Corsair, Charges, Random deal & Card-Zerg.

[fantasticdan]

Dream shroud doesn't stack with memento mori. And I'm not sure where you are getting those MAB trait numbers, but /blm gives you 24 MAB not 30, and /rdm is 20.

[Atreides]

didn't know about memento mori/dream shroud

but it's actually 30 mab for blm sub, which is 24 percent (and 25 mab for /rdm, which is 20%)


edit: fixed my previous post to not include dream shroud

[Suiram]

but it's actually 30 mab for blm sub, which is 24 percent (and 25 mab for /rdm, which is 20%)

This is untrue.

[Atreides]

really? i could have sworn it was a round number increase of 5 mab per trait, which at 0.8% per MAB would give the numbers i said. Hmm...

[Suiram]

Yeah the 0.8% per MAB thing is whack too, I just noticed that.

[Atreides]

so you're saying 1 mab doesnt equal 0.8% increase in dmg? im not saying I don't believe you but that's the first I've heard in 4 years about it not being that way. What's the % increase then per MAB then?

[Suiram]

so you're saying 1 mab doesnt equal 0.8% increase in dmg? im not saying I don't believe you but that's the first I've heard in 4 years about it not being that way. What's the % increase then per MAB then?

It's relative to the MAB you already had. So like, if you have +0 MAB (1.0 in MAB term), adding MAB+10 gives you a 10% increase in damage. But if you have say, 50 MAB already, adding MAB+10 gives you 1.6/1.5 = 6.7% increase in damage. Also this is just useful when comparing the relative bonuses. The term you multiply to figure out the damage itself is just the sum of your total MAB.

In any event, I get 664 damage for your "ideal" case and 436 for your "realistic" case.

[Kylen]

Infinite 2hr loop. COR#1 uses Wild Card on group, resets 2hrs. Group goes and fails and alexander fight. (>.>..)

WAR# changes to lolCORlv1#2 and COR#1 changes to lolWHMlv1. COR#2 uses Wild Card, COR#1's 2hr is reset since he's is now WAR and has mighty strikes as opposed to Wild Card.

Does this work? Maybe? Hmm.

This works, I've done it before. It's mathematically impossible for it to be infinite, though, seeing as the 2-hour reset only happens 1/3 of the time.

[Spekkio]

Infinite 2hr loop. COR#1 uses Wild Card on group, resets 2hrs. Group goes and fails and alexander fight. (>.>..)

WAR# changes to lolCORlv1#2 and COR#1 changes to lolWHMlv1. COR#2 uses Wild Card, COR#1's 2hr is reset since he's is now WAR and has mighty strikes as opposed to Wild Card.

Does this work? Maybe? Hmm.

This works, I've done it before. It's mathematically impossible for it to be infinite, though, seeing as the 2-hour reset only happens 1/3 of the time.

granted, but since you stand to gain 5 2 hrs on a successful use (assuming you have 5 people needing 2 hrs and more people w/ cor unlocked standing by) every roll will cost 1 2 hr with a 1/3 chance of restoring 5. 5/3=1.66... 2 hrs expected return per use of wild card assuming ideal conditions. with a sufficiently large pool of individuals to enable the tendency to predominate over luck, the system will tend towards restoring all but one 2 hr ability. the reason it's not four 2 hrs as the math might suggest (3/3-1=0 net gain for the system) is the granular nature of the restorals. if you fail to hit the restore with 3 people needing, you increase to 4 and try again with an expected net gain. fail again and you're back at the original equation. not entirely sure if this makes a lot of sense, but i'm pretty confident i'm right that said system will tend towards all restored 2 hrs.

[Seyrr]

Infinite 2hr loop. COR#1 uses Wild Card on group, resets 2hrs. Group goes and fails and alexander fight. (>.>..)

WAR# changes to lolCORlv1#2 and COR#1 changes to lolWHMlv1. COR#2 uses Wild Card, COR#1's 2hr is reset since he's is now WAR and has mighty strikes as opposed to Wild Card.

Does this work? Maybe? Hmm.

This works, I've done it before. It's mathematically impossible for it to be infinite, though, seeing as the 2-hour reset only happens 1/3 of the time.

Not mathematically impossible, quite the opposite. A chain of successful 2-hour resets has a non-zero probability no matter how long the chain gets (you're always multiplying by a positive number, so it never actually reaches 0). It is statistically improbable, though. If you only have one person who can Wild Card each time, you're more likely to get a PCC from the BCNM than to get a chain of 3 successful uses.

However, if you've got more than one person with 2-hours left (not that improbable, depending on party makeup, since some 2-hours are less useful and may not get used even during the most climactic battle (and I say this as a THF75)), you get multiple attempts at a reset. Quick calculation of the likelihood of restoring two-hours with the number of Wild Cards available:

1 attempt - 1/3 (about 33.3%)
2 attempts - 5/9 (about 55.5%)
3 attempts - 19/27 (about 70.4%)
4 attempts - 65/81 (about 80.2%)
5 attempts - 211/243 (about 86.8%)

That last one is good to know if there's only one 2-hour you really care about (for instance, making sure the RDM can Chainspell.)

[Kylen]

granted, but since you stand to gain 5 2 hrs on a successful use (assuming you have 5 people needing 2 hrs and more people w/ cor unlocked standing by) every roll will cost 1 2 hr with a 1/3 chance of restoring 5. 5/3=1.66... 2 hrs expected return per use of wild card assuming ideal conditions. with a sufficiently large pool of individuals to enable the tendency to predominate over luck, the system will tend towards restoring all but one 2 hr ability. the reason it's not four 2 hrs as the math might suggest (3/3-1=0 net gain for the system) is the granular nature of the restorals. if you fail to hit the restore with 3 people needing, you increase to 4 and try again with an expected net gain. fail again and you're back at the original equation. not entirely sure if this makes a lot of sense, but i'm pretty confident i'm right that said system will tend towards all restored 2 hrs.

No, assuming you have a finite number of extra COR's to swap in, you will eventually encounter a scenario where all of their Wild Cards fail to restore 2-hours in a row if you're doing an infinite number of iterations. Also, don't forget about the fact that a COR will be restoring a maximum of 4 other COR's failed Wild Cards with their own, assuming that 1 slot in the party is taken up by some other job that needs their 2-hours back (otherwise there's no reason to do this exercise).

I've thought about this some, if it would say be feasible to have a party of 5 level 1 COR's in Tavnazian Safehold during an AV fight whose sole purpose is to reset the 2-hour of any 6th individual that needs it (and reset any COR's whose Wild Card failed after they switch jobs).

In this scenario, the probability of all 5 Wild Cards failing in a row is (2/3)^5, or 13.1%. You will definitely hit this condition if you do it enough times; chances are the whole setup will fail by the time you've reset the 2-hours of 9-10 outside people.

Not mathematically impossible, quite the opposite. A chain of successful 2-hour resets has a non-zero probability no matter how long the chain gets (you're always multiplying by a positive number, so it never actually reaches 0). It is statistically improbable, though. If you only have one person who can Wild Card each time, you're more likely to get a PCC from the BCNM than to get a chain of 3 successful uses.

However, if you've got more than one person with 2-hours left (not that improbable, depending on party makeup, since some 2-hours are less useful and may not get used even during the most climactic battle (and I say this as a THF75)), you get multiple attempts at a reset. Quick calculation of the likelihood of restoring two-hours with the number of Wild Cards available:

1 attempt - 1/3 (about 33.3%)
2 attempts - 5/9 (about 55.5%)
3 attempts - 19/27 (about 70.4%)
4 attempts - 65/81 (about 80.2%)
5 attempts - 211/243 (about 86.8%)

That last one is good to know if there's only one 2-hour you really care about (for instance, making sure the RDM can Chainspell.)

You're talking about a different scenario than the person I was responding to (he was talking about just 2 people resetting each others' 2-hours ad infinitum), but in both cases as long as there is a probability of the 2-hour chain failing which there will be assuming a finite number of COR's, no matter how small that probability is, it will happen if the test is repeated an infinite number of times. Hence it is mathematically impossible for it to not happen in an infinite loop scenario. Lim(n->infinity) (1-1/x)^n = 0 where 1/x is the probability of failure after k tries and k is the number of COR's available.

[Seyrr]

granted, but since you stand to gain 5 2 hrs on a successful use (assuming you have 5 people needing 2 hrs and more people w/ cor unlocked standing by) every roll will cost 1 2 hr with a 1/3 chance of restoring 5. 5/3=1.66... 2 hrs expected return per use of wild card assuming ideal conditions. with a sufficiently large pool of individuals to enable the tendency to predominate over luck, the system will tend towards restoring all but one 2 hr ability. the reason it's not four 2 hrs as the math might suggest (3/3-1=0 net gain for the system) is the granular nature of the restorals. if you fail to hit the restore with 3 people needing, you increase to 4 and try again with an expected net gain. fail again and you're back at the original equation. not entirely sure if this makes a lot of sense, but i'm pretty confident i'm right that said system will tend towards all restored 2 hrs.

No, assuming you have a finite number of extra COR's to swap in, you will eventually encounter a scenario where all of their Wild Cards fail to restore 2-hours in a row if you're doing an infinite number of iterations. Also, don't forget about the fact that a COR will be restoring a maximum of 4 other COR's failed Wild Cards with their own, assuming that 1 slot in the party is taken up by some other job that needs their 2-hours back (otherwise there's no reason to do this exercise).

I've thought about this some, if it would say be feasible to have a party of 5 level 1 COR's in Tavnazian Safehold during an AV fight whose sole purpose is to reset the 2-hour of any 6th individual that needs it (and reset any COR's whose Wild Card failed after they switch jobs).

In this scenario, the probability of all 5 Wild Cards failing in a row is (2/3)^5, or 13.1%. You will definitely hit this condition if you do it enough times; chances are the whole setup will fail by the time you've reset the 2-hours of 9-10 outside people.

[quote:1sfuee87]Not mathematically impossible, quite the opposite. A chain of successful 2-hour resets has a non-zero probability no matter how long the chain gets (you're always multiplying by a positive number, so it never actually reaches 0). It is statistically improbable, though. If you only have one person who can Wild Card each time, you're more likely to get a PCC from the BCNM than to get a chain of 3 successful uses.

However, if you've got more than one person with 2-hours left (not that improbable, depending on party makeup, since some 2-hours are less useful and may not get used even during the most climactic battle (and I say this as a THF75)), you get multiple attempts at a reset. Quick calculation of the likelihood of restoring two-hours with the number of Wild Cards available:

1 attempt - 1/3 (about 33.3%)
2 attempts - 5/9 (about 55.5%)
3 attempts - 19/27 (about 70.4%)
4 attempts - 65/81 (about 80.2%)
5 attempts - 211/243 (about 86.8%)

That last one is good to know if there's only one 2-hour you really care about (for instance, making sure the RDM can Chainspell.)

You're talking about a different scenario than the person I was responding to (he was talking about just 2 people resetting each others' 2-hours ad infinitum), but in both cases as long as there is a probability of the 2-hour chain failing which there will be assuming a finite number of COR's, no matter how small that probability is, it will happen if the test is repeated an infinite number of times. Hence it is mathematically impossible for it to not happen in an infinite loop scenario. Lim(n->infinity) (1-1/x)^n = 0 where 1/x is the probability of failure after k tries and k is the number of COR's available.[/quote:1sfuee87]

Sorry, I get pedantic about the use of "infinity". Infinity is not a number. That's why you have to say the limit as n approaches infinity, that is, the limit as n becomes unboundedly large. I will agree that as the test is repeated, the probability that you can continue the chain approaches 0 (and does so fairly rapidly.) Though if you want to be technical, for any given value of n, the probability is greater than 0 even if it's vanishingly small.

... so really, I don't disagree, I just have a pet peeve about people using the word "infinity" improperly. Told you I was pedantic. And yeah, the AV COR reset party has the problems you mentioned as well. I think your arithmetic is right there, and you're definitely right conceptually.

[Spekkio]

granted, but since you stand to gain 5 2 hrs on a successful use (assuming you have 5 people needing 2 hrs and more people w/ cor unlocked standing by) every roll will cost 1 2 hr with a 1/3 chance of restoring 5. 5/3=1.66... 2 hrs expected return per use of wild card assuming ideal conditions. with a sufficiently large pool of individuals to enable the tendency to predominate over luck, the system will tend towards restoring all but one 2 hr ability. the reason it's not four 2 hrs as the math might suggest (3/3-1=0 net gain for the system) is the granular nature of the restorals. if you fail to hit the restore with 3 people needing, you increase to 4 and try again with an expected net gain. fail again and you're back at the original equation. not entirely sure if this makes a lot of sense, but i'm pretty confident i'm right that said system will tend towards all restored 2 hrs.

No, assuming you have a finite number of extra COR's to swap in, you will eventually encounter a scenario where all of their Wild Cards fail to restore 2-hours in a row if you're doing an infinite number of iterations. Also, don't forget about the fact that a COR will be restoring a maximum of 4 other COR's failed Wild Cards with their own, assuming that 1 slot in the party is taken up by some other job that needs their 2-hours back (otherwise there's no reason to do this exercise).

I've thought about this some, if it would say be feasible to have a party of 5 level 1 COR's in Tavnazian Safehold during an AV fight whose sole purpose is to reset the 2-hour of any 6th individual that needs it (and reset any COR's whose Wild Card failed after they switch jobs).

In this scenario, the probability of all 5 Wild Cards failing in a row is (2/3)^5, or 13.1%. You will definitely hit this condition if you do it enough times; chances are the whole setup will fail by the time you've reset the 2-hours of 9-10 outside people.

[quote:3j4v5cu4]Not mathematically impossible, quite the opposite. A chain of successful 2-hour resets has a non-zero probability no matter how long the chain gets (you're always multiplying by a positive number, so it never actually reaches 0). It is statistically improbable, though. If you only have one person who can Wild Card each time, you're more likely to get a PCC from the BCNM than to get a chain of 3 successful uses.

However, if you've got more than one person with 2-hours left (not that improbable, depending on party makeup, since some 2-hours are less useful and may not get used even during the most climactic battle (and I say this as a THF75)), you get multiple attempts at a reset. Quick calculation of the likelihood of restoring two-hours with the number of Wild Cards available:

1 attempt - 1/3 (about 33.3%)
2 attempts - 5/9 (about 55.5%)
3 attempts - 19/27 (about 70.4%)
4 attempts - 65/81 (about 80.2%)
5 attempts - 211/243 (about 86.8%)

That last one is good to know if there's only one 2-hour you really care about (for instance, making sure the RDM can Chainspell.)

You're talking about a different scenario than the person I was responding to (he was talking about just 2 people resetting each others' 2-hours ad infinitum), but in both cases as long as there is a probability of the 2-hour chain failing which there will be assuming a finite number of COR's, no matter how small that probability is, it will happen if the test is repeated an infinite number of times. Hence it is mathematically impossible for it to not happen in an infinite loop scenario. Lim(n->infinity) (1-1/x)^n = 0 where 1/x is the probability of failure after k tries and k is the number of COR's available.

Sorry, I get pedantic about the use of "infinity". Infinity is not a number. That's why you have to say the limit as n approaches infinity, that is, the limit as n becomes unboundedly large. I will agree that as the test is repeated, the probability that you can continue the chain approaches 0 (and does so fairly rapidly.) Though if you want to be technical, for any given value of n, the probability is greater than 0 even if it's vanishingly small.

... so really, I don't disagree, I just have a pet peeve about people using the word "infinity" improperly. Told you I was pedantic. And yeah, the AV COR reset party has the problems you mentioned as well. I think your arithmetic is right there, and you're definitely right conceptually.[/quote:3j4v5cu4]
as with any system that has a random factor and a point of stable equilibrium, no matter the improbability, given an unlimited number of trials, the result will tend towards that stable equilibrium, i will grant. despite that, due to the extremely remote probability of it with a reasonable sample of corsairs for the above (unfortunately, this "reasonable number" is probably in the 20's so not all that feasable!) the tendency of a bounded number of trials would be towards the unstable equilibrium of resetting all but 1 2 hr. so fundamentally i think we're both right, just not on the same problem!

[Kyreth]

Let's be a mite more reasonable on the original poster.

Endgame COR can reasonably pull around 360-370 damage per QD. Let's say 300 or so. We'll roll 2 of the COR's in a separate PT , then trade them over to the COR 6-pack.

300 x 12 (2 shots per COR) = 3600 damage. COR #1 Random Deals.

Each COR fires off another 2 shots. We're at 7200 damage.

COR #2 Random Deals. Then #3, #4, #5, #6. 3600 x7 = 25,200 total damage if unresisted.

COR #1 Wild Cards. Everyone fires off 12 more shots and Random Deal is reset, so it's time for another 6x Random Deals. Another 25,200.

Repeat Wild Card/Random Deal cycle with the next 5 Corsairs. Yeah, that could get ugly. 176,400 HP delivered. Of course, it'll be less as some QDs will be resisted.

Other optional lulz for fewer COR- add RNG to PT, let him go Barrage/Slugwinder with multiple Random Deals with Hunter's/Chaos roll + possible TP recharge from Wild Cards. Artillery, anyone?

[Darus Grey]

If you wanted to look at it from that perspective I'm *fairly* sure RD/WC refreshes things like Bloodpact timers for SMNs, though they'd run out of MP real fast

[Kirbyprime]

I seriously want to see this done. Only problem is that you'll probably have to merge a few servers just to get enough COR together at the same time to test it.

[Kylen]

as with any system that has a random factor and a point of stable equilibrium, no matter the improbability, given an unlimited number of trials, the result will tend towards that stable equilibrium, i will grant. despite that, due to the extremely remote probability of it with a reasonable sample of corsairs for the above (unfortunately, this "reasonable number" is probably in the 20's so not all that feasable!) the tendency of a bounded number of trials would be towards the unstable equilibrium of resetting all but 1 2 hr. so fundamentally i think we're both right, just not on the same problem!

The problem is, there can be no long-term equilibrium when a single statistical abberation (all Corsairs' Wild Cards failing to reset 2-hours in this case) stops the experiment in its tracks and prevents you from doing any further iterations, at least not for 2 more hours.

I seriously want to see this done. Only problem is that you'll probably have to merge a few servers just to get enough COR together at the same time to test it.

/sea all corsair
>> Result: 7 players found in all areas



While I enjoy the uniqueness of playing a "rare" job, I seriously don't get why more people don't play COR. It's fun as hell, versatile, and has a place in just about every endgame event there is. The cost is a drawback of course, but it's really not that bad.

[Callisto]

While I enjoy the uniqueness of playing a "rare" job, I seriously don't get why more people don't play COR. It's fun as hell, versatile, and has a place in just about every endgame event there is. The cost is a drawback of course, but it's really not that bad.

It depends, if you make all your own ammo/cards it's really not bad at all, if you AH for all of them it can be stupidly expensive just to maintain ammo/QD.

[Wizerd]

I seriously want to see this done. Only problem is that you'll probably have to merge a few servers just to get enough COR together at the same time to test it.

/sea all corsair
>> Result: 7 players found in all areas



While I enjoy the uniqueness of playing a "rare" job, I seriously don't get why more people don't play COR. It's fun as hell, versatile, and has a place in just about every endgame event there is. The cost is a drawback of course, but it's really not that bad.

To address the amount of CORs, I personally know a good amount of 75 CORs who just aren't on the job very often. It's a job that gets leveled, and doesn't get used as much as it could/should.

In terms of why people don't level it, I think the main reasons it gets held back are the previous lack of decent crowd control (with 2 charges now instead of 1, it's better), no form of Haste, no direct form of Slow, and the inability to easily wipe rolls off members of your party (there are several reasons you'd want to do this in a semi-hurry).

It's a very good job, but you have to be cautious if replacing a BRD with it. (Before 75, when 2 BRD parties aren't nearly as common)

[fantasticdan]

I seriously want to see this done. Only problem is that you'll probably have to merge a few servers just to get enough COR together at the same time to test it.

/sea all corsair
>> Result: 7 players found in all areas



While I enjoy the uniqueness of playing a "rare" job, I seriously don't get why more people don't play COR. It's fun as hell, versatile, and has a place in just about every endgame event there is. The cost is a drawback of course, but it's really not that bad.

To address the amount of CORs, I personally know a good amount of 75 CORs who just aren't on the job very often. It's a job that gets leveled, and doesn't get used as much as it could/should.

In terms of why people don't level it, I think the main reasons it gets held back are the previous lack of decent crowd control (with 2 charges now instead of 1, it's better), no form of Haste, no direct form of Slow, and the inability to easily wipe rolls off members of your party (there are several reasons you'd want to do this in a semi-hurry).

It's a very good job, but you have to be cautious if replacing a BRD with it. (Before 75, when 2 BRD parties aren't nearly as common)

I think a bigger reason is the cost/consumables/inventory space required to level and play it properly, especially compared to brd.