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How can I calculate my chance of success given my ability + skill vs a difficulty? How is this complicated when the difficulty is rolled as well?

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Did you pick an answer? – Pureferret Mar 16 '12 at 2:02
up vote 14 down vote accepted

I would recommend using The creator of has written a program on that site to calculate success probabilities for NWoD, but the dice mechanics are very similar and shouldn't need much tweaking. Here is the post. It should give you a rough idea.


I tweeted the creater of and he suggested this:

function: owod N:n tn TN:n {
 if N = 1 { result: -1 }
 if N < TN { result: 0 }
 if N < 10 { result: 1 }
 result: 1 + [owod d10 tn TN]

OWOD: [owod d10 tn 7]

loop N over {1..10} {
 output [highest of [lowest of NdOWOD and 10] and -4] named "[N]d"

Which gives this sort of graph:

enter image description here

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I too recommend if you actually have the time to use it; if you have players who are too lazy it's not going to work, unfortunately. – Kyle Willey Mar 2 '12 at 22:33
What's to use on a regular basis? The OP only wants it for some probabilities. – Pureferret Mar 3 '12 at 1:35
True. It also includes a dice roller, but it's kinda hard to write from scratch. – Kyle Willey Mar 3 '12 at 2:24
I think the "At least" graph option would be better for this answer. After all, he doesn't care about the amount of successes, only his probability of success... so he doesn't care the probability of rolling 1 success, he cares about the probability of rolling at least 1 (i.e., if the roll is successful or not). It's also not very cool that it is reusing colors for 9 and 10 that may make one think that the system is broken at first glance, but you can't do anything about it :P – Yandros Mar 14 '12 at 10:22
@Pureferret Awesome :D – Yandros Mar 16 '12 at 12:34

This is a question that many WoD players have faced and tackled. Here is a good start.

Difficulty shouldn't be rolled, but set at a number (such as 6) and opposing rolls compare successes to determine victor.

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Also discussed here: – Pyrodante Mar 2 '12 at 21:27

Using the AnyDice roller as in PureFerret's answer I've come up with these rules of thumb. You can estimate that in order to get one success it takes, on average,:

2 dice to get 1 success vs difficulty 2-5
3 dice to get 1 success vs difficulty 6
4 dice to get 1 success vs difficulty 7 
5 dice to get 1 success vs difficulty 8
10 dice to get 1 success vs difficulty 9

For multiple successes it's not linear:

Successes   4   5   6   7   8   9
1          2d   2d  3d  4d  5d  10d
2          3d   4d  5d  6d  9d  
3          5d   6d  7d  9d  14d 
4          6d   8d  9d  13d     
5          8d   10d 12d

Note that deviation is approximately +/- 1 success up to 6d and +/- 2 successes for more dice

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A fast and dirty (and not terribly precise) method I recommend for my players of both Shadowrun and V20 is as follows:

Take what percentage of the dice is less than the target number (I.e. 4 on a six-sided die would have a 50% success chance, 6 on a 10-sided die has 50% chance).

For every die past that add half of the chance of the previous die; 1 d6 versus 4 is 50%, 2 is 75%, 3 is 87.5%, so on and so forth, and the same progress applies to d10 on oWoD dice. I'm not familiar with nWoD's system, so I can't help, but I'm assuming if you're playing Masquerade you're using the old system. This doesn't apply so much to V20 since I play with fumbles, so there needs to be a calculation (10% for each die past the first reduced similarly to the normal success chance seems a good rule). This allows them to quickly guess more complex chances, like if the target number were increased (40%; 50%; 55%). It breaks down rather quickly as you throw more dice, however, but usually isn't such a big deal, though it does work a lot better for Shadowrun than it does for V20.

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Here is what I came up with. Chance of 1 success on X dice:

 1 die  2 dice  3 dice  4 dice  5 dice  6 dice  7 dice  8 dice  9 dice  10 dice
30.00%  51.00%  65.70%  75.99%  83.19%  88.24%  91.76%  94.24%  95.96%  97.18%

Someone can check my work in this spreadsheet:

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