Hit probabilities with multiple attacks

[Skrum]

Just for my own amusement, I'm trying to do a comparison of a weapon that adds 1d6 on each hit vs a weapon that adds 2d6 to the first hit made with it each turn.

So far, I've got this -

1d6 per hit vs 2d6 once/turn
Assuming a +8 chance to hit (+3 prof, +4 stat, +1 weapon)
Against AC 16, the average for a CR 8-9 creature

65% chance to hit. The above numbers are obviously dialed in for a level ~7 or 8 character, but I think an overall 65% chance to hit is a pretty reasonable assumption to make throughout a character's career.

If a character has a single attack
35% chance to miss (no damage dealt)
65% chance to hit (7 damage vs 3.5 damage)
This is the trivial case; 2d6 once/turn is strictly superior

If the character has 2 attacks per turn, they have a
12.25% chance of missing both times (no damage is dealt; the abilities are equal)
45.5% chance of hitting once (7 damage vs 3.5 damage; 2d6 once/turn is significantly better)
42.25 % chance of hitting twice (7 damage vs 7 damage, they are equal)

If the character has 3 attacks per turn -
4.29% of missing 3 times
??????????
27.46% of hitting 3 times

==============

So the question marks are where I'm getting stuck :)
How do I do that math? What is the chance to land a single attack, and the chance to land 2 out of 3?
Also, if my math could be checked. Pretty sure it's right for the 2 attack situation, but confirmation would be great.

Thanks

[Talamare]

if the character has 3 attacks per turn -
4.29% of missing 3 times
??????????
27.46% of hitting 3 times
thanks
0.65 * 0.35 * 0.35 = 7.96
0.65 * 0.65 * 0.35 = 14.78

hhh 3
hhm 2
hmh 2
mhh 2
hmm 1
mhm 1
mmh 1
mmm 0

27.46 + 3 * 14.78 + 3 * 7.96 + 4.29 = 99.9

or

27.46 = 27.46% = 3 Hits
3 * 14.78 = 44.34% with 2 hits
3 * 7.96 = 23.88% with 1 hit
4.29 = 4.29% of no hits


Let me know if I need to add words

[Skrum]

0.65 * 0.35 * 0.35 = 7.96
0.65 * 0.65 * 0.35 = 14.78

hhh 3
hhm 2
hmh 2
mhh 2
hmm 1
mhm 1
mmh 1
mmm 0

27.46 + 3 * 14.78 + 3 * 7.96 + 4.29 = 99.9

or

27.46 = 27.46% = 3 Hits
3 * 14.78 = 44.34% with 2 hits
3 * 7.96 = 23.88% with 1 hit
4.29 = 4.29% of no hits


Let me know if I need to add words

Perfection! Tyvm.

[animorte]

hhh 3
hhm 2
hmh 2
mhh 2
hmm 1
mhm 1
mmh 1
mmm 0



Perfection! Tyvm.

I'm going to second this here statement. Nice.

[bid]

General answer. Probability math will use p = hit and q = miss.

(p + q) = 1

(p + q)^2 = (p + q)*(p + q) = p^2 + 2pq + q^2
(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3

For more attacks, you can use https://en.wikipedia.org/wiki/Pascal%27s_triangle

[Amechra]

If you have N attacks, a probability of hitting H, and bonus damage D, here are the two formulas you're looking for:


If the bonus is to each hit, your overall bonus damage is +N * H * D.
If the bonus is to your first hit, however, your damage bonus is +(1 - (1-H)N) * D.


Going through your three cases (using H=0.65, D=3.5 for the "all hits" bonus, and D=7 for the "first hit" bonus):

N=1
1*0.65*3.5 = 2.275 vs. (1 - 0.351) * 7 = 4.55. You want the "first hit" version.

N=2
2*0.65*3.5 = 4.55 vs (1 - 0.352) * 7 = 6.1425. You still want the "first hit" version.

N=3
3*0.65*3.5 = 6.825 vs. (1 - 0.353) * 7 = 6.699875. You want the "all hits" version.

---

If you're wondering why we can simplify the second calculation so much, we're using the fact that there's only one case where you don't get the damage bonus that round, which is when every attack misses, which has probability (1-H)N of happening. Since all of the cases must add up to 1, the collective probability that any of the other cases happen is 1 - (1-H)N.