Oblarg

ME2 had the worst C&C I've seen in a modern RPG, bro.

Well, rhythm guitar, at least. That solo is beyond my ability.

There must be a DLC related to Operation: YEEEEEEEEEEEAAAAAAAAAAAAAAAAAAAAH

Konstantin Brayko- 7 Omen Deng- 44 (-1) How can everyone love a character who has such a tiny role? SIE- 44 (+1) Scarlet Lake- 40 Sis- 33 Steven Heck - 23

Good thing I don't go anywhere near "social games."

Really? Oh come on, everyone should know by now that Volourn is just a troll. Don't waste your time.

My Athlon 64x2 3.2ghz and 8800 gts g92 are holding up admirably, given that the video card sits around 85 degrees under load in this ****ty, undercooled case. Not sure when I'll upgrade, I need a job to pay for it first, and I'd rather buy a new guitar than a new computer at the moment.

You're letting the topic be derailed into a completely worthless and irrelevant debate of definitions (which is where the math started getting shoddy) with no clear "correct" answer.

Going to chime in again, because the math is getting shoddy again. Actually, to be fair, that depends on your definition of probability. This is like asking "what is the probability that your shirt is blue?" It's a perfectly defensible view that in that case there is no probability. Your shirt is either blue or it is not blue. The only reasonable way to pose that question is "if I were to take an infinite number of samples of the shirt color of people effectively identical to you, what percentage would have blue shirts?" Or, for a more relevant example, "If I were to take a sample of an infinite number of groups of four lotteries, in what percentage would the four lotteries be won by the same person?" (Now, if you're paying attention, it should be obvious using that example that the probability of the same person winning four lotteries is the same as the probability of any unique sequence of four people winning the respective lotteries, rendering half of this entire ****ing discussion useless.) You're focusing way too much on that little tidbit of wording, which is a completely irrelevant argument of definitions which changes absolutely nothing about the actual problem, instead of his broader mistakes of "highly improbable events cannot happen" or "I believe an independent event can happen once but not four times."

Here's a *really* easy way to think of it. Same problem, different numbers There are 100 doors. One door has the prize. You are allowed to pick one door. Then, Monty Hall reveals 98 doors that have no prize, leaving one other door. You can either switch or not switch. Now it should be very obvious that you should switch doors. This is the exact same situation.

Not saying it was. Just a word of caution so we don't end up making that same mistake later in the thread. Also, the math nerd in me couldn't resist posting that problem.

Did you just take one of the Arts Faculty's probability courses? Because this statement is wrong in any statistics course, even if it's a course from the school that allegedly invented nanotechnology (which school was this again?) Winning a lottery is an independent event of winning future lotteries. The chances of someone winning with a lottery ticket is not affected by whether or not they have previously won a lottery ticket. I think (or at least hope) that we can all agree on this). That is, for any given lottery, the chances any one individual has of winning it is not affected in any way by that person winning a lottery previously. So lets set this up: P( A | B ) A is the probably of winning two more lotteries B is the probably of winning two lotteries in the past. Since A is independent of B, we know that P( A | B ) = P( A ). Assuming the lotteries in A have equivalent chance to win as the past lotteries in B, then the P( A ) = P( B ). Therefore, since the probability of winning two lotteries in the past is "not surprising" (Dagon's words), then the probability of winning two lotteries in the future must be "not surprising." In order to spell it out a little clearer, lets use coin flips. The probability of the first two coin flips being heads is 0.25 [This is pretty trivial: 0.5 * 0.5 = 0.25] The probability of having two more coin flips that are heads having already had two coin flips that were heads is, I hope is obvious, also 0.25. Because you've already gotten two heads coin flips, the probability of you getting another 2 is irrelevant of what you got previously. Though straight up, the probability of getting 4 heads in a row is 0.0625. People often assume that because the odds of getting heads 4 times in a row is low, the odds of getting that last heads when you've already gotten three is somehow lower. The thing I find very interesting, is that this very concept was what many second year CompSci students struggled with when they took this course at my University (Stats 221 - In fact, Dr. Kouritzin was my Stats 221 teacher back in 2004... he actually worked for Lockheed Martin for a while too /coolfact). For most of us it was our first exposure to random variables and probability. Admittedly, you have to be careful as you can make mistakes the other way, too. Example: I flip two coins. One of the coinflips resulted in heads. What is the probability that the other coinflip resulted in heads? Well, obviously the two coinflips are independent, so many would jump to the conclusion that the probability is 1/2. Unfortunately, that's wrong. Let's set up the problem: P(H1) = probability of the first coin being heads P(H2) = probability of the second coin being heads Now, the incorrect assumption is that we're asking P(H1 and H2|H1) which obviously = P(H2) = 1/2. However, upon closer inspection, it's clear that we're actually asking P(H1 and H2|H1 or H2) = P(H1 and H2 and H1 or H2)/P(H1 or H2) = P(H1 and H2)/P(H1 or H2) = .25/.75 = 1/3.

Going to chime in because I'm a math nerd and the math here is getting really shoddy. An infinitesimal amount is, indeed, rigorously equal to zero. The chance here, however, was not infinitesimal. It was simply very, very small. That said, while it certainly isn't impossible for this to happen, a much better explanation is that there are lurking variables (cheating, for instance) that resulted in not-so-improbable odds of winning. Edit: Upon rereading the thread, I don't think WoD quite understands the concept of infinity. An infinitesimal chance is infinitely less likely to happen than the chance of this happening. Infinity is really, really big. Infinitesimal is not a term you throw around lightly.

That it's a scam. But if you're dead set on not believing that, it even makes more sense to just accept that she was incredibly lucky. You have not presented us with something that actually violates our current understanding of statistics. And, here's the important part: Even if you did, the correct response would be to rethink our current model of statistics than to jump to "god did it." The line of thinking that leads to those large leaps of "we don't know how this works, ergo god" is hopelessly infantile and ultimately detrimental to the betterment of human knowledge. Now, I'm half-convinced that you're just trolling here, so this is my last post in this topic.

Should it make me feel bad, or something? The idea that this (if it is not at a scam) is at all a valid argument in favor of the supernatural is overwhelmingly stupid.

What ****ing terrible graphics. You know, if Alpha Protocol can be ripped for graphics when it looked better than DA:O, this better be completely slaughtered for looking worse than DA:O.

Luck: favorable results given unfavorable odds. Extreme luck: Favorable results given extremely unfavorable odds. It happens. It does not happen often, but mathematics dictates it *must* happen at some point if the sample size is large enough. Regardless, it most likely is a scam. Oh, and that article was idiotic: Anyone who passed high-school statistics can tell you that the only way to win a negative EV game in the long run is to not play.

You left out the possibility of extreme luck. It happens. The chance is nonzero. Still, more likely it's some sort of scam.

Did anyone hear the national anthem for the all-star game? It was terrible. Terrible.

That would be great if the tunnels weren't filled to the brim with swarms of enemies in unpredictable places and if the combat weren't so ****ing clunky.

Vicious Rumors - Abandoned I swear, Carl Albert sounds exactly like Bobby Blitz in this song. I'd probably think this is an Overkill song if I didn't know otherwise, it's so similar.

I care more about whether narrative is good than whether it is reactive. DA:O's narrative was bad. It was bland, uninspired, and vapid, with utterly no character development from the blank slate you started as. The influence system makes your companions seem shallow and robs them of credibility (oh, you gave me a present, let me tell you about my life!). I'd gladly be shoehorned into one character if the result is a better story. And they really have to rework the encounters, because I don't want any more "turn corner -> get swarmed by enemies -> reload save -> plan and then turn corner, killing enemies -> turn next corner, swarmed by enemies again, repeat." It was a chore to get through DA:O's levels.

They could have cut the number of enemies to a third of what it was an have it still be rather grating at points. The level design was absolutely horrendous in DA:O.

No, certainly the tree cluttered by completely worthless sustained abilities is not a clunky tree. No way. Oh, and arrow of death and scattershot are in no way overbudget.