Discussion Strongest Upgrade for a Warbrig

[WilyJaymes]

I'm not sure if that's the case since I got 12x from a warship. I'll test it some more on different types of ships to see if it comes out the same.

 

[TheramoreIsTheBomb]

Poll has been added because you guys are confusing me.

 

[Sky Kiwi]

Poll has been added because you guys are confusing me.

All you need to know from that discussion is copperhead has the highest broadside damage, followed by storm chaser, and then skull and bones.

That's not the only thing to consider though, it depends on what you want to do with the ship, so answering what the strongest is like that isn't so straight forward.

 

[WilyJaymes]

Lol, just wants to know which one dah strongest, but me and brotha mane Kiwi be over here like...

In all seriousness though, if by strongest you mean the most damage output, then copperhead is the way to go. You may not be going very fast, but who cares when you have 180% armor? Brig turning radii are already on a dime, so not much of a problem there.

 

[Sky Kiwi]

@WilyJaymes
Confirming explosive is 12x round shot for both warships and hunters, armoured.

 

[WilyJaymes]

Awesome. Using round shot as a point of reference turned out to be pretty useful. The game seems to build on top of it, and so we get nice, even multipliers.

 

[Sky Kiwi]

Awesome. Using round shot as a point of reference turned out to be pretty useful. The game seems to build on top of it, and so we get nice, even multipliers.

Well I always knew that for thunderbolt it was 3x and fury is 2x. Didn't know explosive was a clean multiple though haha. And firebrand isn't exactly as straightforward, since it's supposed to have burn damage (and in POTCO the base shot did less damage than round)

 

[WilyJaymes]

Yeah, I hadn't considered firebrand because it's, well...firebrand! lol

 

[Jack Ropefury]

Wait, can someone confirm that the damages dealt by various ammunitions are always just a fixed multiple of each other? That is that the ratio between damage dealt by any ammo type and round shot in particular is always fixed regardless of the level of enemy armor? That would make comparing the expected damage on broadsides much easier since you could get a fixed ratio on those (of course as a binomial random variable you can only compare expectations).

But yea, I did the math and had very similar numbers to you Sky Kiwi. Also I believe we spoke in game, nice to see others interested in the math as well!

 

[Sky Kiwi]

Wait, can someone confirm that the damages dealt by various ammunitions are always just a fixed multiple of each other? That is that the ratio between damage dealt by any ammo type and round shot in particular is always fixed regardless of the level of enemy armor? That would make comparing the expected damage on broadsides much easier since you could get a fixed ratio on those (of course as a binomial random variable you can only compare expectations).

But yea, I did the math and had very similar numbers to you Sky Kiwi. Also I believe we spoke in game, nice to see others interested in the math as well!

The base damage values of each shot seem to be, yeah. Level 5 round vs level 5 thunderbolt is 3x, 5 round vs 5 explosive is 12x. According to the loading tip, skill points just raise damage by 25% (up to 244% damage if multiplicative)

Also I saw your name in the alert and got excited! lol

 

[Jack Ropefury]

Hey, so thought I'd just elaborate on some of the math that was made to reach the conclusions here, and hopefully clear a few things up. Please let me know if one of my assumptions is incorrect, or worse, if I have made some mistake. There's going to be some math here, but hopefully nothing too messy (I wish there was some way to display Latex here, but I guess there's no need for that on most forums ), and perhaps this will help some learn some basic Probability along the way (most of the probability terms are overkill here, but hopefully someone might find it interesting).

I have somewhat unnecessarily broken this into points, just so its easier to read and reply to:

1. Your base damage is not a random variable. By base damage, I just mean a number that represents the damage you would deal to some fixed enemy, here lets say an unarmored Navy Ferret. If you attack an unarmored Navy Ferret, regardless of your ammo or skills, your damage will be the same every time (this is for a single ammo shot of a fixed ammo type) . This is in contrast to say Sword, where your damage with every hit is a Random Variable, which can take on many different values with different probabilities.
2. All damage dealt will be some multiple of base damage. This is not really saying anything (since any number is some multiple of another), but what I mean to emphasize here, is that your base damage is all you need to make comparisons. While you might deal some multiple of your base damage to an unarmored Tally Ho for instance, the multiplier is independent of what the initial base damage is. So you only need to calculate the base damage to make comparisons.
3. A single ammo shot of a variable ammo type is a (Bernoulli) Random Variable. All this means is that when you fire a single ammo shot from your broadsides (with upgraded hulls of course), there are two possibilities-- it is a round shot, or it is the special ammo type. It can only be one of these two outcomes, and the base damage will just be the base damage of either ammo (which depends on your skill too, but more on that later). The probabilities of each outcome is also fixed, for instance for Copperhead VI, the probability is 0.85 and 0.15 for round shot and explosive respectively (lets call it failure and success respectively), while for Stormchaser VI, the probabilities are 0.55 and 0.45 respectively.
4. Each round of broadsides is a set of n iid trials of the Bernoulli Random Variable. Iid stands for Independent Identically Distributed, and means that each ammo from the broadside is identical in that it is the same random variable (not same outcome, but that it has the same possible outcomes with the same attached probabilities), and is independent in that the outcome of a single ammo shot from a broadside does not effect the outcome of any other. The broadside damage is simply the sum of these n (say 10 broadsides as in War Frigates) variables. This is known as a Binomial Random Variable. Unlike the Bernoulli Random Variable, it has several possible outcomes, corresponding to how many of the trials were successes and failures (in Broadsides terms, you could get all round shots, or all but one round shot and one special shot, or all but two round shot and so on).
5. Adding skill points just increases the base damage of the respective ammo type by 25%.

Now we can do some simply math, leaving things abstract since we are interested in damage ratios. Lets say the base damage for round shot (which depends on your skill obviously) is x, while the base damage for your special ammo (which depends on what the ammo is, and the skill level too) is y. Then if p is the probability of firing the special ammo, the average shot will have damage (also called expectation of the, here Bernoulli, random variable):
(1-p)*x + p*y
Expectations of Bernoulli Random Variables always have this form. Now if perform n iid trials of this (as in a broadside), by linearity, the average of the broadside is:
n*[(1-p)*x + p*y]
Now, we have the nice fact that (assuming constant skill levels across ammo), the damage of special ammo, is just some multiple of damage of round shot. For Copperhead VI, that multiple is 12 (for explosive), and p is 0.15, so putting that in, the base damage (still in terms of x, the base damage of a round shot), is:
n*[0.85*x + 0.15*12*x]=n*(2.65)*x
While for Stormchaser VI, that multiple is 3 (for lightning), and p is 0.45, so the parallel expression is:
n*[0.55*x + 0.45*3*x]=n*(1.9)*x

In other words, the ratio of base damage of Copperhead vs Stormchaser VI is 2.65:1.9, in other words, Copperhead VI deals on average 40% more damage than Stormchaser VI. This did not assume any enemy type or armor level, nor any ship type (broadsides are left as a variable).

The only thing it assumes is that all skill levels are 1. Now what if that changes. Well, if you were to keep the skill level of round shot and special ammo the same (both 5, which is the common case), then they both increase by 125%, and the two numbers above, just have an extra scalar factor of 2.25, which will cancel in the ratio, giving the same ratio. However, if you do not keep both skills at the same level, the ratio can actually change. I won't get into this (although its simple) but basically, the more probably a special ammo, the more you should prioritize spec'ing into it (very intuitive). If you spec round shot high but special ammo low, the ratio from above will lean even more heavily in Copperhead's favor. However, if you spec the special ammo type more, and keep round shot low, then it will lean in Stormchasers favor. This can easily be scene by taking derivatives, but is also intuitive enough.

Lets compare the best case for Copperhead VI, when your special ammo skill is 1, but round shot skill is 5. This will also help see where keeping things abstract in x terms helps. That is lets compare the average base damage between a broadside of a Copperhead VI ship, with the captain having round shot 5, and explosive 1, to that of a Stormchaser VI ship, with the captain having round shot 5, and lightning shot 1. In this case the base damage for round shot will be 2.25*x while that for the special ammo will be 12*x and 3*x respectively
Copperhead VI:
n*[0.85*x*2.25 + 0.15*12*x]=n*(3.7125)*x
Stormchaser VI:
n*[0.55*x*2.25 + 0.45*3*x]=n*(2.5875)*x

For a ratio of 3.7125:2.5875, or 43% more base damage to Copperhead VI, better than the 40% above.

Finally, lets compare the worst case for Copperhead VI, when your special ammo skill is 5, but round shot skill is 1. That is lets compare the average base damage between a broadside of a Copperhead VI ship, with the captain having round shot 1, and explosive 5, to that of a Stormchaser VI ship, with the captain having round shot 1, and lightning shot 5. We have:
Copperhead VI:
n*[0.85*x + 0.15*12*x*2.25]=n*(4.9)*x
Stormchaser VI:
n*[0.55*x + 0.45*3*x*2.25]=n*(3.5875)*x

For a ratio of 4.9:3.5875 or 36%, lower than the 40% from above. So we see that form the best to worst case, Copperhead VI deals 36 to 44% more damage on average than Stormchaser VI, with 40% being the typical case. This assumes that the skill distribution is the same in either case (that is the same skills in round shot and the respective ammo type). I've done the math for the other broadsides before, and Stormchaser VI is second best, but Copperhead VI dominates that, as you can see here. It truly is beastly.

 

[Sky Kiwi]

Hey, so thought I'd just elaborate on some of the math that was made to reach the conclusions here, and hopefully clear a few things up. Please let me know if one of my assumptions is incorrect, or worse, if I have made some mistake. There's going to be some math here, but hopefully nothing too messy (I wish there was some way to display Latex here, but I guess there's no need for that on most forums ), and perhaps this will help some learn some basic Probability along the way (most of the probability terms are overkill here, but hopefully someone might find it interesting).

I have somewhat unnecessarily broken this into points, just so its easier to read and reply to:

1. Your base damage is not a random variable. By base damage, I just mean a number that represents the damage you would deal to some fixed enemy, here lets say an unarmored Navy Ferret. If you attack an unarmored Navy Ferret, regardless of your ammo or skills, your damage will be the same every time (this is for a single ammo shot of a fixed ammo type) . This is in contrast to say Sword, where your damage with every hit is a Random Variable, which can take on many different values with different probabilities.
2. All damage dealt will be some multiple of base damage. This is not really saying anything (since any number is some multiple of another), but what I mean to emphasize here, is that your base damage is all you need to make comparisons. While you might deal some multiple of your base damage to an unarmored Tally Ho for instance, the multiplier is independent of what the initial base damage is. So you only need to calculate the base damage to make comparisons.
3. A single ammo shot of a variable ammo type is a (Bernoulli) Random Variable. All this means is that when you fire a single ammo shot from your broadsides (with upgraded hulls of course), there are two possibilities-- it is a round shot, or it is the special ammo type. It can only be one of these two outcomes, and the base damage will just be the base damage of either ammo (which depends on your skill too, but more on that later). The probabilities of each outcome is also fixed, for instance for Copperhead VI, the probability is 0.85 and 0.15 for round shot and explosive respectively (lets call it failure and success respectively), while for Stormchaser VI, the probabilities are 0.55 and 0.45 respectively.
4. Each round of broadsides is a set of n iid trials of the Bernoulli Random Variable. Iid stands for Independent Identically Distributed, and means that each ammo from the broadside is identical in that it is the same random variable (not same outcome, but that it has the same possible outcomes with the same attached probabilities), and is independent in that the outcome of a single ammo shot from a broadside does not effect the outcome of any other. The broadside damage is simply the sum of these n (say 10 broadsides as in War Frigates) variables. This is known as a Binomial Random Variable. Unlike the Bernoulli Random Variable, it has several possible outcomes, corresponding to how many of the trials were successes and failures (in Broadsides terms, you could get all round shots, or all but one round shot and one special shot, or all but two round shot and so on).
5. Adding skill points just increases the base damage of the respective ammo type by 25%.

Now we can do some simply math, leaving things abstract since we are interested in damage ratios. Lets say the base damage for round shot (which depends on your skill obviously) is x, while the base damage for your special ammo (which depends on what the ammo is, and the skill level too) is y. Then if p is the probability of firing the special ammo, the average shot will have damage (also called expectation of the, here Bernoulli, random variable):
(1-p)*x + p*y
Expectations of Bernoulli Random Variables always have this form. Now if perform n iid trials of this (as in a broadside), by linearity, the average of the broadside is:
n*[(1-p)*x + p*y]
Now, we have the nice fact that (assuming constant skill levels across ammo), the damage of special ammo, is just some multiple of damage of round shot. For Copperhead VI, that multiple is 12 (for explosive), and p is 0.15, so putting that in, the base damage (still in terms of x, the base damage of a round shot), is:
n*[0.85*x + 0.15*12*x]=n*(2.65)*x
While for Stormchaser VI, that multiple is 3 (for lightning), and p is 0.45, so the parallel expression is:
n*[0.55*x + 0.45*3*x]=n*(1.9)*x

In other words, the ratio of base damage of Copperhead vs Stormchaser VI is 2.65:1.9, in other words, Copperhead VI deals on average 40% more damage than Stormchaser VI. This did not assume any enemy type or armor level, nor any ship type (broadsides are left as a variable).

The only thing it assumes is that all skill levels are 1. Now what if that changes. Well, if you were to keep the skill level of round shot and special ammo the same (both 5, which is the common case), then they both increase by 125%, and the two numbers above, just have an extra scalar factor of 2.25, which will cancel in the ratio, giving the same ratio. However, if you do not keep both skills at the same level, the ratio can actually change. I won't get into this (although its simple) but basically, the more probably a special ammo, the more you should prioritize spec'ing into it (very intuitive). If you spec round shot high but special ammo low, the ratio from above will lean even more heavily in Copperhead's favor. However, if you spec the special ammo type more, and keep round shot low, then it will lean in Stormchasers favor. This can easily be scene by taking derivatives, but is also intuitive enough.

Lets compare the best case for Copperhead VI, when your special ammo skill is 1, but round shot skill is 5. This will also help see where keeping things abstract in x terms helps. That is lets compare the average base damage between a broadside of a Copperhead VI ship, with the captain having round shot 5, and explosive 1, to that of a Stormchaser VI ship, with the captain having round shot 5, and lightning shot 1. In this case the base damage for round shot will be 2.25*x while that for the special ammo will be 12*x and 3*x respectively
Copperhead VI:
n*[0.85*x*2.25 + 0.15*12*x]=n*(3.7125)*x
Stormchaser VI:
n*[0.55*x*2.25 + 0.45*3*x]=n*(2.5875)*x

For a ratio of 3.7125:2.5875, or 43% more base damage to Copperhead VI, better than the 40% above.

Finally, lets compare the worst case for Copperhead VI, when your special ammo skill is 5, but round shot skill is 1. That is lets compare the average base damage between a broadside of a Copperhead VI ship, with the captain having round shot 1, and explosive 5, to that of a Stormchaser VI ship, with the captain having round shot 1, and lightning shot 5. We have:
Copperhead VI:
n*[0.85*x + 0.15*12*x*2.25]=n*(4.9)*x
Stormchaser VI:
n*[0.55*x + 0.45*3*x*2.25]=n*(3.5875)*x

For a ratio of 4.9:3.5875 or 36%, lower than the 40% from above. So we see that form the best to worst case, Copperhead VI deals 36 to 44% more damage on average than Stormchaser VI, with 40% being the typical case. This assumes that the skill distribution is the same in either case (that is the same skills in round shot and the respective ammo type). I've done the math for the other broadsides before, and Stormchaser VI is second best, but Copperhead VI dominates that, as you can see here. It truly is beastly.

Oh now we're really getting all up in the maths here!

It seems weird to me that putting the skill points into round shot is better than putting it into explosives. Generally speaking, with copperhead at both skills the same level (we'll say 5), the explosives does more damage than the round shot. You can figure this out per broadside cannon by simply multiplying the damage by the chance.
Round Shot: 1 x 0.85 = 0.85
Explosive: 12 x 0.15 = 1.8
So wouldn't it make more sense that increasing the 12 by the 2.25 factor increases damage more than the 1?

Maybe it's just misleading in that you're comparing it to another ship's damage in percentage if the skill points are allocated there too. Point is, most of the damage is done with the special broadsides in Copperhead's case (actually all of them I think, except perhaps Firestorm because firebrand), so wouldn't speccing into the special broadsides be the best case scenario in general?

 

[Jack Ropefury]

Oh now we're really getting all up in the maths here!

It seems weird to me that putting the skill points into round shot is better than putting it into explosives. Generally speaking, with copperhead at both skills the same level (we'll say 5), the explosives does more damage than the round shot. You can figure this out per broadside cannon by simply multiplying the damage by the chance.
Round Shot: 1 x 0.85 = 0.85
Explosive: 12 x 0.15 = 1.8
So wouldn't it make more sense that increasing the 12 by the 2.25 factor increases damage more than the 1?

Oh no you are right, for damage output, increasing the explosive ammo is more important. By best and worst, I was referring to ratio. By best I meant the situation where Copperhead boasts the greatest lead over Stormchaser. The parallel calulation for Stormchaser to yours is:
Round Shot: 1 x 0.55 = 0.55
Lightning: 3 x 0.45 = 1.35

So Lightning is 2.45 more important than round shot in Stormchasers, while Explosive is 2.11 times more important than round shot in Copperhead. So Copperhead would have the greatest edge over Stormchaser in the situation when Special Ammo is not prioritized. Does that make sense? You are correct as this shows, in both cases, the Special Ammo is more than twice as important as round shot. I was just referring to something else.

 

[Sky Kiwi]

Oh no you are right, for damage output, increasing the explosive ammo is more important. By best and worst, I was referring to ratio. By best I meant the situation where Copperhead boasts the greatest lead over Stormchaser. The parallel calulation for Stormchaser to yours is:
Round Shot: 1 x 0.55 = 0.55
Lightning: 3 x 0.45 = 1.35

So Lightning is 2.45 more important than round shot in Stormchasers, while Explosive is 2.11 times more important than round shot in Copperhead. So Copperhead would have the greatest edge over Stormchaser in the situation when Special Ammo is not prioritized. Does that make sense? You are correct as this shows, in both cases, the Special Ammo is more than twice as important as round shot. I was just referring to something else.

Yep that makes sense. I realized that kinda with my edit there lol.

However, I have another question for you! You seem to know math better than I do, so hopefully you know this one.
I know how you calculate the chance of getting at least one special broadside per attack:
1 - (p ^ n) where p is 1 - (special chance) and n is the number of cannons on one side. So a Copperhead War Frigate is:
1 - (0.85 ^ 10) = 80.3% chance for at least one explosive.

What I can't figure out (and googling this is hard without a name, because of the way numbers work) is how to calculate the chance of getting any number above one in a single broadside attack. Whether it's "two", "three", "two or more", etc.

Also don't judge me for that little formula there, I kinda built it myself.

 

[Jack Ropefury]

Yep that makes sense. I realized that kinda with my edit there lol.

However, I have another question for you! You seem to know math better than I do, so hopefully you know this one.
I know how you calculate the chance of getting at least one special broadside per attack:
1 - (p ^ n) where p is 1 - (special chance) and n is the number of cannons on one side. So a Copperhead War Frigate is:
0.85 ^ 10 = 80.3% chance for at least one explosive.

What I can't figure out (and googling this is hard without a name, because of the way numbers work) is how to calculate the chance of getting any number above one in a single broadside attack. Whether it's "two", "three", "two or more", etc.

Also don't judge me for that little formula there, I kinda built it myself.

The formula is perfect! What you are trying to compute is actually just the probability of various outcomes of a Binomial trial. So there are two things to note here. The first, is simply what is the probability of 1 special ammo in a given configuration (say the first ammo is special and rest round shot). This has probability p^1*(1-p)^(n-1), as the prob for first being special is p, and the prob of rest being 1-p is (1-p) for each of the rest, so (n-1) of them. Similarly, the prob for first k being special and rest being round is p^k*(1-p)^(n-k). In the case k is 0 (which is what you calculated the inverse of), you get (1-p)^n as prob for 0 specials, what you found yourself!

However, that's not the end, as you only care about the number of specials that occur, not which trials they occur, ie if the first broadside shot is special and the rest regular, is the same as if the first n-1 are regular and last special, and so on. So, when you are finding the prob of k successes, there are several different probabilities you want to add up. If k is 1, you'd want to add:
Prob of special then normal normal..., to Prob of normal then special then normal, and so on. There are n of these. If k is 2 then there turn out to be n(n-1)/2 of them.

The number is basically the number of ways of selecting k successes from a series of n objects (which is actually exactly what you're counting-- how many ways are there to get k specials from a set of n shots in a broadsides). This is called n choose k (or the kth binomial coefficient), and has the formula:
n!/[(n-k)!*k!]
(read up on factorials if you don't know what "!" is, I can give a small explanation or proof of this formula if you want)

For k=0 this is 1 (1 way to choose 0 objects from a set of n), for k=1 this is n (n ways to choose a single object from a set of n-- you either choose first, or second or third, ... or nth), and for k=1, this is n(n-1)/2.

Here's a brief explanation. You first choose a success from the n trials (there are n choices), then you choose where you want the second success, there are (n-1) spots left. But this is overcounting by a factor of 2, since you could have chosen x,y or y,x for instance, which are the same (ie, your first could have been 1 and second 2, or first 2 and second 1, but those are the same!), so you divide by 2 to get rid of the over counting. Similarly, for k=3, this is n(n-1)(n-2)/2*3, for k=4 this is n(n-1)(n-2)(n-3)/2*3*4 and so on...

Back to the question. Each of those k success (k special shot) configurations has probability p^k*(1-p)^(n-k). Since there are n!/[(n-k)!*k!] possibilities of those configurations, the prob of k successes is:

n!/[(n-k)!*k!] * p^k*(1-p)^(n-k).

For k=0 the whole thing is:
1*(1-p)^10=0.19 (you calculated the prob of not 0 in your post, which is this subtracted from 1)

For k=1 it is:
n*p*(1-p)^9=10*0.15*0.85^9=0.34

For k=2 it is:
n*(n-1)/2*p^2*(1-p)^8=10*9/2 * (0.15)^2*0.85^8=0.27

and so on. The prob will sharply fall as you increase the exponent on the 0.15 term. Hope this helps!

Look up binomial random variable probabilities online for more info. As mentioned before, the number of specials fired is a binomial random variable, with n=10 and p=0.15. You'll find nice plots of the prob of 0, 1, 2...k successes, as well as calculator (on wolfram alpha for instance) that just find the probabilities for you.

See this link:
http://www.wolframalpha.com/input/?i=binomial(n=10,p=0.15)
You can click to see table of values next to the plot of PDF (probability density function) for the probs of k=0,1,2...n

 

[WilyJaymes]

Man...it's fascinating how hard it is to put into words the systems that go into answering questions a little bit beyond mere averages. I understand the answers you're looking for completely, but admittedly the math behind it is hard to wrap my head around. It could just be because the two of us have different ways of perceiving the big picture.

I'd like to know a mathematical method to predict the probability of an explosive round actually making contact with a target. As you know, their position within that blunderbuss-spread of a broadside volley is totally random. And enemy ship behavior - especially hunters - means cycling through various speeds that throw aiming calculations off. Your precious payload of 1 or 2 explosives might be located in the sub-group of the volley that falls behind and misses. Or it might be part of the sub-group that hits. With brigs this is even more of an issue, because typically there's a significant length of time between when the first shot of broadside is fired, and the when the last one is fired. So that means when the last shot is fired, for it the aiming calculation is outdated. Pull out the graph paper, this mess is getting crazy!

 

[Sky Kiwi]

Man...it's fascinating how hard it is to put into words the systems that go into answering questions a little bit beyond mere averages. I understand the answers you're looking for completely, but admittedly the math behind it is hard to wrap my head around. It could just be because the two of us have different ways of perceiving the big picture.

I'd like to know a mathematical method to predict the probability of an explosive round actually making contact with a target. As you know, their position within that blunderbuss-spread of a broadside volley is totally random. And enemy ship behavior - especially hunters - means cycling through various speeds that throw aiming calculations off. Your precious payload of 1 or 2 explosives might be located in the sub-group of the volley that falls behind and misses. Or it might be part of the sub-group that hits. With brigs this is even more of an issue, because typically there's a significant length of time between when the first shot of broadside is fired, and the when the last one is fired. So that means when the last shot is fired, for it the aiming calculation is outdated. Pull out the graph paper, this mess is getting crazy!

That one seems strongly irrelevant though since the captain's sailing style totally messes up the numbers! But lol I get what you're saying.

 

[tlonge]

The formula is perfect!...

Holy heck, what did I just read?? That's a lot of maths for a casual game lol. .

 

[Jack Ropefury]

Man...it's fascinating how hard it is to put into words the systems that go into answering questions a little bit beyond mere averages. I understand the answers you're looking for completely, but admittedly the math behind it is hard to wrap my head around. It could just be because the two of us have different ways of perceiving the big picture.

I'd like to know a mathematical method to predict the probability of an explosive round actually making contact with a target. As you know, their position within that blunderbuss-spread of a broadside volley is totally random. And enemy ship behavior - especially hunters - means cycling through various speeds that throw aiming calculations off. Your precious payload of 1 or 2 explosives might be located in the sub-group of the volley that falls behind and misses. Or it might be part of the sub-group that hits. With brigs this is even more of an issue, because typically there's a significant length of time between when the first shot of broadside is fired, and the when the last one is fired. So that means when the last shot is fired, for it the aiming calculation is outdated. Pull out the graph paper, this mess is getting crazy!

Yea lol, this seems almost impossible to do purely theoretically. One thing I will say is that there is probably some gradual decrease in probability of each shot of the broadside hitting the ship. The first broadside is most likely, because there is a chance the ship you're firing on moves out of the way (especially if its small or fast).

Regardless, you would need empirical data to come up with those numbers, and I don't think it will be very useful in any case. It will very strongly depend on too many factors (ship you're firing on, distance from ship, captain skill, broadside count etc).

Also in any case, such considerations won't change the broadside ratio by too much. This is due to the linearity of expectation (its just a linear sum for discrete distributions, so everything commutes nicely). In particular, if you redo all the calculations I did in my first post, but with n=1, the ratio stays the same (as its independent of n, which just got canceled in either case). So the ratio of a single broadside cannon damage in expectation is the same. You can multiply that by the probability of it hitting, and do the same for the other broadsides and add them up. When you take the ratio of those sums, the ratio will actually be unchanged, so it doesn't really matter! It does effect higher order moments though, like the variance in particular.

 

[Jack Cannonskull]

It depends solely on what your doing with it.

Fortune Hunter (which you didn't mention) is by far the best for cargo and going on mat runs.

If you want firepower for SvS or whatever then your best choice is to buy a War Sloop which are op, and upgrade it to Stormchaser or Skull and Bones. Maybe copperhead.