# How do you calculate the odds of success for an ability contest

I'm having a hard time understanding how to figure out the odds of success from an ability contest. I originally thought that each +1 would be 5% but anydice suggests 4% but only after a +1 difference?

http://anydice.com/program/4097

Also, given the nature of bounded accuracy, am I correct in believing that there will be an extreme of possible mods from -7 to +17?

Whether you adding or subtract results from dice, it tends to produce similar-looking distributions with different offsets. For instance 2d6 looks very like 1d6-1d6, except the former is centred on 7 and the latter on 0.

So, an ability contest where two contestants roll 1d20+bonus, and highest wins, can be seen as looking as 1d20 - 1d20 + (difference in bonus) > 0 and has a similar "triangular" probability distribution shape as 2d20 > 21 + (difference in bonus).

This means that initial small differences between the opponents have a larger impact on percentage points. The difference between a +0 and +1 advantage is around a 5% step, whilst the last bit of "lock out" where you go from a +19 to a +20 higher bonus than an opponent gives you a measly 0.25% step (but importantly goes from 1 in 400 chance of losing to no chance of losing whatsoever).

This can be turned into a relatively simple formula. If $P(n)$ is the percentage chance of winning an opposed contest, when you have a bonus $n$ better than your opponent:

$P(n) = 100 - 0.125 \times ( 20 - n ) \times (21 - n )$

(for n from 0 up to 21 ) OR

$P(n) = 0.125 \times (n + 19) \times (n + 20)$

(for n from 0 down to -20)

• Is there a formula? – GMNoob Jul 13 '14 at 12:04
• What is the meaning of 0.125? What does it represent? Thanks for all your effort. – GMNoob Jul 13 '14 at 18:02
• It is 1/8. It is actually 1/400 * 1/2 * 100 combined for (1 in 20 * 20 for all 400 combinations of the 2 dice) times (factor of 1/2 in triangular number formula) times (100 to convert probability to percentage) – Neil Slater Jul 13 '14 at 18:45