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82

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 ...


41

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% ...


36

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 ...


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

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 ...


27

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 ...


26

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 ...


21

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 ...


21

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]" } }


18

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 that RedBrick wrote on their website. The guy uses basic ...


17

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 ...


16

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 ...


15

To preserve the probabilities exactly, the new DC should be "14 + monster defense." How I got that number So, you want to convert this: d20 + monster_save vs. 8 + caster_modifiers Into this: d20 + caster_modifiers vs. ?? + monster_save Here's how to figure out the "??" using a bit of intuition about probability: Ignore the modifiers for a ...


15

Trying to do this old school (no programs, just statistics and probability 101), it won't be short, but should be very informative (I'll add a summery later on). To help making this more vivid, let's consider 3 characters: "Fumbles" - he is really unlucky or unskilled, so he gets a -5 modifier. "Average Joe" (or just "Joe") - no modifiers. "Rambo" - he is ...


14

There's two parts to this answer - range, and distribution. I'm going to talk about dice rolls as rolling a total number of dice, choosing a number of results to keep (most usually all of them), summing them, and optionally adding/subtracting a fixed number. Too many people assume that the range of any roll is 1-max, but remember that each die can roll a ...


13

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, ...


13

The answer depends on what you're rolling, what your target number is, and what modifiers are already in place. For D20s, the answer is about five, but less if you need to roll really high or really low. See this answer. In general, the formula for probability of success is: 100% - (chance of failure per roll)^(number of attempts). This applies to Choose ...


12

Is this what you were looking for? http://axiscity.hexamon.net/users/isomage/rpgmath/qualify/ It covers the probability of rolling various classes based on minimum requirements, for a couple of different dice rolling methods, e.g. 3d6 any order vs 3d6 ordered.


12

If you want your curves flat, use a single die. If you start using two dice, the curve becomes a bell and the more dice you use, the highest the middle peak. This is the graph for a d20 roll This is the graph for 2d8+1d6-2 One easy way to get curves slanted to one side is to roll some dice and then remove the best or the worst ones before totalling. ...


11

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 ...


11

If they have Safecracking 5 and there aren't any safes around, then that 5 doesn't matter; they're going to be falling back on their broader, lower-dice skills. They may get new skills out of that, but they'll be something different again. In order for them to actually progress beyond the point where they have made Skullduggery 2 obsolete, they'd have to ...


10

Bryant is right about the Bonuses. In GURPS I had be to be careful about bonuses as beyond a certain point success (or failures) is all but certain. I recently returned to D&D, playing Swords & Wizardry, and one thing I noticed over GURPS is how more variable the the results seems. With the d20 numbers were all over the place and even character with ...


8

I'm not sure what the real question is. I'm thinking at least part of it is "Help me understand the mechanics of exploding dice" so I'll start there. Where N is the size of the die. Average # of rolls = N/(N - 1) Average result of an exploding die = N*(N+1) / (2 * (N-1)) Die Average Average Rolls Value d4 1.3333334 ...


8

Using 3dX was a fun idea for me more because of skills than combat; everyone's perspectives feel very combat-oriented to me. (bear in mind, I have a 3.5 mindset) The problem I had which was solved with 3dX, was that some people had super cool character concepts which were just impossible or unsatisfying with a d20... I can't explain the effect on ...


8

A better solution is to use Mid 3d20 (3M20). That is to select the middle roll from three d20. This has the advantage of creating a bell curve but still giving you the full range of a d20. The probs are: mid20 Prob % of TN Prob Eq or higher % 1 0.725 100 2 2.075 99.275 3 3.275 97.2 4 4.325 93.925 ...


8

Ubiquity basically uses coin flips for rolls--heads is a success, tails isn't--except that it uses dice instead of coins. (You could say odd is failure, even is success, or that the low half is failure and the high half is success, or you could use a 4-sided die that had sides reading 0, 1, 1, 2, for fail-fail, fail-succeed, succeed-fail, succeed-succeed, ...


8

Since you are statistics-capable, I would recommend that you take a look at Troll, a dice-roll language and calculator. It's capable of loops and conditionals and has a statistical bent. There's also R, the statistical computing environment. Being a Python programmer myself, I would probably gravitate towards SAGE instead, which provides a unified ...


8

What your friend may be referencing is a quirk of the original World of Darkness rules, in which each one rolled cancelled a success, and TNs were the number needed for a die to count as a success. If the TN was very high (10 or greater) the more dice you rolled, the more likely you would be to roll more ones than successes, and therefore the higher the ...


8

I've already commented but here's a quick writeup of the probabilities in case you don't find the other answer to your taste. Assuming a perfectly random die, yadda yadda... The probability of rolling a 8 or more on a d10 is p(8,9,10 | 1-10) = 3/10 Adding dice, it actually becomes easier to determine by calculating the probabilities for 0 successes (which ...



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