Hit Chance, Probability and Sneak Attack

[Bjarkmundur]

I'm looking at Two-Weapon Fighting in relation to Sneak Attack. More notably the idea that a higher number of attacks makes me more likely to apply my Sneak Attack any given turn.

If we make a hypothetical situation where my rogue has a 50% hit Chance (no advantage), does this mean that my chances to apply Sneak Attack with two attack rolls is 75%?

Is it right that the chances of at least one attack landing, out of three attacks made, is 87%.?

Is it right that out of four attacks, my probability to apply sneak attack that turn goes to a whopping 94%?

I'm pretty sure I'm worse at math than I thought I was xD

[PhantomSoul]

For calculating your likelihood to hit at all, you're essentially figuring out your likelihood of not [not hitting at all], so the chance of not-hitting each time. You can then multiply probabilities to get the result.

Using a 50% hit rate:

1. One attack
- Chance of not hitting at all: 0.5
- Chance of hitting at least once: 0.5

2. Two attacks
- Chance of not-hitting both times: 0.25 (0.5*0.5)
- Chance of hitting at least once: 0.75 (1-0.25)

3. Three attacks
- Chance of not-hitting all three times: 0.125 (0.5*0.5*0.5)
- Chance of hitting at least once: 0.875 (1-0.125)

4. Four attacks
- Chance of not-hitting all four time: 0.0625 (0.5*0.5*0.5*0.5)
- Chance of hitting at least once: 0.9375 (1-0.0625)

In short, you've got it! You can use { 1-(miss_probability ^ number_of_attacks) } as your likelihood of hitting at all.


_____
EDIT: Fixed brain fart word swap... thanks jh12!

[jh12]

You can use { 1-(hit_probability ^ number_of_attacks) } as your likelihood of hitting at all.

Shouldn't it be { 1-(miss_probability ^ number_of_attacks) }?

[Bjarkmundur]

Now, how does this translate into DPR when

No sneak attack, 2d4+3 with 2, 3 and 4 attacks.
No Sneak attack, 2d4+6 with 2, 3 and 4 attacks.

Is it just 50% of 32, since each attack has 50% hit Chance and the average total of the dice is 32?

[PhantomSoul]

Shouldn't it be { 1-(miss_probability ^ number_of_attacks) }?

Oops, yes, wrote the wrong word there despite literally everything in my post! Fixed, thanks!

[Tanarii]

Assuming you use it on the first hit, you can calculate DPR as roughly (1-MissAll)*SA. That's not completely accurate though, because it doesn't take into account your chance of the first hit also being a critical.

[LudicSavant]

The breakdown of the formulas for all of these questions (and a tool that will do the calculations for you if you want) can be found here: http://www.giantitp.com/forums/showthread.php?582779-Comprehensive-DPR-Calculator-(v2-0)


Now, how does this translate into DPR when

No sneak attack, 2d4+3 with 2, 3 and 4 attacks.
No Sneak attack, 2d4+6 with 2, 3 and 4 attacks.

Is it just 50% of 32, since each attack has 50% hit Chance and the average total of the dice is 32?

Assuming 50% hit chance, default d20 rolls, and crits on rolls of 20:

2d4+3/2 attacks: 8.5 DPR
2d4+3/3 attacks: 12.75 DPR
2d4+3/4 attacks: 17 DPR
2d6+6/2 attacks: 13.7 DPR
2d6+6/3 attacks: 20.55 DPR
2d6+6/4 attacks: 27.4 DPR

Formula would be ((hit chance * damage) + (crit chance * extra damage on crits)) * (number of attacks). So for example, for 2d4+3 with 2 attacks: (8*.5 + 5*.05) * 2 = 8.5 DPR