Math question... help please

[gr8artist]

So, I've been working on this for about an hour, and I'm not sure I'm doing the math right... or really, how to do the math at all.

The scenario is pretty straightforward.
Guy A (good-guy), is rolling 2d6+1
Guy B (bad-guy), is rolling 2d6
Guy A succeeds with his roll if his result is greater than or equal to Guy B's result. Obviously, the +1 bonus he has will help.

Question: What is the average increase in Guy A's success chance caused by the +1 bonus?



Roll
Rate
Odds
A ≥ B
Odds
A+1 ≥ B
Success
Increase
Increase
x Rate


2
3%
3%
8%
+6%
+0.18%


3
6%
8%
16%
+8%
+0.48%


4
8%
16%
27%
+11%
+0.88%


5
11%
27%
41%
+14%
+1.54%


6
14%
41%
58%
+17%
+2.38%


7
17%
58%
72%
+14%
+2.38%


8
14%
72%
83%
+11%
+1.54%


9
11%
83%
91%
+8%
+0.88%


10
8%
91%
97%
+6%
+0.48%


11
6%
97%
100%
+3%
+0.18%


12
3%
100%
100%
0
0


Everything up to that final column seems to make sense... but now what do I do with it? Do I total that column for a sum? Do I average it?

I... realize now that what I'm effectively asking is "what're the chances that someone else's roll will be only 1 higher than your own roll" which is a pretty slim amount.
I'm just at a bit of a loss, so hopefully someone with a better head for mathematics can help me out.

[Sian]

http://anydice.com/program/8d2e

From how i read the table it creates, A have at or about ~55.63% of winning, 10.81% of drawing and 33.56% of losing

[danzibr]

First, odds (against) are things like like 2 to 1. Probability is stuff like 8%.

I guess I'd do it pairwise. Like... if good guy rolls a 2, he wins only if the bad guy rolls a 2. Good guy rolls a 3, bad guy has to roll a 2 or 3.

So... .0277*.0277+.0555(.0277+.0555)+.0833(.0277+.0555+. 0833)+.1111(.0277+.0555+.0833+.1111)+.1388(.0277+. 0555+.0833+.1111+.1388)+.1666(.0277+.0555+.0833+.1 111+.1388+.1666)+.1388(.0277+.0555+.0833+.1111+.13 88+.1666+.1388)+.1111(.0277+.0555+.0833+.1111+.138 8+.1666+.1388+.1111)+.0833(.0277+.0555+.0833+.1111 +.1388+.1666+.1388+.1111+.0833)+.0555(.0277+.0555+ .0833+.1111+.1388+.1666+.1388+.1111+.0833+.0555)+. 0277(.0277+.0555+.0833+.1111+.1388+.1666+.1388+.11 11+.0833+.0555+.0277), looks like around 55.6%. I rounded a bit. Might've made a mistake.

EDIT: Yeah, Sian's way was way faster.

[Lethologica]

The 10.8% figure Sian mentions is precisely the one you're looking for. Worth noting that counting the number of opposed 2d6 rolls where A would lose by 1 without the bonus, and then dividing by the total number of possible outcomes, is fairly simple:

A's roll - # lost by 1
2 - 1*2
3 - 2*3
4 - 3*4
5 - 4*5
6 - 5*6
7 - 6*5
8 - 5*4
9 - 4*3
10 - 3*2
11 - 2*1
12 - 1*0
(The first number in each product is the number of ways A can roll the given number, the second is the number of ways B can roll one higher.)

Total lost by 1:
(2+6+12+20+30)*2 = 140

Divide by all possible outcomes for (2d6)^2:
140 / 36^2 ~ 10.8%