I check the Auction House most days, and I can never understand why anyone would post them for more than 25 and 100k, but they do. And why on Earth anyone would buy them at that price is even more of a mystery.
Same here. I have recently begun buying blue marks for my other characters and I always check the AH first for those listed below Bazaar price. If I don't find any, I buy from the Wondrous Bazaar instead. I'd also rather use the "AD Sink" rather than keep my AD in circulation, rather than buy those listed at base price. There's no good reason to pay more for something on the AH than it is worth and offered at a base price, buyable anywhere in the game through the remote Wondrous Bazaar screen.
Yes! Someone asked in Zone where to get them, so I said, "in your currency screen, next to your Astral Diamonds, click SPEND and it will open the bazaar. Check the AH as well, but if they are not less than 25k, do not buy them there, get them from WB."
"THANKS!" came the reply as a PM.
Must have been a new player?
0
ionvnegativoMember, NW M9 PlaytestPosts: 119Arc User
The Wondrous Bazaar sells blue Marks of Potency for 25k AD, purple Greater Marks of Potency for 100k.
I check the Auction House most days, and I can never understand why anyone would post them for more than 25 and 100k, but they do. And why on Earth anyone would buy them at that price is even more of a mystery.
Five Greater marks for 450,000 AD, I will buy, as I am saving 50,000 AD.
Five at 499,999 I will not buy and I prefer to used the AD Sink.
Five at 550,000, I want to e-mail the poster and tell him he's going to lose his posting fee.
But I would be more than happy to buy 100 Marks of Potency for 100k on the AH rather than 2,475,000 AD at the Wondrous Bazaar. But I have NEVER seen them at that price. I doubt they would stay up for much longer than a nano-second, anyway.
Or do you mean 100k EACH, in stacks of 99? So that is 9.9 MILLION AD? I have never seen them at that price, either. Only a fool would ask that much, and only a bigger fool would pay it, when Rhix's mistress is asking only 2.475 million.
But bot-farmed gear tends to drive the price down, not up. People tend to drive the price up.
marks of power (blue).....current price is about 100k in AH.....so.. if u want to upgrade an artifact of power to lvl 60 u need 100k from greater mark of potency + ~100k from mark of power + etc, etc
marks of power (blue).....current price is about 100k in AH
Ah, I see. Are they Power Stones, or Marks of Power? These are two different things, IIRC.
Where else do they come from, otherwise? I have quite a pile of blue and purple Marks of Power, Stability and Union from opening lock boxes over the last nine months. I decided to keep them instead of trading them. As I recall, the Power Stones and Marks were always more expensive on the AH than Stability and Union.
I seem to recall getting Greater Marks of Power/Stability/Union from Dread Ring dungeons.
But I do agree that there should be an AD Sink for all of these things and they should never come only from Lock Boxes. Especially if the Intermediate level marks and stones are so rare from an in-game source through normal play. Lock Boxes should be optional extras, like the special mounts etc, not the main or sole source.
The chance to succeed ON the 4th try is only about 10%.
The chance to need four or less tries is about 68% (25% + 19% + 14% + 11%).
>8o
!!
:rolleyes:
A 25% chance is a 1/4 chance, so with 4 trials you would expect to succeed once, on average, if you made a statistically significant number of independent trials, like 4000.
A 25% chance is a 1/4 chance, so with 4 trials you would expect to succeed once, on average, if you made a statistically significant number of independent trials, like 4000.
You do not understand what you are calculating. Every "roll" is completely independent from the others; from those premises it is self-evidently impossible for say three prior failures to somehow magically reduce the probability of the fourth down to 11%, as you claimed.
What one *could* calculate is the probability of requiringN number of "rolls", though.
0
thefabricantMember, NW M9 PlaytestPosts: 5,248Arc User
Actually, the probability of succeeding, on any attempt, ignoring the results of any previous attempts will always be 1/4. However, if you want to know the probability of succeeding BY try N, including try N, it is in fact different. Assuming to succeed you only need to succeed once and assuming you have a 1/4 chance of succeeding, the mathematics is as follows:
N
sigma((3^I-1)/(4^I))
I=1
Unfortunately I don't actually have an available sigma function to stick in here in order to make that more understandable but basically this is how it works:
Whatever number of times you attempt, you input as n. So say you attempted 4 times, then n=4. Every time you run the function, you add 1 to I until it equals n and you add the result you get from every run up to n to 1/4.
Heres an example of the function working:
If n=4
P (success)=1/4+0+3/16+9/64+27/256=175/256 or a 68.359% chance of succeeding.
If you struggle seeing how this works, then think of it like this:
The first time you make an attempt there are 4 possible outcomes, one of which are a success. Now, assuming you succeed and you only care about succeeding once, then it doesn't matter wether you succeed or not any of the subsequent tries, you have already succeeded. Thats where the base 1/4 comes from. Now lets look at the 2nd time you try. The 2nd time you try, you have to look at how many branches there already are. Well, from the start of the 2nd try, there are the 4 possible outcomes from the previous try, multiplied by the 4 possible outcomes from the next try. This leads to there being 16 possible outcomes....or in otherwords, 4^2. However, remember, we don't care about 1/4 of the possible outcomes, because we already know down at least 4 of the outcomes there is a success. (The one time you succeeded, multiplied by 4) that means that every subsequent time you do the calculation, you can ignore that branch and just work with the remainder. As there are 12 remaining branches and 1/4 of the time out of those 12 you succeed, you can then succeed 3 out of those 12 times. Now, every time you make an addtional attempt, it doesn't matter if you succeed or fail down the branches you have succeeded in, you once again, only have to consider the branches where you fail. The stuff between the sigma sign accounts for adding all the probabilities which you have not already counted.
Suffice to say, each time you make an attempt, the odds of you succeeding get slightly better. However, you would have to make an infinite number of attempts for the odds to ever be exactly one, which is, as we know, impossible. Of coarse, most of the time we interested in more then just succeeding once, but assuming we were only interested in succeeding once, that is how you would calculate it. And to make the calculation universal, not just working for 1/4 it would be:
N
sigma (((numerator of 1-Pin)^I-1)/((denominator of 1-Pin)^I))
I=1
Pin=the initial probability for success on the first attempt
If you want to know how to calculate the probability of succeeding a certain number of times and not just a single time, then pm me and I will explain.
The chances of succeeding specifically on the nth try, rather then the probability of succeeding by the nth try, however, is still 1/4, as the chances are not dependent on the previous attempts.
My Neverwinter gaming:
- Log in. Kill 5 dragons. Log out.
- Log through all toons (invoke, leadership).
- Exit Neverwinter.
That´s it. All old dungeons including CN dont´t drop anything I could sell. New Dungeon and Skirmish bore me to death cause I ran them a hundred times and the AD gain per time, the BoP drops ... dont work. PvE players dont need the new equipment but everybody wants it. If a casual player wants legendary level equip... he or she has only a few options. $, illegal stuff or toon-farm. No one needs the game... only for a every-module-short-visit, dailies, ugly pvp and equipment posing.
Enough.
//Bellistor
0
chemboy613Member, NW M9 PlaytestPosts: 1,521Arc User
edited November 2014
Let's spend more on the AH because... oh, what is that little diamond icon? XD XD
My interpretation of why. New players don't' know everything and sometimes rush through and will use the AH instead of the bazaar because they don't know any better.
Now, we've had discussions about exploiting new players and I really think offering to sell things HIGHER than the bazaar is to induce them to be exploited due to lack of knowledge XD XD. I obviously don't do such a thing.
In terms of marks of power - the price is whatever the market will bear. They are used to upgrade artifacts and artifact weapons, and they are not very common drops. The reason power marks are higher than union or stability is because power artifacts are (for whatever reason) more popular than union artifacts, which are in turn more popular than stability artifacts. Hence more demand and less supply means a higher price.
Marks drop during sharandar, dread ring, icewind dale (likely from skill nodes), randomly from bosses in dungeons (shattersnape in VT is a farming favorite), and in elol about once every two runs SOMEONE randomly gets a random blue mark in their inventory, so you will get a mark about once every ten ELOL runs. However if you die at the boss, you do not get a mark, so remember dying is bad
I don't know how farmable they are, but as i get them in my travels i just put them in the bank and wait until i need them.
I don't wanna sound like a doombringer, but in my opinion Cryptic gave up on making the game better and instead chose to try to milk whoever's left. They may have shifted their attention to newer games.
I do not need to read that, thank you very much. I am already perfectly aware that a 25% chance of success is a 1/4 chance of success, so if I make four attempts and it is a fair test, I fully expect to succeed once; on average.
Actually, the probability of succeeding, on any attempt, ignoring the results of any previous attempts will always be 1/4. However, if you want to know the probability of succeeding BY try N, including try N, it is in fact different. Assuming to succeed you only need to succeed once and assuming you have a 1/4 chance of succeeding, the mathematics is as follows:
N
1/4+sigma((3^I-1)/(4^I))
I=1
Unfortunately I don't actually have an available sigma function to stick in here in order to make that more understandable
Cheers! I am going to out that into Excel and try it out!
0
thefabricantMember, NW M9 PlaytestPosts: 5,248Arc User
I do not need to read that, thank you very much. I am already perfectly aware that a 25% chance of success is a 1/4 chance of success, so if I make four attempts and it is a fair test, I fully expect to succeed once; on average.
:rolleyes:
If you roll a 6 sided die 6 times, there will likely be at least 1 side that didn't land.
Flip a coin 4 times and there's a very real chance of not getting 2 heads and 2 tails. Have 1000 people flip a coin 4 times, and some of them will probably not get a single head or a single tail despite it still being 50% chance to land.
It's also a gamblers weakness. "I've lost so many times, it's about time my luck changed" despite having the same chance to win.
0
thefabricantMember, NW M9 PlaytestPosts: 5,248Arc User
Except in this case, the rolls aren't exactly independent, since you only roll until you succeed, and the question is which roll in a series will succeed. It's also a matter of what perspective you adopt and what question you put forward. If you pose the question "at what roll will I be most likely to finally succeed?" then the first roll is the correct answer, since that has the greatest chance of negating subsequent rolls. So if you are 25% likely to succeed at rolling a 1 at each given roll, then you are also 25% likely to negate the necessity of additional rolls with each roll, reducing the likelihood of subsequent rolls being the one that will actually succeed (even if each roll independently always has a 25% chance of success, that's not what he was calculating).
Just try and imagine it in gambling terms: Someone is rolling a four-sided polyhedron, and you are to bet on which particular roll in a series will come up a 1. Once the die comes up a 1, there will be no more rolls. The first roll has a 25% chance of being a 1. Each subsequent roll has the same chance of being a 1, but each subsequent roll is also less likely to occur because it's increasingly likely that a 1 has already come up. Thus, the chance of the first successful roll being the third roll is less than the chance of the first successful roll being the first one.
Those are the numbers he came up with; they add up to the cumulative chance of rolling a 1 with that many rolls, it's just how the chance is distributed between each roll being the roll to succeed.
The first roll is 25% likely to be the one to succeed.
The second roll is also 25% likely to succeed, but it's also only 75% likely to exist, if the first roll succeeded. That makes it 18.75% likely to be the successful roll.
The third roll is also 25% likely to succeed, but it's also only 56.25% (.75 × .75) likely to exist, since either the first or second roll could have succeeded already. That makes it 14.0625% likely to be the successful roll.
The problem you are making is based off the way you making your calculations. You are arguing 2 specific things which end up contradicting each other. The first thing you are doing is you are saying that if you succeed, you aren't going to keep trying and so all further attempts down that path no longer exist. However, you are then basing your calculations, the 1/4*3/4*3/4*3/4... on the fact that they do. Why do I say that? Simple, every time you divide by 4, you are multiplying the number of probabilities by 4. In doing so, you are accounting for the existence of every single branch you have said to ignore. The reason this causes problems and causes your probability to incorrectly diminish is because you are dividing the number of successful outcomes on whichever attempt assuming you have failed up till then by the actual total number of outcomes along all branches including the successful ones which you have claimed we ignoring. What you are doing, for example, is this:
number of successful chances first time=1
number of possible outcomes on first time=4
number of successful chances on second time on failed branch=3
number of possible outcomes on second time=16
Rather then:
number of successful chances first time=1
number of possible outcomes on first time=4
number of successful chances on second time on failed branch=3
number of possible outcomes on second time on failed branch=12
If you want to count the outcomes from the successful branch, then you also have to count the fact that as that branch succeeded, every SINGLE one of those probabilities also has to be counted as a success. Just because you are infinitesimally unlikely to have not succeeded by the 1000th try, doesn't change the fact that on the 1000th try you still just as likely to succeed. Go back and read my post on probability which shows how to calculate the chance of succeeding by the nth try and stop arguing that independent events are dependent using flawed logic. Remember with independent events P(AB)=P(A)*P(B) which is what we are seeing in this case.
If you roll a 6 sided die 6 times, there will likely be at least 1 side that didn't land.
Flip a coin 4 times and there's a very real chance of not getting 2 heads and 2 tails. Have 1000 people flip a coin 4 times, and some of them will probably not get a single head or a single tail despite it still being 50% chance to land.
It's also a gamblers weakness. "I've lost so many times, it's about time my luck changed" despite having the same chance to win.
I hated this stuff in school.
But yeah a 25% chance doesn't ever mean you will win 1 out of 4, after all you have a 75% chance on every roll not to succeed.
Although, I wouldn't mind if the drop rates took this into account and raised the drop% after every unsuccessful attempt.
So that at the 4th unsuccessful attempt the chance was 68% to drop instead of 25% still for that roll.
A progressive drop rate would definitely help out in IWD *hint hint devs hint*.
My Neverwinter gaming:
- Log in. Kill 5 dragons. Log out.
- Log through all toons (invoke, leadership).
- Exit Neverwinter.
That´s it. All old dungeons including CN dont´t drop anything I could sell. New Dungeon and Skirmish bore me to death cause I ran them a hundred times and the AD gain per time, the BoP drops ... dont work. PvE players dont need the new equipment but everybody wants it. If a casual player wants legendary level equip... he or she has only a few options. $, illegal stuff or toon-farm. No one needs the game... only for a every-module-short-visit, dailies, ugly pvp and equipment posing.
Enough.
//Bellistor
Sounds like you should move to another game that will hold your interest better.
0
thefabricantMember, NW M9 PlaytestPosts: 5,248Arc User
edited November 2014
Interestingly enough assuming that the chances of a belt drop in SoT are all independent and assuming you had 0.01% chance of seeing a belt drop, you would have, by the 100th try roughly a 10% chance of having a single belt drop... Then remember, there being 5 people in the party, you then need to multiply the chances of a belt dropping by the chances of you winning it which is 1/5. So in SoT with a realistic belt drop chance, you have roughly a 2% chance of winning a belt in 100 attempts. If you were to make 500 attempts, you would be approaching a 48% chance of seeing a belt drop and a 48/5% chance of winning one... Unfortunately my computer doesn't handle big enough fractions for me to be able to tell you when the probability of seeing a belt drop with this realistic drop chance is greater then 50%, but I would estimate it to be between 500 and 600 tries.
There are times I wish this forum would allow us to vote on how well a given post was written (good or bad). I agree with all of your points; and it is those points along with problems in matchmaking with PVP (not class performance in general, but more so with matchmaking) that I don't continue to invest money in this game (they received $20 from me). I've only been playing since late August / early September of this year (I didn't write the start date down), and it is starting to weigh on me for how grindy it is to get anywhere once you get to 60.
By the way, I do appreciate your guides and input; between you and Kalec, I think a lot of ground is covered.
I made a silly mistake in that post, which I have now corrected. The probability is actually:
N
sigma((3^I-1)/(4^I))
I
Hope that helps I forgot that as I was running from 1 it auto includes the 1/4 (been a while since I worked with probability)
Thanks for that!
I am not much of a Maths buff, but I do enjoy learning new things with Excel.
I have a table of all the Refining Points needed for Artifacts, their RP value for a matching and non-matching Artifact and what level you get by putting a certain Rank into a Rank 1.
[B][COLOR="#000000"]
A B C D E F G
Rank Tot RP To Next To Rank 100 RP No Match RP Matched Added Ranks
1 0 20 3,847,635 1,500 7,500 17
2 20 90 3,847,615 1,510 7,550 18
[/COLOR][/B]
Column G has:
=INDEX($A$2:$A$101,MATCH(F2,$B$2:$B$101))
This is VERY useful for calculating what I will get if I have two Bloodcrystals, make 1 level 59 and put it into the first, then put that into my Rank 89 etc and whether it is worth it or not.
If I just put one Rank 1 into another Rank 1, I'll get a Rank 17 (assuming no Critical "hit" for refining).
But yeah a 25% chance doesn't ever mean you will win 1 out of 4.
No one said it did.
But if I make four attempts and it is a fair test, I fully expect to succeed once; on average.
"On Average" means with a statistically valid number of trials. So if you roll 4,000,000 times, you'd expect to succeed 1,000,000 times. The deviation from a precise 1/4 success rate will not be statistically significant if it is a fair and unbiased test with no artificial constraints, such as taking into account the number of previous successes or failures, hypothesising that you stop when you succeed, or taking into account any attempts that will not be made etc.
Interestingly enough assuming that the chances of a belt drop in SoT are all independent and assuming you had 0.01% chance of seeing a belt drop, you would have, by the 100th try roughly a 10% chance of having a single belt drop...
.01% x 100 = 1%. But that doesn't matter. It doesn't make a difference if you run once or 1 million times, the chances of a drop remain the same as they are not cumulative. If the belt drops at a rate of .01% chance and assuming everyone chooses need, you'll have a .002% chance of winning it.
My Neverwinter gaming:
- Log in. Kill 5 dragons. Log out.
- Log through all toons (invoke, leadership).
- Exit Neverwinter.
That´s it. All old dungeons including CN dont´t drop anything I could sell. New Dungeon and Skirmish bore me to death cause I ran them a hundred times and the AD gain per time, the BoP drops ... dont work. PvE players dont need the new equipment but everybody wants it. If a casual player wants legendary level equip... he or she has only a few options. $, illegal stuff or toon-farm. No one needs the game... only for a every-module-short-visit, dailies, ugly pvp and equipment posing.
Comments
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"THANKS!" came the reply as a PM.
Must have been a new player?
marks of power (blue).....current price is about 100k in AH.....so.. if u want to upgrade an artifact of power to lvl 60 u need 100k from greater mark of potency + ~100k from mark of power + etc, etc
Ah, I see. Are they Power Stones, or Marks of Power? These are two different things, IIRC.
Where else do they come from, otherwise? I have quite a pile of blue and purple Marks of Power, Stability and Union from opening lock boxes over the last nine months. I decided to keep them instead of trading them. As I recall, the Power Stones and Marks were always more expensive on the AH than Stability and Union.
I seem to recall getting Greater Marks of Power/Stability/Union from Dread Ring dungeons.
But I do agree that there should be an AD Sink for all of these things and they should never come only from Lock Boxes. Especially if the Intermediate level marks and stones are so rare from an in-game source through normal play. Lock Boxes should be optional extras, like the special mounts etc, not the main or sole source.
That's nonsense... Here are the chances of success on the n-th try for a 25% base success chance:
1: 25%
2: 19% (75% * 25%)
3: 14% (75% * 75% * 25%)
4: 11% (75% * 75% * 75% * 25%)
The chance to succeed ON the 4th try is only about 10%.
The chance to need four or less tries is about 68% (25% + 19% + 14% + 11%).
You Fail Math Forever.
>8o
!!
:rolleyes:
A 25% chance is a 1/4 chance, so with 4 trials you would expect to succeed once, on average, if you made a statistically significant number of independent trials, like 4000.
Where is your proof? Just claiming something without proofing me wrong is lame.
We are talking about something called Bernoulli/Binomial distribution here (http://en.wikipedia.org/wiki/Bernoulli_distribution).
Go read the article and proof me wrong. Otherwise you just showed that you do not know what you are talking about.
Also read http://en.wikipedia.org/wiki/Binomial_distribution.
Go read http://en.wikipedia.org/wiki/Bernoulli_distribution and
http://en.wikipedia.org/wiki/Binomial_distribution.
What one *could* calculate is the probability of requiring N number of "rolls", though.
Actually, the probability of succeeding, on any attempt, ignoring the results of any previous attempts will always be 1/4. However, if you want to know the probability of succeeding BY try N, including try N, it is in fact different. Assuming to succeed you only need to succeed once and assuming you have a 1/4 chance of succeeding, the mathematics is as follows:
N
sigma((3^I-1)/(4^I))
I=1
Unfortunately I don't actually have an available sigma function to stick in here in order to make that more understandable but basically this is how it works:
Whatever number of times you attempt, you input as n. So say you attempted 4 times, then n=4. Every time you run the function, you add 1 to I until it equals n and you add the result you get from every run up to n to 1/4.
Heres an example of the function working:
If n=4
P (success)=1/4+0+3/16+9/64+27/256=175/256 or a 68.359% chance of succeeding.
If you struggle seeing how this works, then think of it like this:
The first time you make an attempt there are 4 possible outcomes, one of which are a success. Now, assuming you succeed and you only care about succeeding once, then it doesn't matter wether you succeed or not any of the subsequent tries, you have already succeeded. Thats where the base 1/4 comes from. Now lets look at the 2nd time you try. The 2nd time you try, you have to look at how many branches there already are. Well, from the start of the 2nd try, there are the 4 possible outcomes from the previous try, multiplied by the 4 possible outcomes from the next try. This leads to there being 16 possible outcomes....or in otherwords, 4^2. However, remember, we don't care about 1/4 of the possible outcomes, because we already know down at least 4 of the outcomes there is a success. (The one time you succeeded, multiplied by 4) that means that every subsequent time you do the calculation, you can ignore that branch and just work with the remainder. As there are 12 remaining branches and 1/4 of the time out of those 12 you succeed, you can then succeed 3 out of those 12 times. Now, every time you make an addtional attempt, it doesn't matter if you succeed or fail down the branches you have succeeded in, you once again, only have to consider the branches where you fail. The stuff between the sigma sign accounts for adding all the probabilities which you have not already counted.
Suffice to say, each time you make an attempt, the odds of you succeeding get slightly better. However, you would have to make an infinite number of attempts for the odds to ever be exactly one, which is, as we know, impossible. Of coarse, most of the time we interested in more then just succeeding once, but assuming we were only interested in succeeding once, that is how you would calculate it. And to make the calculation universal, not just working for 1/4 it would be:
N
sigma (((numerator of 1-Pin)^I-1)/((denominator of 1-Pin)^I))
I=1
Pin=the initial probability for success on the first attempt
If you want to know how to calculate the probability of succeeding a certain number of times and not just a single time, then pm me and I will explain.
The chances of succeeding specifically on the nth try, rather then the probability of succeeding by the nth try, however, is still 1/4, as the chances are not dependent on the previous attempts.
My Neverwinter gaming:
- Log in. Kill 5 dragons. Log out.
- Log through all toons (invoke, leadership).
- Exit Neverwinter.
That´s it. All old dungeons including CN dont´t drop anything I could sell. New Dungeon and Skirmish bore me to death cause I ran them a hundred times and the AD gain per time, the BoP drops ... dont work. PvE players dont need the new equipment but everybody wants it. If a casual player wants legendary level equip... he or she has only a few options. $, illegal stuff or toon-farm. No one needs the game... only for a every-module-short-visit, dailies, ugly pvp and equipment posing.
Enough.
//Bellistor
My interpretation of why. New players don't' know everything and sometimes rush through and will use the AH instead of the bazaar because they don't know any better.
Now, we've had discussions about exploiting new players and I really think offering to sell things HIGHER than the bazaar is to induce them to be exploited due to lack of knowledge XD XD. I obviously don't do such a thing.
In terms of marks of power - the price is whatever the market will bear. They are used to upgrade artifacts and artifact weapons, and they are not very common drops. The reason power marks are higher than union or stability is because power artifacts are (for whatever reason) more popular than union artifacts, which are in turn more popular than stability artifacts. Hence more demand and less supply means a higher price.
Marks drop during sharandar, dread ring, icewind dale (likely from skill nodes), randomly from bosses in dungeons (shattersnape in VT is a farming favorite), and in elol about once every two runs SOMEONE randomly gets a random blue mark in their inventory, so you will get a mark about once every ten ELOL runs. However if you die at the boss, you do not get a mark, so remember dying is bad
I don't know how farmable they are, but as i get them in my travels i just put them in the bank and wait until i need them.
Everything you need to know about CW:
http://nw-forum.perfectworld.com/showthread.php?780981-Chem-s-CW-Compendium-Everything-you-need-to-know
+1 for thefabricant for a very good explanation of statistics. Also well said.
+1 for b3llist0r for even though it depresses me to see a description of exactly what NW has become for me.
I do not need to read that, thank you very much. I am already perfectly aware that a 25% chance of success is a 1/4 chance of success, so if I make four attempts and it is a fair test, I fully expect to succeed once; on average.
:rolleyes:
Cheers! I am going to out that into Excel and try it out!
I made a silly mistake in that post, which I have now corrected. The probability is actually:
N
sigma((3^I-1)/(4^I))
I
Hope that helps
If you roll a 6 sided die 6 times, there will likely be at least 1 side that didn't land.
Flip a coin 4 times and there's a very real chance of not getting 2 heads and 2 tails. Have 1000 people flip a coin 4 times, and some of them will probably not get a single head or a single tail despite it still being 50% chance to land.
It's also a gamblers weakness. "I've lost so many times, it's about time my luck changed" despite having the same chance to win.
The problem you are making is based off the way you making your calculations. You are arguing 2 specific things which end up contradicting each other. The first thing you are doing is you are saying that if you succeed, you aren't going to keep trying and so all further attempts down that path no longer exist. However, you are then basing your calculations, the 1/4*3/4*3/4*3/4... on the fact that they do. Why do I say that? Simple, every time you divide by 4, you are multiplying the number of probabilities by 4. In doing so, you are accounting for the existence of every single branch you have said to ignore. The reason this causes problems and causes your probability to incorrectly diminish is because you are dividing the number of successful outcomes on whichever attempt assuming you have failed up till then by the actual total number of outcomes along all branches including the successful ones which you have claimed we ignoring. What you are doing, for example, is this:
number of successful chances first time=1
number of possible outcomes on first time=4
number of successful chances on second time on failed branch=3
number of possible outcomes on second time=16
Rather then:
number of successful chances first time=1
number of possible outcomes on first time=4
number of successful chances on second time on failed branch=3
number of possible outcomes on second time on failed branch=12
If you want to count the outcomes from the successful branch, then you also have to count the fact that as that branch succeeded, every SINGLE one of those probabilities also has to be counted as a success. Just because you are infinitesimally unlikely to have not succeeded by the 1000th try, doesn't change the fact that on the 1000th try you still just as likely to succeed. Go back and read my post on probability which shows how to calculate the chance of succeeding by the nth try and stop arguing that independent events are dependent using flawed logic. Remember with independent events P(AB)=P(A)*P(B) which is what we are seeing in this case.
Agreed, and just because you have a 1/4 chance of succeeding, doesn't mean that you should on average succeed once every 4 times. What it means is that you have a 68.359% chance of succeeding at least once out of those 4 attempts:p Also, for anyone interested, here is how you calculate the probability of succeeding K number of times given N attempts:
http://www.askamathematician.com/2010/07/q-whats-the-chance-of-getting-a-run-of-k-successes-in-n-bernoulli-trials-why-use-approximations-when-the-exact-answer-is-known/
I hated this stuff in school.
But yeah a 25% chance doesn't ever mean you will win 1 out of 4, after all you have a 75% chance on every roll not to succeed.
Although, I wouldn't mind if the drop rates took this into account and raised the drop% after every unsuccessful attempt.
So that at the 4th unsuccessful attempt the chance was 68% to drop instead of 25% still for that roll.
A progressive drop rate would definitely help out in IWD *hint hint devs hint*.
Sounds like you should move to another game that will hold your interest better.
There are times I wish this forum would allow us to vote on how well a given post was written (good or bad). I agree with all of your points; and it is those points along with problems in matchmaking with PVP (not class performance in general, but more so with matchmaking) that I don't continue to invest money in this game (they received $20 from me). I've only been playing since late August / early September of this year (I didn't write the start date down), and it is starting to weigh on me for how grindy it is to get anywhere once you get to 60.
By the way, I do appreciate your guides and input; between you and Kalec, I think a lot of ground is covered.
Thank you.
Thanks for that!
I am not much of a Maths buff, but I do enjoy learning new things with Excel.
I have a table of all the Refining Points needed for Artifacts, their RP value for a matching and non-matching Artifact and what level you get by putting a certain Rank into a Rank 1.
Column G has:
=INDEX($A$2:$A$101,MATCH(F2,$B$2:$B$101))
This is VERY useful for calculating what I will get if I have two Bloodcrystals, make 1 level 59 and put it into the first, then put that into my Rank 89 etc and whether it is worth it or not.
If I just put one Rank 1 into another Rank 1, I'll get a Rank 17 (assuming no Critical "hit" for refining).
I'll have to make one for Artifact Gear as well!
No one said it did.
But if I make four attempts and it is a fair test, I fully expect to succeed once; on average.
"On Average" means with a statistically valid number of trials. So if you roll 4,000,000 times, you'd expect to succeed 1,000,000 times. The deviation from a precise 1/4 success rate will not be statistically significant if it is a fair and unbiased test with no artificial constraints, such as taking into account the number of previous successes or failures, hypothesising that you stop when you succeed, or taking into account any attempts that will not be made etc.
~
.01% x 100 = 1%. But that doesn't matter. It doesn't make a difference if you run once or 1 million times, the chances of a drop remain the same as they are not cumulative. If the belt drops at a rate of .01% chance and assuming everyone chooses need, you'll have a .002% chance of winning it.
SNAP
/10char