jaysc wrote:yes yes... very interestiong now can you put that in simple ?

Okay, let's start very basic. We are going to assume that a person's chance of getting into the beta is a uniform random distribution. This means that every person has exactly the same chance of getting in. This is the kind of distribution that we would use to model the flipping of a coin (heads and tails are equally likely), the rolling of a die (each number is equally likely to turn up), or the drawning of a name out of a hat (every name is just as likely to be drawn as any other).

In a uniform random distribution, the probability (P) of a specific event occuring (x) is 1 divided by the total possible number of events (n):

P(x) = 1/n

For instance, when you roll a six sided die, there are six possible outcomes (you could roll a 1, 2, 3, 4, 5, or 6). Each outcome is equally likely. Therefore, the chance of rolling any particular number (a 6, for instance) is equal to 1 divided by the total number of possible outcomes (in this case, six possible outcomes). Thus, the probability of rolling a 6 is 1/6, about 17 out of 100.

In the case of getting into the beta, there is one other factor that we must keep in mind -- we are not looking for a single outcome from one trial, but a single outcome from 100 trials. We are not drawing one name out of a hat, but 100. Furthermore, we are not using replacement -- i.e. once a name is drawn from the hat, it is not put back in the hat. Thus, if there were 10 names in the hat, you would have a 1/10 chance of being drawn with the first pick, and a 1/9 chance of being drawn second (assuming you were not drawn in the first pick. However, for sample spaces that are large enough, this is largely irrelevant. We will model this probability with replacement, just to make the math easier.

So, we now have something that looks like this: we have n people that have applied for the beta, and we want to know what the probability of a specific person being drawn is. We are going to draw 100 names, and we are going to replace them (we are not actually going to replace them, but, because there are so many names in the hat, the difference is minimal). The probability of a specific person being drawn in the first round (x1) is 1/n:

P(x1) = 1/n

The probability of the person being drawn in the second round (x2) is 1/n:

P(x2) = 1/n

And so on:

P(x3) = P(x4) = ... = P(x100) = 1/n

Again, as stated above, we can do this because we are replacing the names (or, in actuallity, because the sample space is large enough that we can assume that it behaves as though we are replacing names -- in fact, there is some minor error introduced by this assumption, but it is not important), and each event is independent; i.e. the result of the first draw will have no effect on the next draw, or any subsequent draw.

Now, to find the total probability of being picked (P(X)), we can simple add the probabilities together:

P(X) = P(x1) + P(x2) + ... + P(x100) = 1/n + 1/n + ... + 1/n = 100/n

So, assuming that a person has applied to beta test, the probability of being picked is 100/n, where n is the total number of people who have applied.

So, that brings us to Phrage's chances of getting into the beta:

Generally speaking, his chance is 100/n. The small n is, the better his chances. Let us say that n is effected by the way in which the betas were advertised. If they are simply advertised on the IV news forum, n1 people will apply. If they are advertised on the IV news forum and on the Defcon forum, we can assume that the same number of people will apply from the news forum (n1), but that more people will apply from the Defcon forum (n2). Thus, we get something like this:

Total number of applicants:

News forum only: n1

News and Defcon forums: n1+n2

Now, remember how I said that the chance of getting picked was inversly proportional to the number of people who applied (i.e. as n gets bigger, the chance of being selected gets smaller)? If the beta had been announced on both forums, more people would have applied, and any particular person's likelyhood of being selected would be smaller.

Now, I misread NeoThermic's post, and thought that he was implying that Phrage's chances would be better if he had not applied at all (I missed the part about him not seeing the announcement at all). I thought that his statement was based on this fact: if Phrage had applied, one more person would have applied, and everyone else's chances would be decreased, including Phrage's:

P(x0) = 100/n (before Phrage applies)

P(x1) = 100/(n+1) (after Phrage applies -- there is one more person to draw)

P(x0) > P(x1)

Thus, with one fewer person applying, the probability of being drawn is higher, if only slightly so. However, the problem with that is that the probability of being drawn if you haven't applied is 0 (i.e. it is impossible to draw a particular name out of a hat if it hasn't been put into the hat in the first place).

So, my reply we in response to misunderstanding what NeoThermic was saying (and, to some extent, assuming that he was an idiot, for which I am deeply sorry).

So, basically, we have three different scenerios:

1) betas are announced on IV News section only, and Phrage applies

2) betas are announced on IV News and Defcon news sections, and Phrage applies

3) Phrage does not apply.

Phrage's best chance to get into the beta would have been case (1), where n is small. His next best shot would have been case (2), but this would have been much less likely, as Phrage is likely not the only person who did not check the IV News section. Phrage's worst chance to get into the beta is what happened -- he didn't apply.

I still maintain the pedant that Phrage's chances would have been better if the betas had been announced in the Defcon section, but only because any probability, however small, is greater than zero. Thus while his chances may have been better in case (1), he didn't apply because he didn't see it, and the probability of being selected in case (2) is still greater than case (3).

xander