Editing Probability (section)

==General Polyhedral Dice==
Relevant for: Dungeons and Dragons
===What are the odds that dX beats dY?===
Want to know if a d6 is going to beat a d12? First, there are 72 ways this could come out (6 times 12, from the two dice), so the answer will be X/72. Next, look at it this way: 

If the d12 lands on a 6 or higher, then the d6 will not be higher than it. If it lands on a 5, then one of the results on the d6, the 6, would be higher than it. So that's 1/72 so far. If it lands on a 4 then there are two results, if it lands on a 1 then there are five. There's a total of 1+2+3+4+5 results that would make the d6 the higher die, 15 in total. The odds of a d6 being higher than a d12 are 15/72.

d10 vs d12? That's 9+8+etc / 10*12, so 45/120.

It turns out, if ''Y'' is larger than ''X'', the formula for the probability is very simple:
[[File:Probability Eq10.png|x60px]]

===What are the odds that adX beats bdY?===
step 1: AdX (sum the results) yields a number between A (when you roll all ones) and A*X (when you roll all X's (i.e. 6's).  The trick here is that there are more results than sums, and so you need to work out how often these sums occur for your results.

lets look at 2d6 : there are 36 unique unique combinations of 1 through 6 (taken twice) 

2 can occur 1 way (1/36)<br/>
3 can occur 2 ways (2/36)<br/>
4 can occur 3 ways (3/36)<br/>
5 can occur 4 ways (4/36)<br/>
6 can occur 5 ways (5/36)<br/>
7 can occur 6 ways (6/36)<br/>
8 can occur 5 ways (5/36)<br/>
9 can occur 4 ways (4/36)<br/>
10 can occur 3 ways (3/36)<br/>
11 can occur 2 ways (2/36)<br/>
12  can occur 1 way (1/36)<br/>

lets look at the corresponding 2d10: there are 100 unique combinations of 1 through 10 (taken twice) 
these can sum up to 

2 can occur 1 way (1/100)<br/>
3 can occur 2 ways (2/100)<br/>
4 can occur 3 ways (3/100)<br/>
5 can occur 4 ways (4/100)<br/>
6 can occur 5 ways (5/100)<br/>
7 can occur 6 ways (6/100)<br/>
8 can occur 7 ways (7/100)<br/>
9 can occur 8 ways (8/100)<br/>
10 can occur 9 ways (9/100)<br/>
11 can occur 8 ways (8/100)<br/>
12  can occur 7 ways (7/100)<br/>
13-20 can occur 36 ways (36/100)<br/>

So for 2d6 to beat 2d10:<br/>
one way is 2d6 =3 and 2d10 = 2 (2/36)*(1/100)<br/>
one way is 2d6 =4 and 2d10 = 2 (3/36)*(1/100) or one way is 2d6 =4 and 2d10 = 3 (3/36)*(2/100)<br/>
one way is 2d6 =5 and 2d10 = 2 (4/36)*(1/100) or one way is 2d6 =5 and 2d10 = 3 (4/36)*(2/100)or one way is 2d6 =5 and 2d10 = 4 (4/36)*(3/100)<br/>

add up all the results and you have your answer!<br/>

====In general====
Now, I am a stats nerd with an undergrad education. I have spent several hours trying to figure out the exact probabilities for the sums of dice, which is probably more than any sane person would spend on this article. So I'm sure you don't care either. Plus, the math involves Greek letters, and I'm quite attached to my kneecaps.

Fortunately, other stats nerds get tired of this shit too, so they discovered a good approximation: the normal distribution. This works well for 3dX and gets better the more dice you roll.

Looking at my notes, for an adX the mean (average) is [[File:Probability Eq11.png|x50px]], and the "standard deviation" is [[File:Probability Eq12.png|x50px]]. That's all you need [https://www.wolframalpha.com/input/?i=normal+distribution+calculator&assumption=%7B%22F%22%2C+%22NormalProbabilities%22%2C+%22z%22%7D+-%3E%22-1%22&assumption=%7B%22F%22%2C+%22NormalProbabilities%22%2C+%22mu%22%7D+-%3E%220%22&assumption=%7B%22FVarOpt%22%7D+-%3E+%7B%7B%22NormalProbabilities%22%2C+%22sigma%22%7D%2C+%7B%22NormalProbabilities%22%2C+%22z%22%7D%2C+%7B%22NormalProbabilities%22%2C+%22mu%22%7D%7D to get an answer from Wolfram Alpha], but with a few caveats: You can't ask it for the probability that you'll roll ''exactly'' a number (it will pedantically tell you 0), but you can ask it the probability of rolling ''less than or equal'' or ''more than or equal'' to some number, and it doesn't respect the range of the dice pool (it will say it's possible, if unlikely, to roll a negative number, for example).

So if, for whatever reason, you're doing a check roll for 12 or better on a 4d6, know that the mean is (4+4*6)/2 = 14 and the standard deviation is the square root of 182/3 (about 7.8), so the chance of passing the check is about [https://www.wolframalpha.com/input/?i=probability+that+x+%3E%3D+12+given+x+is+normally+distributed+mean%3D14+standard+deviation%3D7.8 60%].