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Pandaren
2010-02-18, 09:30 PM
I am free of being a DM for the first time in four years, which is greatly fortunate, unfortunately, I know feel the need to munchkin the hell out of my character. Not anything major, no min maxing or anytihng, just increasing efficiency as much as possible.

Which is where the problem comes in, we are running a d20 Call of Cthulhu campaign, and I have been trying to figure out if my character takes multishot and shoots tiwice a round, if he will have greater accuracy than once a round, or at least a close enough percent where the double shot will be more preferable.

All the info:
-All the guns are standard fire (meaning two shots at a penalty)
-There is a -6/-6 penalty when shooting twice in a round(with the same gun):
-Them multishot feat reduces penalty of double shot by two:
-Character has an attack bonus ranging from +2 to +3 depending on shotgun or pistol (not calculating distance here, I'll save that for later)
-I am currently using the equation of

(Happ x Not Happ) + (Not Happ X Happ) + (Happ x Happ)

As in if there is an 11+ to hit, it is 50% chance, so

(.5 x .5) + (.5 x .5) + (.5 x .5) = 75% Vs the single shots 7+ = 70%

....that was a horrible example, they are all .5's....

Bleh, need help, if I continue using this equation I can sometimes end up with percentages above 100, which does not work for me, I need exactness. I'm certain you are not supposed to just "add up" percentages.

sofawall
2010-02-18, 09:41 PM
I know feel the need to munchkin the hell out of my character. Not anything major, no min maxing or anytihng, just increasing efficiency as much as possible.

Just a minor nitpick, but what you are wanting to do is Min/Maxing. What you want to avoid is Munchkining.

Siosilvar
2010-02-18, 09:47 PM
Actually, adding up percentages works quite well for chances to hit (or at least calculating average damage).

-4 to hit with two shots means you break even at 19+ hits and 13+ hits, better with the double at 20+ and less than 13 needed. 14-18, single shot is better.

quiet1mi
2010-02-18, 09:49 PM
Munchkin's Idea of combat... Use Dynamite... lots of Dynamite and a flamethrower to light the fuses of dynamite at your opponent's feet

Min/maxing Idea of combat... take a gun, aim, fire... look for things that give you circumstance bonuses, while finding any bonus you can find for your soldier.

Pandaren
2010-02-18, 09:57 PM
But at first level, assuming AC average is say, 11

And using the 3+ hit.
Then 8+ hits with single with is 13/20 which is 65%
And 12+ for double hits is 9/ 20 which is 45%

If two 45% hits equal
(.45 x .55) + (.55 x .45) + (.45 x .45)

Equals an average of...69.75 which is greater and offers a chance of doing twice the damage

But my main dilema is whether equation works or not, I got it from a friend earlier, and as you go higher in level, the percentages go above 100, which is not possible, it is not exact enough for my speculation


Also, on that note, I always imagine min maxing as say, MinMax in Goblins, sacrifice say, you CHA and DEX for STR. Which is kind of cheap. Maxing out ranks in move silently is not min maxing, it's not even unusual, especially for rouge-ish classes.

sonofzeal
2010-02-18, 10:02 PM
You can just straight add percentages here. Three shots at 50% hit chance each, gives an expected value of E(X) = .5 + .5 + .5 = 1.5, meaning you'll hit 1.5 times on average.

1stEd.Thief
2010-02-18, 10:19 PM
You should do some basic statistics research. You'll need to know The Law of Total Probability. Basically, everything needs to add up to 1, or you know you're doing it wrong.

Another hint: sometimes you need to determine the probability of something NOT happening.

Because it is easier to Do something than Explain it, your break even % hit (with a -4/-4 with 2 attacks) is 40% probability (or a roll of 12 13 hits) on a SINGLE attack. (ie: if you need a 13/20 you'll hit 40% of the time)

ie: 1 attack at 40% = 0.4 units of damage (1 being 1 definite hit)

2 attacks at 20% (-4 on d20 = -20%) means 4% of the time you hit 2x (0.08 units damage) , 32% you'll hit with one attack (.2 * .8)+(.8 * .2), 64% you hit never (.8 * .8), for (.08 + .32) 0.4 units damage. Note: all possibilities add up to 1.

Any higher percentage "to hit" and you'll do more damage with 2 attacks

FirebirdFlying
2010-02-18, 10:29 PM
The equation works for if you'll hit at all; it's just the standard

p(A or B) = p(A) + p(B) - p(A and B)

worded differently. But it only works that way for two events.

For three, and I'm sure there's a simpler way to do this but it's just something I hashed out, there's this;

p(A)*p(B)*p(C) + (this is the probability they will all happen)
p(A)*p(1-B)*p(1-C) + (the probability of only A)
p(1-A)*p(B)*p(1-C) + (the probability of only B)
p(1-A)*p(1-B)*p(C) + (the probability of only C)
p(A)*p(B)*p(1-C) + (the probability of A and B but not C)
p(A)*p(1-B)*p(C) + (the probability of A and C but not B)
p(1-A)*p(B)*p(C) (the probability of B and C but not A)
= p of hitting anything.

If you add p(1-A)*p(1-B)*p(1-C), it will simplify down to one.

Since all the hit chances are the same, this ends up being

p^3+ 3p((1-p)^2) + 3(p^2)(1-p).

Basically? The number of favorable outcomes (hits) over the number of possible outcomes (shoots).

Prolly ninja'ed by a person who actually knows things about probability.

EDIT: Of course, it's much easier to just calculate the probability that you won't hit.

Using this, anyway, with your .45 chance -

.45^3 = .091125
3*.45(.55^2) = .408375
3*(.45^2)(.55) = .334125
total = .8333625, or about an 83% chance, with a 9% chance of hitting thrice, a 33% chance of hitting twice, and a 41% chance of hitting once.

Or simply 1 - .55^3.

huttj509
2010-02-18, 11:58 PM
Basically you want your average damage.

If the bullet does the same damage each regardless, we can call it a single unit.

One bullet: (21-single target number)/20 = average damage per round (so if the target number is 7, you get 14/20 = .7 bullets of damage per round).

Two bullets: ((21-dual target number)/20) * 2 = Avg damage per round (so for the above example it's (10/20)*2 = 1 bullet of damage per round)

Take the larger damage per round.

Since it seems that the dual target number = single target number + 4, we can change the second equation to:

Two bullets: ((21-(single target number+4))/20)*2 = (17-single target number)/10


So, a brief chart:

"Target" - roll this or higher for a single shot, nat 1s assumed to miss, so start with 2
"Single" - your avg damage/round in units of bullet hits for a single shot
"Double" your damage per round in units of bullet hits for double shot, bottoms out at .1 since a 20 always hits, and .05 per shot * 2 shots = .1

The indented part is where 1 shot is better than 2.



Target Single Double
2 .95 1.5
3 .90 1.4
4 .85 1.3
5 .80 1.2
6 .75 1.1
7 .70 1.0
8 .65 .9
9 .60 .8
10 .55 .7
11 .50 .6
12 .45 .5
13 .40 .4
14 .35 .3
15 .30 .2
16 .25 .1
17 .20 .1
18 .15 .1
19 .10 .1
20 .05 .1

So, if your single shot target number is less than 13 (if you need to roll a 12 or less to hit your target), shoot twice. Otherwise, shoot once, UNLESS you need a 20 to hit with a single shot. Might as well double your chances then and shoot twice, 20 hits either way.

Pandaren
2010-02-19, 12:09 AM
Wow, thanks all. Chart is going to prove very handy, and hutt's equation looks a bit less complicated than the one I was using.


I'm printing those things out as I write this.

Edit: Also, the chart, for every +1 to my attack, I just move the start of the indentations one down, or for every +1 to their AC, I begin the indentations 1 earlier, correct?

huttj509
2010-02-19, 12:16 AM
The indentation does not change. You are changing what target number you need, not the chances of rolling a particular number on the die.

If you get +1 to hit, you now have a lower target number, move up a line.

If the enemy gets +1 AC, you now have to roll higher to hit him, move to the appropriate line.

If you have +5 to hit, and the enemy has AC 15, you need to roll a 10 or better, so you go to that line.

If you have a +6 to hit, and the enemy has 16 AC, same line. Still need to roll a 10 or better.

ericgrau
2010-02-19, 12:17 AM
Dunno about that system but in D&D even medium BAB classes usually hit on a 10. Since you want 12 or lower, I'd double shot against most enemies unless they're particularly tough. Especially if you're a full BAB class b/c then you often hit on a 5 and it may be worth it even on some of the tougher opponents.

faceroll
2010-02-19, 03:29 AM
I'd take the feat that let's me run faster, but then I don't like being eaten by deep ones.

Gralamin
2010-02-19, 03:39 AM
Dunno about that system but in D&D even medium BAB classes usually hit on a 10. Since you want 12 or lower, I'd double shot against most enemies unless they're particularly tough. Especially if you're a full BAB class b/c then you often hit on a 5 and it may be worth it even on some of the tougher opponents.

Heck, some Half BAB classes can hit on a 2, reliably.

Iceforge
2010-02-19, 05:25 AM
I am starting this from scratch, as I don't like to continue on other peoples math, as then if they have any errors, I'll be prone to repeating those same ones.

So basicly, the mechanic is that you get -4 on each shoot, if you are firing 2 shots, as opposed to -6 without the multishot ability.

You want to know if that is an investment that is worth doing, so we need numbers for

*single shots
*double shots WITH multishot
*double shots WITHOUT multishot

When I below talk about difficulty, I mean the target number you need to naturally roll on your D20, IF you had been shooting a single shot, meaning that on the "double shots with multishot" table, you actually need to roll a 5 to hit when the table says difficulty 1.

Chance of hit is the chance that at least 1 of the fired shots hit, this is calculated by reserving the chance of both missing, so for difficulty 1 on the "double shot with multishot" the calculation is "(1-(0,2*0,2))*100" where 0,2 is the chance of rollling a miss on any dice (you need to roll 5+ to hit, so a roll of 1, 2, 3 or 4 would be miss, which means 20% miss-chance or 0,2)

Chance of double hit is the chance that BOTH of them will hit

Damage pr. round is a percentage of total damage dealt, this is done by adding the numbers of chance of hit and chance of double hit together.
You do NOT add the double hit twice, as the damage of the first bullet is already part of "chance of hit" table.

Single Shot:
{table]Difficulty| Chance of hit| Chance of double hit| average damage
1|100%|0%|100%
2|95%|0%|95%
3|90%|0%|90%
4|85%|0%|85%
5|80%|0%|80%
6|75%|0%|75%
7|70%|0%|70%
8|65%|0%|65%
9|60%|0%|60%
10|55%|0%|55%
11|50%|0%|50%
12|45%|0%|45%
13|40%|0%|40%
14|35%|0%|35%
15|30%|0%|30%
16|25%|0%|25%
17|20%|0%|20%
18|15%|0%|15%
19|10%|0%|10%
20|05%|0%|05%
[/table]

Double Shot WITH Multishot
{table]Difficulty| Chance of hit| Chance of double hit| average damage
1|96%|64%%|160%
2|93,75%|56,25%|150%
3|91%|49%|140%
4|87,75%|42,25%|130%
5|84%|36%|120%
6|79,75%|30,25%|110%
7|75%|25%|100%
8|69,75%|20,25%|90%
9|64%|16%|80%
10|57,75%|12,25%|70%
11|51%|9%|60%
12|43,75%|6,25%|50%
13|36%|4%|40%
14|27,75%|2,25%|30%
15|19%|1%|20%
16|9,75%|0,25%|10%
17|0%|0%|0%
18|0%|0%|0%
19|0%|0%|0%
20|0%|0%|00%
[/table]

Double shot without multishot
{table]Difficulty| Chance of hit| Chance of double hit| average damage
1|91%|49%|140%
2|87,75%|42,25%|130%
3|84%|36%|120%
4|79,75%|30,25%|110%
5|75%|25%|100%
6|69,75%|20,25%|90%
7|64%|16%|80%
8|57,75%|12,25%|70%
9|51%|9%|60%
10|43,75%|6,25%|50%
11|36%|4%|40%
12|27,75%|2,25%|30%
13|19%|1%|20%
14|9,75%|0,25%|10%
15|0%|0%|0%
16|0%|0%|0%
17|0%|0%|0%
18|0%|0%|00%
19|0%|0%|00%
20|0%|0%|00%
[/table]

A table to compare all 3's average damage output based on difficulty for a single shot to hit:
Difficulty| Av. Dam Single Shot| Av. Dam Double Shot w/ multishot| Av. Dam Double Shot w/o multishot
1|100|160|140
2|95|150|130
3|90|140|120
4|85|130|110
5|80|120|100
6|75|110|90
7|70|100|80
8|65|90|70
9|60|80|60
10|55|70|50
11|50|60|40
12|45|50|30
13|40|40|20
14|35|30|10
15|30|20|0
16|25|10|0
17|20|0|0
18|15|0|0
19|10|0|0
20|05|0|0


Conclusions:You average damage when doing double shots increase by 20% of standard damage when you have multishot compared to when you do not have multishot, meaning that over 5 rounds, you will in average have landed 1 more hit than you would without multishot.

If you do not have multishot, when the difficulty is above 9, you will do more damage in average by firing a single shot

If you do have multishot, the difficulty has to be above 13 before double-shooting becomes less effective than firing a single shot, which corresponds with the 20% difference found in average damage output between having and not having multishot.

Average Damage against an unknown difficulty between 2-20 (1 is unimportant as I assume the system includes the 1=failure system?) yields interesting results.

Averaging the damage of a single shot gives 47,5% of average damage done, when the target number is random between 2-20 (random substituting for unknown)
Without multishot, firing 2 shots gives 45,5% of average damage done, meaning that without multishot, firing 2 shots when the difficulty is unknown is not in your advantage.
With multishot, however, firing 2 shots gives 60% of average damage, meaning that when the target number is unknown, it is to your advantage to fire 2 shots and hope for the best. Do note that the moment a natural roll of 17 doesn't hit, you should switch to single shots.

Known Issues with these stats:

This table does NOT take into accounts that you can face targets in which you actually need to roll LOWER than a natural 1 to hit with a single shot, but the lack of this being taken into account works in FAVOUR for multishot.

Also, it doesn't take into account the rule of 20 always being a success, which again would work in favour of multishot, but not more so than for average damage without multishot.
The numbers of the conclusion change as follows when taking the rule of natural 20 into account:

"Averaging the damage of a single shot gives 47,5% of average damage done, when the target number is random between 2-20 (random substituting for unknown)
Without multishot, firing 2 shots gives 46,1% of average damage done, meaning that without multishot, firing 2 shots when the difficulty is unknown is not in your advantage.
With multishot, however, firing 2 shots gives 60,4% of average damage, meaning that when the target number is unknown, it is to your advantage to fire 2 shots and hope for the best. Do note that the moment a natural roll of 17 doesn't hit, you should switch to single shots."