Ability points PC vs. NPCs

[jmucchiello]

That probability is found by the simple trick of subtracting from 1 the probability of not having any 18s. Probabilities are always between and including 0 and 1, incidentally; 1 represents the probability that the NPC will have 6 attributes, each between and including 3 to 18 (well, duh).

Or you just look for the probability that a character rolled with 3d6 6 times has 1 or more 18s. Since he has 6 tries to get an 18 in any ability, that is 6 * 1/216 or 1/36. Much easier.

 

[frisbeet]

Or you just look for the probability that a character rolled with 3d6 6 times has 1 or more 18s. Since he has 6 tries to get an 18 in any ability, that is 6 * 1/216 or 1/36. Much easier.

Remarkably close, but not exactly. And as with anything in mathematics, not exactly is not correct. Here's why it seems correct, but isn't.

You're summing probabilities.

6 * 1/216 = 1/216 + 1/216 + 1/216 + 1/216 + 1/216 + 1/216

Summing probabilities is appropriate when you've got an OR condition. The P(something happening) which depends on either thing 1 happening or thing 2 happening = P(thing1) + P(thing2). (Compare to the AND probability theorem in my original post, which multiplies. These two laws of probability are somewhat intuitive but are hard to prove here; take them on faith).

So your sum reflects the P(getting an 18 Str, OR getting an 18 Dex, OR getting an 18 Con, etc.). It doesn’t consider the P(two 18s), or P(three 18s), etc. It’s pretty close, however, because the probabilities for two or more 18s are more remote.

A more detailed analysis (please god, stop now) Let

P(one 18, but not more) = P(rolling an 18) x P(rolling a 17 or less)^5
P(two 18s, but not more) = P(rolling an 18)^2 x P(rolling a 17 or less)^4
P(three 18s, but not more) = P(rolling an 18)^3 x P(rolling a 17 or less)^3
P(four 18s, but not more) = P(rolling an 18)^4 x P(rolling a 17 or less)^2
P(five 18s, but not more) = P(rolling an 18)^5 x P(rolling a 17 or less)^1
P(six 18s) = P(rolling an 18)^6

From the AND theorem, we see it’s appropriate to multiply P(18)^n by P(<18)^(6-n) for any particular P(n 18s, but not more).

From the OR probability theorem, P(at least one 18) = P(one 18, b.n.m.) + P(two 18s, b.n.m.) + P(three 18s, b.n.m.) + P(four 18s, b.n.m) + P(five 18s, b.n.m.) + P(six 18s), since any one of these results satisfies “at least one 18”, but to the exclusion of any of these other results.

There’s just one little problem. While P(one 18, b.n.m.) is correct as written for the case of rolling an 18 Str, <18 all other attributes (= 1/216 x (215/216)^5), it doesn’t factor in other ways of rolling one 18, b.n.m. The case of P(at least one 18), because of the OR condition, must take into account all possible results which satisfy “at least one 18”. I could have also rolled 18 Dex, <18 all other attributes. Since we’re summing all individual P(n 18, b.n.m.)s, it’s a bit quicker multiplying P(n 18, b.n.m.) by the number of ways of combining n 18s with 6 - n < 18 results. Introducing Combinatorics. I won’t derive anything, but take it on faith that the set of P(n 18, b.n.m.)s above is better written as:

P(one 18, but not more) = P(18) x P(<18)^5 x COMB(6,1)
P(two 18s, but not more) = P(18)^2 x P(<18)^4 x COMB(6,2)
P(three 18s, but not more) = P(18)^3 x P(<18)^3 x COMB(6,3)
P(four 18s, but not more) = P(18)^4 x P(<18)^2 x COMB(6,4)
P(five 18s, but not more) = P(18)^5 x P(<18)^1 x COMB(6,5)
P(six 18s) = P(rolling an 18)^6 x COMB(6,6)

where COMB(a,b) = a!/(b! x (a-b)!).

Then P(at least one 18) = P(18) x P(<18)^5 x 6 + P(18)^2 x P(<18)^4 x 15 + P(18)^3 x P(<18)^3 x 20 + P(18)^4 x P(<18)^2 x 15 + P(18)^5 x P(<18) x 6 + P(18)^6

Factor out P(18) x 6. Substitute 1/216 for P(18) and (215/216) for P(<18). Then

P(at least one 18) = P(18) x 6 x [(215/216)^5 + 1/216 x (215/216)^4 x 2.5 + (1/216)^2 x (215/216)^3 x 3.33 + (1/216)^3 x (215/216)^2 x 2.5 + (1/216)^4 x (215/216) + (1/216)^5]

Yuck. Simplify. Let 215/216 = 1, which it almost does. Let 1/216 = 0.0046, which it almost does. Then

P(at least one 18) = P(18) x 6 x [1 + 0.0046 x 1 x 2.5 + 0.000021 x 1 x 3.33 + 0.000000099 x 1 x 2.5 + 0.00000000046 x 1 + 0.0000000000021]

I hope you can see that the dominant term is 1. Then P(at least one 18) very nearly = P(18) x 6 x 1, or 6 x 1/216, as you’ve suggested.

Wow, that was fun.

Edit: And if you add everything up:

P(at least one 18) = 1/216 x 6 x [(215/216)^5 + (1/216) x (215/216)^4 x 2.5 + (1/216)^2 x (215/216)^3 x 3.33333 + (1/216)^3 x (215/216)^2 x 2.5 + (1/216)^4 x (215/216) + (1/216)^5] = 0.0274582534226626,

which I've shown before is exactly = 1 - (215/216)^6.

I'd say that's simpler.

 

[Nifft]

So, you're saying that the average stat is above average?

The average member of group Y is above average for group (X union Y).

-- N (not currently that funny)