Yeah, that's not a review. That's a re-visitation of the same single post. The mistake was on your part in not only failing to, but blatantly refusing to, read the post in the context of the larger discussion. My affirmative response is indeed correct. It is likewise connected to the Gambler's Fallacy. As stated several times, there cannot be any "maturity of chances" (which, by the way is another name for the Gambler's Fallacy) or "Cumulative Probability" as mathematicians like to try to euphemistically rename it in an effort to cover the fact that they're guilty of the Gambler's Fallacy when they spew nonsense like this, in any case involving more than two possible outcomes. Even mathematicians will admit this much, but really, there is no "cumulative probability" in ANY trial where the attempts are not in any way affected by past attempts. Now, if the code allows previous attempts to affect subsequent attempts, that's another matter, and the "cumulative probability is coded, but otherwise, random is random (Chaos Theory notwithstanding), and any attempt to "predict" an outcome involving Probability is fallacious.
^^ Lots of anger here.
Your 'cumulative probability' argument is a bit of a strawman, really. as I see *no one* seriously argue that the mathematical chance of getting a JHAS increases which each new box you open. Fact remains, however, if you open a 1,000 lockboxes, chances are you got a JHAS by now. If you open just 1, chances are you don't. Simple as that.
Listen carefully: when more than two possible outcomes exist for a given act, there can be no cumulative probability for any possible outcome of that given act. You can check that in books on Mathematics as well as in books on Logic. See, I've had this discussion before elsewhere. Mathematicians' "Cumulative Probability" is the same mistake known by Logicians as "Maturity of Chances." Good heavens, the names even obviously mean the same thing. Renaming the fallacy in an effort to insist that it's sound reasoning doesn't obfuscate the issue, particularly when the meaning of the two names is so obviously identical.
Yeah, that's not a review. That's a re-visitation of the same single post. The mistake was on your part in not only failing to, but blatantly refusing to, read the post in the context of the larger discussion. My affirmative response is indeed correct. It is likewise connected to the Gambler's Fallacy. As stated several times, there cannot be any "maturity of chances" (which, by the way is another name for the Gambler's Fallacy) or "Cumulative Probability" as mathematicians like to try to euphemistically rename it in an effort to cover the fact that they're guilty of the Gambler's Fallacy when they spew nonsense like this, in any case involving more than two possible outcomes. Even mathematicians will admit this much, but really, there is no "cumulative probability" in ANY trial where the attempts are not in any way affected by past attempts. Now, if the code allows previous attempts to affect subsequent attempts, that's another matter, and the "cumulative probability is coded, but otherwise, random is random (Chaos Theory notwithstanding), and any attempt to "predict" an outcome involving Probability is fallacious.
Actually, cumulative probability is perfectly compatible with the Gambler's Fallacy/ not being any maturity of chances.
Let's take 10 trials, assuming binomial distribution, with probability of success being 0.5% = 0.005. Let a success be S and a failure be F.
Now the probability of at least one success is P(at least one S) = 1-P(S=0), which in this case is about 5%. Now, let's take a classic case of the Gambler's Fallacy - the person has carried out 9 out of the 10 trials so far, and they have all been failures.
Results:
F F F F F F F F F ?
Someone with an imperfect knowledge of probability would think that the probability of the ? being a success is 5%, and not 0.5%. After all, he says, probability of there being a success was 5%! This is an error, as P(at least one S) was calculated for any one of the trials being a success. That is, the first trial could have been it, the second could have been it, etc.
However, in the scenario above, the results of the first 9 trials have been completed and fixed. The only two options available are FFFFFFFFS (obviously, P=0.5%) and FFFFFFFFFF(P=99.5%). In other words, the probability here is the conditional probability of the final trial being a success, conditional on the first 9 trials being failures, which is equal to probability of success in a single trial, 0.5%, and not the probability of success in 10 trials any longer. The fallacy that the gambler has in mind is indeed a fallacy.
The full binomial distribution cumulative probability formula only predicts cases where no trials have been carried out yet. Once you have carried out a trial, that trial is fixed - P = 1. Anyone who had a proper understanding of cumulative probabilities would understand this fact.
In summary, both of you are wrong - cumulative probabilities only apply to trials that have not been completed. therealfluffy, it does not apply once you've opened packs, and protogoth, cumulative probability is not exactly what you think it is. The fundamental basis is this: cumulative probability distributions like this only apply to things that have not happened yet - once something (a trial) has already happened, the previously calculated figure is invalidated. Honestly, what are schools teaching people these days?
The full binomial distribution cumulative probability formula only predicts cases where no trials have been carried out yet. Once you have carried out a trial, that trial is fixed - P = 1. Anyone who had a proper understanding of cumulative probabilities would understand this fact.
Not exactly sure what you're trying to say here. There is no previous trial's 'memory.' If you get to throw a dice 10 times, and have already done so 9 times, on your last trial, the chance to, say, throw '6' is still 1 in 6, irrespective of what happened before.
Not exactly sure what you're trying to say here. There is no previous trial's 'memory.' If you get to throw a dice 10 times, and have already done so 9 times, on your last trial, the chance to, say, throw '6' is still 1 in 6, irrespective of what happened before.
He's trying to reduce all possible outcomes to 2 and only 2 (there are more than two, and reducing them to S = "get super ship" and F= "fail to get super ship" ignores the reality of the situation in question), AND trying to ignore the fact that all of the possible outcomes continue to exist at each attempt, which is why this "Cumulative Probability" stuff is fallacious.
But hey, Logician vs. Mathematician, nothing new here, since we both work with a Pure Science and generally get into tiffs like this when our fields overlap, as they do in the matter of Probability and Statistics. Pro tip: Wittgenstein, Uber Gewissheit (*shakes fist at forum which dislikes special characters*), s.v. "Sprachspiele."
Not exactly sure what you're trying to say here. There is no previous trial's 'memory.' If you get to throw a dice 10 times, and have already done so 9 times, on your last trial, the chance to, say, throw '6' is still 1 in 6, irrespective of what happened before.
That's part of what I'm trying to say. In that bit, I meant that let's say you calculate the probability for number of successes in 10 trials. This figure is only valid for 10 trials. Once you've done one trial, to get probability of results for the remaining trials, we must calculate for 9 trials on their own. The first trial does not affect the rest. Oh, and by trial, I mean attempt.
He's trying to reduce all possible outcomes to 2 and only 2 (there are more than two, and reducing them to S = "get super ship" and F= "fail to get super ship" ignores the reality of the situation in question), AND trying to ignore the fact that all of the possible outcomes continue to exist at each attempt, which is why this "Cumulative Probability" stuff is fallacious.
But hey, Logician vs. Mathematician, nothing new here, since we both work with a Pure Science and generally get into tiffs like this when our fields overlap, as they do in the matter of Probability and Statistics. Pro tip: Wittgenstein, Uber Gewissheit (*shakes fist at forum which dislikes special characters*), s.v. "Sprachspiele."
Cumulative probability simply predicts the chances of getting a specific macro-configuration out of the many that may exist. And I do think it's not exactly fallacious, or statistical thermodynamics would fall apart in predicting macrostates.
Could you elaborate on "ignoring the reality of the situation in question"? Yes, all possible outcomes exist - F is simply a form of notation to represent all possible outcomes. You know, since P(X or Y or Z) = P(X) + P(Y) + P(Z) - P(X and Y) - P(X and Z) - P(Y and Z) - P(X and Y and Z). And even if I didn't distill things down, I could still get the same results. It would just be a bunch more clunky. And that book you linked doesn't help in this case either - could you explain?
Edit: Darnit, typo in the last paragraph of my previous post. Supposed to be this:
cumulative probabilities only apply to trials that have not been completed.
So yeah, cumulative probability is fine for these purposes. You apply it making certain that "number of trials" is only the amount of packs you have not opened yet, and things will be fine. However, packs you've already opened have nothing whatsoever to do with chances of getting the ship out of the next packs. Just want to make that clear. Oh, and protogoth, sorry for being so vague on that last part. I meant that cumulative probability, when properly applied, proves that the Gambler's Fallacy really is a fallacy, like you said.
He's trying to reduce all possible outcomes to 2 and only 2 (there are more than two, and reducing them to S = "get super ship" and F= "fail to get super ship" ignores the reality of the situation in question), AND trying to ignore the fact that all of the possible outcomes continue to exist at each attempt, which is why this "Cumulative Probability" stuff is fallacious.
But hey, Logician vs. Mathematician, nothing new here, since we both work with a Pure Science and generally get into tiffs like this when our fields overlap, as they do in the matter of Probability and Statistics. Pro tip: Wittgenstein, Uber Gewissheit (*shakes fist at forum which dislikes special characters*), s.v. "Sprachspiele."
LOL. I should probably stay defuq out of it then, as math and me, well.. we're not friends. :P
For all you saying "just buy it in the exchange", if every single player had that attitude there wouldn't be any ships at all for sale. SOMEBODY has to open the stuff to get them.
As for me, I don't get doff packs from the c-store. I get all the free doffs I need from all sorts of assignments, including purple quality. If I have a new toon that needs doffs I just buy some packs in the exchange that come from lockboxes. If I need a specific type of doff, I just buy it from the exchange.
As for super duper ship? I couldn't care less. If I ever get one, great, if not, so what? I do like to open lockboxes occasionally, but it's mainly to get lobi crystals (or doff packs to save for later since I get so damned many of them and I sell most of them in the exchange if none of my toons need any).
For all you saying "just buy it in the exchange", if every single player had that attitude there wouldn't be any ships at all for sale. SOMEBODY has to open the stuff to get them.
True. And admirable if you do it. Me, I don't. Spend $200 for an 800 Lobi ship, or just get off Exchange for ca. 65 Mil? (like my Jem'Hadar Dreadnought) Then I choose the latter, sorry.
Also, only time I ever opened lockboxes was for the Temporal Warfare set. And Jem'Hadar space set upgrade, and Rule 62 Console, etc. The Temporal ships themselves I bought off exchange too.
You're mistaken. Read the earlier posts in the conversation. He was questioning if that was what I had already said. I responded in the affirmative. Please don't "correct" me on matters pertaining to a discipline which I was trained to teach, especially without the context of the larger conversation. I wasn't saying that what I was replying to at that moment was an example of the Gambler's Fallacy. Quite the contrary.
I think you mis understand what I was getting at about a math formula, as there is one.
is a % chance, even if its .001% chance, its a chance, and that means there is a finite number of tickets, or boxes in this case, till that chance happens, not a infinite one.
It not cumulative correct, otherwise at 1% you'd see a prize at around the 100th ticket, However the more times you play, the more chance that 1% would come up, but since its not liner.
it is a yes, no, 1 or 2 system, win loose. at .5% it would be 99.5% no .5% yes, every box. but its still not a infinite system, its still finite, at some point a opened box will be the .5% yes. The more times you play, the more likely you are you get a win, but not in the idea that its cumulative, but in the idea that you have moved closer to that finite number of opening before the .5% happens.
take a d100 roll by example, people would think that the odds of getting a 1 or a 100 are 1%, thus that in 100 rolls you should get at least 1 and 100 once. but the reality is that it could take less or more rolls.
I don't know the formula my self, I understand the concept, but I'm not a mathematician my self. that's what he was trying too provide.
And its sort of a if chance is X then opening Y boxes causes chance to approach Z. but its not a cumulative formula, its a way of figuring that if you do something enough times, how many times would it likely take to get x result.
yes if you open 1 box the chance is the same, if you open 10 boxes the chance is the same, if you open 100 boxes, same, 1000, same, 10,000 same. however the difference is if you open 1 box, its one chance, if you open 10,000 its 10,000 chances. you've changed the number of times you open boxes, which while it doesn't change the chance per box, it does change the number of times say .5% could come up. that's what the formula I was referring to does, it works out, based on prior experiments, or hard numbers, a rough value of how many times something would need to be done before X result should happen, its not a guarantee, just a way of figuring that if I do X this many times I should get Y result at least once.
your still dealing with a RNG in this case though, which isn't truly random, no matter how well its programed.
That's part of what I'm trying to say. In that bit, I meant that let's say you calculate the probability for number of successes in 10 trials. This figure is only valid for 10 trials. Once you've done one trial, to get probability of results for the remaining trials, we must calculate for 9 trials on their own. The first trial does not affect the rest. Oh, and by trial, I mean attempt.
Cumulative probability simply predicts the chances of getting a specific macro-configuration out of the many that may exist. And I do think it's not exactly fallacious, or statistical thermodynamics would fall apart in predicting macrostates.
Could you elaborate on "ignoring the reality of the situation in question"? Yes, all possible outcomes exist - F is simply a form of notation to represent all possible outcomes. You know, since P(X or Y or Z) = P(X) + P(Y) + P(Z) - P(X and Y) - P(X and Z) - P(Y and Z) - P(X and Y and Z). And even if I didn't distill things down, I could still get the same results. It would just be a bunch more clunky. And that book you linked doesn't help in this case either - could you explain?
Edit: Darnit, typo in the last paragraph of my previous post. Supposed to be this:
So yeah, cumulative probability is fine for these purposes. You apply it making certain that "number of trials" is only the amount of packs you have not opened yet, and things will be fine. However, packs you've already opened have nothing whatsoever to do with chances of getting the ship out of the next packs. Just want to make that clear. Oh, and protogoth, sorry for being so vague on that last part. I meant that cumulative probability, when properly applied, proves that the Gambler's Fallacy really is a fallacy, like you said.
Ummm......yeah, I was agreeing with that, and I was saying that cumulative probability agrees with that when applied properly. I think my typo might have caused a misunderstanding, and I got a bit long-winded - sorry about that. I was saying that the person you were arguing with misapplied the theory and got a wrong answer, and so trying to prove that the theory actually agrees that the gambler's fallacy really is a fallacy. I think we're getting some channels crossed here, so never mind.
I saw the webpage, but I also remembered seeing that as a book in my university library, but now that I think back, it was more pamphlet-sized than book-sized. Oh well.
actually having read that they don't invalidate what we are saying.
their coin flip example, if you flipped the coin till you got heads, then flipped it again till you go heads, and again, and so on, and recorded the results. you could work out roughly how many flips before heads will likely happen. Lets arbitrarily say 50 flips, so what that would tell is that for every 50 flips of the coin, 1 flip is likely to come up heads, it could be the first flip, it could be the last flip, but in a set of 50 trys 1 flip should result in a heads. the key there is should, since the odds per flip remain the same, there is a chance even once you have a set that should result in a single heads, that any given set might contain no heads result, just that the vast majority of the sets should contain at least one heads result, some might even contain more then one.
Comments
^^ Lots of anger here.
Your 'cumulative probability' argument is a bit of a strawman, really. as I see *no one* seriously argue that the mathematical chance of getting a JHAS increases which each new box you open. Fact remains, however, if you open a 1,000 lockboxes, chances are you got a JHAS by now. If you open just 1, chances are you don't. Simple as that.
Actually, cumulative probability is perfectly compatible with the Gambler's Fallacy/ not being any maturity of chances.
Let's take 10 trials, assuming binomial distribution, with probability of success being 0.5% = 0.005. Let a success be S and a failure be F.
Now the probability of at least one success is P(at least one S) = 1-P(S=0), which in this case is about 5%. Now, let's take a classic case of the Gambler's Fallacy - the person has carried out 9 out of the 10 trials so far, and they have all been failures.
Results:
F F F F F F F F F ?
Someone with an imperfect knowledge of probability would think that the probability of the ? being a success is 5%, and not 0.5%. After all, he says, probability of there being a success was 5%! This is an error, as P(at least one S) was calculated for any one of the trials being a success. That is, the first trial could have been it, the second could have been it, etc.
However, in the scenario above, the results of the first 9 trials have been completed and fixed. The only two options available are FFFFFFFFS (obviously, P=0.5%) and FFFFFFFFFF(P=99.5%). In other words, the probability here is the conditional probability of the final trial being a success, conditional on the first 9 trials being failures, which is equal to probability of success in a single trial, 0.5%, and not the probability of success in 10 trials any longer. The fallacy that the gambler has in mind is indeed a fallacy.
The full binomial distribution cumulative probability formula only predicts cases where no trials have been carried out yet. Once you have carried out a trial, that trial is fixed - P = 1. Anyone who had a proper understanding of cumulative probabilities would understand this fact.
In summary, both of you are wrong - cumulative probabilities only apply to trials that have not been completed. therealfluffy, it does not apply once you've opened packs, and protogoth, cumulative probability is not exactly what you think it is. The fundamental basis is this: cumulative probability distributions like this only apply to things that have not happened yet - once something (a trial) has already happened, the previously calculated figure is invalidated. Honestly, what are schools teaching people these days?
Obtain ZEN. Buy about 50 packs.
Sell on exchange for 6.5-8mil a piece
50 packs sold... buy JHAS for about 300-350mil depending on the day
Profit?
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Lifetime Subscriber since 2012 == 17,200 Accolades = RIP PvP and Vice Squad
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Not exactly sure what you're trying to say here. There is no previous trial's 'memory.' If you get to throw a dice 10 times, and have already done so 9 times, on your last trial, the chance to, say, throw '6' is still 1 in 6, irrespective of what happened before.
He's trying to reduce all possible outcomes to 2 and only 2 (there are more than two, and reducing them to S = "get super ship" and F= "fail to get super ship" ignores the reality of the situation in question), AND trying to ignore the fact that all of the possible outcomes continue to exist at each attempt, which is why this "Cumulative Probability" stuff is fallacious.
But hey, Logician vs. Mathematician, nothing new here, since we both work with a Pure Science and generally get into tiffs like this when our fields overlap, as they do in the matter of Probability and Statistics. Pro tip: Wittgenstein, Uber Gewissheit (*shakes fist at forum which dislikes special characters*), s.v. "Sprachspiele."
That's part of what I'm trying to say. In that bit, I meant that let's say you calculate the probability for number of successes in 10 trials. This figure is only valid for 10 trials. Once you've done one trial, to get probability of results for the remaining trials, we must calculate for 9 trials on their own. The first trial does not affect the rest. Oh, and by trial, I mean attempt.
Cumulative probability simply predicts the chances of getting a specific macro-configuration out of the many that may exist. And I do think it's not exactly fallacious, or statistical thermodynamics would fall apart in predicting macrostates.
Could you elaborate on "ignoring the reality of the situation in question"? Yes, all possible outcomes exist - F is simply a form of notation to represent all possible outcomes. You know, since P(X or Y or Z) = P(X) + P(Y) + P(Z) - P(X and Y) - P(X and Z) - P(Y and Z) - P(X and Y and Z). And even if I didn't distill things down, I could still get the same results. It would just be a bunch more clunky. And that book you linked doesn't help in this case either - could you explain?
Edit: Darnit, typo in the last paragraph of my previous post. Supposed to be this:
So yeah, cumulative probability is fine for these purposes. You apply it making certain that "number of trials" is only the amount of packs you have not opened yet, and things will be fine. However, packs you've already opened have nothing whatsoever to do with chances of getting the ship out of the next packs. Just want to make that clear. Oh, and protogoth, sorry for being so vague on that last part. I meant that cumulative probability, when properly applied, proves that the Gambler's Fallacy really is a fallacy, like you said.
LOL. I should probably stay defuq out of it then, as math and me, well.. we're not friends. :P
As for me, I don't get doff packs from the c-store. I get all the free doffs I need from all sorts of assignments, including purple quality. If I have a new toon that needs doffs I just buy some packs in the exchange that come from lockboxes. If I need a specific type of doff, I just buy it from the exchange.
As for super duper ship? I couldn't care less. If I ever get one, great, if not, so what? I do like to open lockboxes occasionally, but it's mainly to get lobi crystals (or doff packs to save for later since I get so damned many of them and I sell most of them in the exchange if none of my toons need any).
True. And admirable if you do it. Me, I don't. Spend $200 for an 800 Lobi ship, or just get off Exchange for ca. 65 Mil? (like my Jem'Hadar Dreadnought) Then I choose the latter, sorry.
Also, only time I ever opened lockboxes was for the Temporal Warfare set. And Jem'Hadar space set upgrade, and Rule 62 Console, etc. The Temporal ships themselves I bought off exchange too.
I'm sorry; but I'm not made of pure money.
I think you mis understand what I was getting at about a math formula, as there is one.
is a % chance, even if its .001% chance, its a chance, and that means there is a finite number of tickets, or boxes in this case, till that chance happens, not a infinite one.
It not cumulative correct, otherwise at 1% you'd see a prize at around the 100th ticket, However the more times you play, the more chance that 1% would come up, but since its not liner.
it is a yes, no, 1 or 2 system, win loose. at .5% it would be 99.5% no .5% yes, every box. but its still not a infinite system, its still finite, at some point a opened box will be the .5% yes. The more times you play, the more likely you are you get a win, but not in the idea that its cumulative, but in the idea that you have moved closer to that finite number of opening before the .5% happens.
take a d100 roll by example, people would think that the odds of getting a 1 or a 100 are 1%, thus that in 100 rolls you should get at least 1 and 100 once. but the reality is that it could take less or more rolls.
I don't know the formula my self, I understand the concept, but I'm not a mathematician my self. that's what he was trying too provide.
And its sort of a if chance is X then opening Y boxes causes chance to approach Z. but its not a cumulative formula, its a way of figuring that if you do something enough times, how many times would it likely take to get x result.
yes if you open 1 box the chance is the same, if you open 10 boxes the chance is the same, if you open 100 boxes, same, 1000, same, 10,000 same. however the difference is if you open 1 box, its one chance, if you open 10,000 its 10,000 chances. you've changed the number of times you open boxes, which while it doesn't change the chance per box, it does change the number of times say .5% could come up. that's what the formula I was referring to does, it works out, based on prior experiments, or hard numbers, a rough value of how many times something would need to be done before X result should happen, its not a guarantee, just a way of figuring that if I do X this many times I should get Y result at least once.
your still dealing with a RNG in this case though, which isn't truly random, no matter how well its programed.
@justinmathieu
Book? It was a rather short webpage. Here, maybe someone putting it in Math-speak will get the point across:
http://mathforum.org/library/drmath/view/62485.html
Ummm......yeah, I was agreeing with that, and I was saying that cumulative probability agrees with that when applied properly. I think my typo might have caused a misunderstanding, and I got a bit long-winded - sorry about that. I was saying that the person you were arguing with misapplied the theory and got a wrong answer, and so trying to prove that the theory actually agrees that the gambler's fallacy really is a fallacy. I think we're getting some channels crossed here, so never mind.
I saw the webpage, but I also remembered seeing that as a book in my university library, but now that I think back, it was more pamphlet-sized than book-sized. Oh well.
actually having read that they don't invalidate what we are saying.
their coin flip example, if you flipped the coin till you got heads, then flipped it again till you go heads, and again, and so on, and recorded the results. you could work out roughly how many flips before heads will likely happen. Lets arbitrarily say 50 flips, so what that would tell is that for every 50 flips of the coin, 1 flip is likely to come up heads, it could be the first flip, it could be the last flip, but in a set of 50 trys 1 flip should result in a heads. the key there is should, since the odds per flip remain the same, there is a chance even once you have a set that should result in a single heads, that any given set might contain no heads result, just that the vast majority of the sets should contain at least one heads result, some might even contain more then one.