I'm certain that you, Veeky Forums, of all people are capable of solving this simple problem

I have only rudimentary high school math knowledge, but I think both of the swords have the same damage/s.

For the blue sword,
probability of 1 attack/s = 0.4
therefore average rate of attacks = 0.4 attacks/s

Using the notation nCk for the binomial coefficient, for the red sword,
probability of 4 attacks/s = 0.1^4
probability of 3 attacks/s = 4C3*0.1^3*0.9
probability of 2 attacks/s = 4C2*0.1^2*0.9^2
probability of 1 attack/s = 4C1*0.1*0.9^3
probability of 0 attacks/s = 0.9^4
So the average rate of attack = 4*0.1^4+3*4C3*0.1^3*0.9+2*4C2*0.1^2*0.9^2+1*4C1*0.1*0.9^3+0*0.9^4 = 0.4 attacks/s

So the average number of attacks per second is the same for both swords.

how many hits does it take to defeat an opponent, is the one attack / second deal its damage at the start end or middle of a second.

If we assume both opponents have one hp and the damage is dealt early in the swing (draw back to striking pose takes longer than the initial strike) then we could assume red hits once, then blue hits then red hits three more times. This would give the advantage to blue as red will be dead a statistically significant amount of time before red gets the rest of his swings in, this advantage decreases as you add hp to both sides.

there's no need to bash your mind against a square problem, especially if you can let computers do the work. what you should have asked is what general algorithm will do this. even then, save that for class, and discuss more beautiful concepts here. =-:3

You're right but there's an easier way to do it for the red sword.

The point is that

(Average # of Hits)=(Average # of hits on first attack)+...+(Average # of hits on 4th attack)

But the average number of hits on the first attack is .1(1)+.9*0=.1, and same for the other attacks. So the average is just .1+.1+.1+.1

Probability for the first sword to hit once per second: [math]0.4[/math]

Probability for the second sword to hit at least once within a second: [math]1 - (1 - \frac{1}{10})^4 = 0.3439[/math]

So the first is better.

You have to actually count the multiplicity of extra hits per turn to calculate the DPS.

My answer is correct. Let me break this to you:

Probability for some single hit to succeed (second sword): [math]\frac{1}{10}[/math]
Probability that it doesn't succeed: [math]1 - \frac{1}{10}[/math]
Probability that none of the 4 hits succeed: [math](1 - \frac{1}{10})^4[/math]
Probability that at least one of the 4 hits succeed: [math]1 - (1 - \frac{1}{10})^4[/math]

And "at least one of the 4 hits succeed" includes the scenarios "1 hit", "2 hits", "3 hits" and "4 hits".

Again; The second sword is better.

First sword is better*

But that doesn't take into account that the cases where there are 2 hits, 3 hits, or 4 hits do more damage than just the 1 hit.

aaaand you still didnt catch on to the difference between "at least one hit" and DAMAGE per unit time.

So go back, and tally up the different damage amounts according to their normalized probabilities, and sum that shit up, and LO AND BEHOLD both swords have the same average damage per unit time.

wow. math sure is fun when you do it right.

>he's an idiot, but I can't be bothered to explain why
>mostly because I can dish out criticism, but I don't want anyone to criticize my ideas

This is why nobody likes you, user.

poesp.com/tools/attackdps/

Plugging the information for both into this, assuming the damage is 100 says that both do 41 damage per second.

What the fuck are you going on about? Assuming the damage per hit is the same, isn't this as simple as a weighted average?

Sword 1 = 0.4
Sword 2 = 0.1 + 0.1 + 0.1 + 0.1

Blue Red
Second_1 .4 .34
Second_2 .64 .56
Second_3 .78 .71
Second_4 .87 .81
Second_5 .92 .87

This is chance to hit after x seconds. Sorry my table got clobbered. So The blue sword is strictly better if you assume one hit one kill

> If we assume both opponents have one hp and the damage is dealt early in the swing
Conversely, if you assume that it's dealt at the end (warm-up rather than cool-down), red has the advantage.

Essentially, expected total damage as a function of time is a stair-step with a mean slope of 0.4 hits/sec in both cases. If the damage occurs right at the beginning (cool-down model), the red curve will always be below the blue. If it occurs right at the end (warm-up model), the red curve will always be above the blue. If it's half way, there's no net advantage.

But unless you can one-shot your opponent, the absolute minimum time to kill will always be longer for the blue (which can do at most 1 hit/sec) than the red (which can do up to 4). As the opponent's HP increases, statistical variation becomes less less significant.

tl;dr: "better" is subjective. If you want an unambiguous answer, you need an unambiguous question.

Hmm, but this doesn't take into account the possibility you got killed... Let me revisit

It's 50/50 y'all know why

.1 + ( .9 * .1 ) + (.9^2 * .1) + (.9^3 * .1) + (.9^4 * .1) is the chance the red sword has of killing after one second. which is 0.40951

Red sword.

This is wrong.

Chance red kills after quarter second

1/4 sec : .1 = .1
2/4 sec : .1 + (.9 * .1) = .19
3/4 sec : .1 + (.9 * .1) + (.9^2 * .1) = 0.271
4/4 sec : .1 + (.9 * .1) + (.9^2 * .1) + (.9^3 +.1) = .3439

Prob Blue kills after 1 second is prob they were alive after 3/4 second times their prob to hit

(1-0.271) * .4 = 0.2916

Red sword is better after one second. Seconds are independent, so Red is strictly better.

this is a shit thread

I saw this thread and predicted it would be full of Veeky Forumsentists arguing over wich is the better sword.
Also, they're completely the same. It's only when you would add variables like life per hit (benefit for the red one ofc) that the would diverge in quality.

You people are retarded and are overthinking this. You just calculate over a time period.

Let's do 40 seconds because round and demoninators:

40 seconds:

blue sword: 40 attacks, 16 attacks hit.

Red sword: 160 attacks, 16 hit.

They are equal.

Yeah, same dmg/s but Red has higher attack speed, which would be better if you have high luck or any stacking effects that work over times/hit, indirectly tied to your hp!(the lower yout hp the higher the damage) or any stun,paralysis per hit effect might give you the edge.

if you don't, blue gets the same result in less time.

chance of blue sword not hitting in one second = .9 * .9 * .9 * .9 = .651

chance of red sword hitting once in any given second: 1-.651 = 35%
Chance of blue sword once in any given second: 40%

Take the blue sword.

that's poor reasoning

checkmate atheists

This guy gets it.

Why?

Sorry, again.

Why? Is't it reasonable that in any second you would've laid down your hit much sooner than the other sword, since it's slower?

The blue sword can hit more than once

It's simple statistics, just calculate expected values of attacks.
[latex]E[X] = x_{1}y_{1}+x_{2}y_{2}+...+x_{k}y_{k}\par
E[X_{b}]=1*0.4=0.4\par
E[X_{r}]=1*0.1=0.1\par
aE[X_{r}]=E[aX_{r}]\par
4E[X_{r}]=E[4X_{r}]=
4*0.1=0.4\par
4E[X_{r}]=E[X_{b}][\latex]
The damage per second is the same. But blue sword is clearly better. Red sword wastes more attacks as greater percentage of those miss. All attack you attempt costs you time. While red needs to attack constantly blue can make use of increase in time meanwhile attacks, like casing spells, moving or dodging.

ok how do I latex?

> Is't it reasonable that in any second you would've laid down your hit much sooner than the other sword, since it's slower?
Games generally have two distinct mechanisms for limiting the attack rate: warm-up and cool-down. Warm-up is the delay between issuing an attack command and the attack occurring. Cool-down is the delay between an attack occurring and a subsequent attack commencing.

"One attack per second" could mean zero warm-up and one second cool-down, or one second warm-up and zero cool-down, or half a second warm-up and half a second cool-down, or any other combination which sums to one second.

Short warm-up and long cool-down favours blue, the reverse favours red. But regardless of the balance between the two, neither is unambiguously better. You'd need to state the complete set of factors which affect win or lose to determine which sword has the higher probability of winning.

In case it wasn't obvious, the question itself is a troll. The intent is to get people to make assumptions without stating them (and often without even realising them) then have people working off different assumptions call each other idiots.

[math]E[X] = x_{1}y_{1}+x_{2}y_{2}+...+x_{k}y_{k}\par
E[X_{b}]=1*0.4=0.4\par
E[X_{r}]=1*0.1=0.1\par
aE[X_{r}]=E[aX_{r}]\par
4E[X_{r}]=E[4X_{r}]=
4*0.1=0.4\par
4E[X_{r}]=E[X_{b}][\math]

[math]E[X] = x_{1}y_{1}+x_{2}y_{2}+...+x_{k}y_{k}\\
E[X_{b}]=1*0.4=0.4\\
E[X_{r}]=1*0.1=0.1\\
aE[X_{r}]=E[aX_{r}]\\
4E[X_{r}]=E[4X_{r}]=
4*0.1=0.4\\
4E[X_{r}]=E[X_{b}][/math]
ok got it

the hit rate for both weapons follow a symmetric distribution centered at the same location.

your chance for missing x amount of times over y seconds is greater for the red sword REGARDLESS of how large or small x is.

ie you're equally likely to miss more as you are to hit more

>insult people on Veeky Forums albeit they make perfectly valid arguments
No matter what you say, you will be the most prominent retard of this thread

>Which sword has the highest Damage Per Second?
The question is invalid. A correct question would be "Which sword has the highest expected Damage Per Second?". Since [math]E[Red_{dps}] = 4 * .1 = 1 * .4 = E[Blue_{dps}][/math] we conclude that both swords have equally high DPS and as such both of them have the highest DPS.

Damage per second = expected damage per second
It's like saying speed or expected speed
But blue sword is better. Red sword wastes more attacks as greater percentage of those miss. All attack you attempt costs you time. While red needs to attack constantly blue can make use of increase in time meanwhile attacks, like casing spells, moving or dodging.

He asked about dps

For all intents and purposes you can assume hero is fighting a monster with infinite hp

Ill bite:
>Damage per second = expected damage per second
Go open a book on probability theory and statistics. Specifically, look into chapters on stochastic variables and expected value.

>But blue sword is better. Red sword wastes more attacks as greater percentage of those miss. All attack you attempt costs you time. While red needs to attack constantly blue can make use of increase in time meanwhile attacks, like casing spells, moving or dodging.
OP asked which sword has the highest DPS. He didn't ask what sword is best to use. When I was 15y I always answered questions before I understood them too, senpai

user pls. >In case it wasn't obvious, the question itself is a troll. The intent is to get people to make assumptions without stating them (and often without even realising them) then have people working off different assumptions call each other idiots.

just assumed extra cooldown after a successful hit, which is a common game mechanic.

It doesn't necessarily mean cooldown. It could be larger weapon which you swing slowly like an axe. Red weapon would be like a dagger, with you just swinging around missing most of your attacks. While swinging you can also move and dodge.

Well yeah, it's a troll, but people who are making assumptions outside of the given information _are_ idiots.

This isn't that hard, the right answer is this: