and use the highest or lowest value, ...

19

That's fine. Everything you described is exactly equivalent: Rolling one d10 five times and writing down the results each time. Rolling two to four d10, writing down the results, then picking up however many and rolling them again until you have five results written down. Rolling five d10 all at once. All of them result, overall, in five dice rolls being ...

18

If you actually get a standard distribution from the dice in the 3d6 x12 method, it will be slightly better than a standard distribution of results from the 4d6 method. The more samples you take, the more likely it is that you will get something approaching average or a standard distribution. The fewer samples you take, the more likely the results will just ...

18

If you're using 1 die as a hundreds digit, 1 die as a tens digit, and 1 die as a ones digit, then every number between 1 and 1000 has a 0.1% chance of occurring. If by doubles you mean 155, 944, etc. and by triples you mean 333, 777, etc. then those have the same probability as any other number in the range. Think about it: each d10 should have a 10% chance ...

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