Crafting 2x0.05 interval G16 Daggers

Neodaystar - Sanctuary

I've been trying to get a pair of G16 netherworld guidance with 2x 0.05. I've rerolled 15 times and only one of those time has interval on it. The chances of getting 0.05 interval is 5%, and I've seen about 3 unique G16 daggers on my server with 2x interval, so it's not near impossible. Anyone know whats the average times you have to reroll to get it? Thank you.


Also, I want to sell my current daggers, it's

Interval 0.05
Attack level +1
Max attack +130


How much is a crafted one with interval and good offensive stats sell for usually?

Jarkhen - Archosaur

I won't claim to be the best when it comes to market prices (and I'm not on Sanc anyway), so I won't comment on that. As for the crafting %s, though:

Each random addon has a 5% chance of being -int, so you can expect 1-(0.95^3) = 14.3% of rolls to have at least one -int addon.

Breaking that down, we find that there is a:

13.5% chance of having exactly one -int addon (one every 7.4 rolls)
0.7% chance of having exactly two -int addons (one every 143 rolls)
0.04% chance of having exactly three -int addons (one every 2500 rolls)

Fissile - Archosaur

Not everything in Jarkhen's post is exactly right (until he ghost edits it after the fact).

Chance of having 3/3 addons all -0.05 interval (with 5% chance of each) is exactly 0.05 * 0.05 * 0.05 = 0.0125%.

Chance of having 2/3 addons is 3!/(2! * (3-2)!) * 0.05 * 0.05 * 0.95 = 3 * 0.05 * 0.05 * 0.95 = 0.7125%

Chance of having 1/3 addons is 3!(1! * (3-1)!) * 0.05 * 0.95 * 0.95 = 13.5375%

Chance of 0/3 addons is 0.95 * 0.95 * 0.95 = 85.7375%

---

Chance of exactly 3 is 1 in 8000.0
Chance of exactly 2 is 1 in 140.35
Chance of exactly 1 is 1 in 7.3869
Chance of exactly 0 is 1 in 1.1664


Chance of at least 2 is 1 in 137.93
Chance of at least 1 is 1 in 7.0114

Jarkhen - Archosaur

Fissile - Archosaur wrote: »

Not everything in Jarkhen's post is exactly right (until he ghost edits it after the fact).

Chance of having 3/3 addons all -0.05 interval (with 5% chance of each) is exactly 0.05 * 0.05 * 0.05 = 0.0125%.

Chance of having 2/3 addons is 3!/(2! * (3-2)!) * 0.05 * 0.05 * 0.95 = 3 * 0.05 * 0.05 * 0.95 = 0.7125%

Chance of having 1/3 addons is 3!(1! * (3-1)!) * 0.05 * 0.95 * 0.95 = 13.5375%

Chance of 0/3 addons is 0.95 * 0.95 * 0.95 = 85.7375%

---

Chance of exactly 3 is 1 in 8000.0
Chance of exactly 2 is 1 in 140.35
Chance of exactly 1 is 1 in 7.3869
Chance of exactly 0 is 1 in 1.1664


Chance of at least 2 is 1 in 137.93
Chance of at least 1 is 1 in 7.0114

Bah. It's been too long since I had my discrete math class.

HrunsPanda - Archosaur

Fissile - Archosaur wrote: »

Chance of having 2/3 addons is 3!/(2! * (3-2)!) * 0.05 * 0.05 * 0.95 = 3 * 0.05 * 0.05 * 0.95 = 0.7125%

Could you explain what this 3!/(2! * (3-2)!) exactly means ? Its been too long... Wasnt 3! something like 1+2+3 ? And do you also have a linky (or can you explain) why and how to get to this formula ? (or does it have a name that i should wiki?)

Jarkhen - Archosaur

HrunsPanda - Archosaur wrote: »

Could you explain what this 3!/(2! * (3-2)!) exactly means ? Its been too long... Wasnt 3! something like 1+2+3 ? And do you also have a linky (or can you explain) why and how to get to this formula ? (or does it have a name that i should wiki?)

3! is three factorial: 3*2*1. Basically, the point of the formula is to take the number of possible permutations of the result you're looking for (2 -int, one not -int) and multiply it by the probability of one of those permutations to get the total probability.

In this case, potential permutations are:

-int, -int, non-int
-int, non-int, -int
non-int, -int, -int

So there are 3 permutations (which is what the formula works out to be. Specifically, the formula is n!/(r!*(n-r)!) where n is the total number of "attempts", slots, whatever it is, and r is the number of "successful" attempts.

You then multiply that value by the chance of one of those permutations happening. Since each addon slot has a 5% chance of being -int, this works out to be 0.05*0.05*0.95 (chance of the first slot being -int * chance of second slot being -int * chance of third slot not being -int).

So this gives us 3!/(2!*(3-2)!) * 0.05*0.05*0.95 = 0.007125 = 0.7125%


If you wanted to avoid the formula for whatever reason, you could just add the probabilities of each permutation and you would arrive at the same result (though for larger n/r values, this would be a pain):

(0.05*0.05*0.95)+(0.05*0.95*0.05)+(0.95*0.05*0.05) = 0.007125

HrunsPanda - Archosaur

cool thanx

soundslegit

Here's my math:

3 int = hahaha
2 int = fat chance
1 int = cake

Zanryu - Dreamweaver

I want quadruple -int daggers.

Mr_Swiss - Heavens Tear

KingSeeke, on my server rolled 1 time and got triple int on a pair of daggers and then rolled his sisters to get 3x again. Super rare. We do have tons of 2x int on my server.

Get on my server and ask him. He will link both to you.

Zsw - Dreamweaver

Mr_Swiss - Heavens Tear wrote: »

KingSeeke, on my server rolled 1 time and got triple int on a pair of daggers and then rolled his sisters to get 3x again. Super rare. We do have tons of 2x int on my server.

What is the probability of 1 person getting 3x int two times in 2 tries?

Mr_Swiss - Heavens Tear

Zsw - Dreamweaver wrote: »

What is the probability of 1 person getting 3x int two times in 2 tries?

I don't have a clue. but like I said, you can get on my server and he will link them to you. A set of fists and daggers. When he first made them, he was selling at 2bil for the daggers, then decided to keep them. He is the only person I know that has been able to roll 2 weps with 3x. Does anyone on any other servers have someone with 3x?

HrunsPanda - Archosaur

Zsw - Dreamweaver wrote: »

What is the probability of 1 person getting 3x int two times in 2 tries?

Much smaller than either of the following:

-this person telling bull**** and showing the same weapon that he borrowed from someone else on 2 different characters .
-this person actually rerolling a zillion times. (several 1000 if he is just lucky, several 100 if he is insanely lucky)
-this person hacking the PWI servers.

The exact answer is not so hard though: 0.05 ^ 6 *100% = 0.0000015625 % chance. or 1 in 64 million

Mr_Swiss - Heavens Tear

HrunsPanda - Archosaur wrote: »

Much smaller than either of the following:

-this person telling bull**** and showing the same weapon that he borrowed from someone else on 2 different characters .
-this person actually rerolling a zillion times. (several 1000 if he is just lucky, several 100 if he is insanely lucky)
-this person hacking the PWI servers.

The exact answer is not so hard though: 0.05 ^ 6 *100% = 0.0000015625 % chance. or 1 in 64 million

Or he could have just been lucky. it's random. Just because someone puts out a % of a drop, it doesn't nesessarily mean that's the %. I bought 20 dragon fire packs and got a 12*, But yet the % of getting a 12*, says I would have to open 1000+ packs to even have a chance. Maybe I **** the system too, cuz god forbid someone has some luck, it's a zombie apocolypse. I'm ready to eat some face.

HrunsPanda - Archosaur

Mr_Swiss - Heavens Tear wrote: »

Or he could have just been lucky. it's random.

LOL you remind of some guy who in a thread about genie training wouldnt give up on his claim that the chance of getting 10 points is 50% because you either get it or you dont get it. Too lazy to find it now, but im sure some remember...

I think i did chose my words carefull enough though. The chance is 1 in 64 million and that this chance is smaller than those things i mentioned.
I did give some numbers of needed rerolls that i linked with normal english words used by the non mathemathician to give some indication. Never did i claim it was not random. Never did i claim that you need to reroll 8000 times because that is how many rerolls are on average needed for 1 3x int weapon.

the chance you get a dragon orb out of 20 packs is about 1 in 500. That is what we call really lucky. 1 in 64 million is indeed so unlikely that the chances of this guy not telling the entire truth or him being a very skilled hacker are bigger. Of course as i said, if he did several 100 to 1000 rerolls, this changes and the chances of him just being one lucky mfer are becoming very much within reason. Note that the question i answered was to the statement of them both being 3x int on the first roll.

ovenusarmanio

The odds of winning the some lotteries are like 1 in 135 million, so it's not outside the realms of possibility that someone got incredibly lucky in a video game. Stranger things have happened. I mean the odds of getting struck by lightning are better but still. :P Easy solution is just ask him, so can we get back on topic please? Thanks.

Knownase - Heavens Tear

HrunsPanda - Archosaur wrote: »

Much smaller than either of the following:

-this person telling bull**** and showing the same weapon that he borrowed from someone else on 2 different characters .
-this person actually rerolling a zillion times. (several 1000 if he is just lucky, several 100 if he is insanely lucky)
-this person hacking the PWI servers.

The exact answer is not so hard though: 0.05 ^ 6 *100% = 0.0000015625 % chance. or 1 in 64 million

Supposedly this guy just farms Warsong daily, and pays people to solo pavillions so he can solo the final boss. He probably uses alts to horde up on badges.

Mr_Swiss - Heavens Tear

Knownase - Heavens Tear wrote: »

Supposedly this guy just farms Warsong daily, and pays people to solo pavillions so he can solo the final boss. He probably uses alts to horde up on badges.

I don't blame him, he found a system. I'm sure you know who i'm talking about, and most likely have seen the 3x int.