Quote:
Originally Posted by Amy Awien
No, no, no, you were being nasty, if you'd truly believed I didn't know the word you'd have used half a sentence to explain it.
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This is the Internet. One tries to avoid using many words when few will do, especially since there are such excellent resources out there enabling you to resolve the issue yourself.
Quote:
Originally Posted by Amy Awien
No, you ignore the point when you start to argue about 'average' waiting time, where the point is that an extremely long waiting time will have a more impact on perception then the 'average' time. You then claim, without any evidence, that the chance of (extremely) long waiting times is negligible.
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If it's the case that the variance increases but the mean doesn't move much on a skewed distribution like this, then it follows that the chance of experiencing the result out in the tail, many timers in succession, is very small.
You continue to insist that I lack evidence, but we don't need to induct from empirical evidence to solve this problem. We can use logic and a bit of math and deduct the odds of failing to match. You are essentially insisting that we have to do things Darwin's way, but Newton's mechanics and Einstein's relativity are also useful theories, no?
The following is a rigorous proof of my claims:
Start by defining two natural numbers: N and q.
Let's assume that ANet's coders are smart, and that the pairing system always matches every player possible. If that's the case, then every 30 seconds the game creates N games with 8N players in them, and from 0 to 7 players are left out in the cold. (This assumes away the problem of leavers; doing so reduces the hypothetical number of players and improves the result for your side.) Assume that the number of players that does not match is truly random, yielding an equivalent probability of 1/8 for each case. (This should be a truly random process).
If the above is true, then if we implement Lemming's proposal, your odds of drawing a single NOP are given by 1/(8N+3.5), and the odds of drawing q NOPs in sequence are given by 1/(8N+3.5)^q. However, your risk for drawing a single NOP has not changed from the status quo (since you don't know where the timer stands when you click it), so all we are really concerned about is the cases where q>1.
We'll make a final assumption that is extremely charitable to your side. Since we almost never experience an NOP, much less multiple NOPs at present, we can infer that N>0 almost all of the time. The worst case scenario would therefore be N=1.
Given N=1, it follows that the odds of drawing two NOPs are 1/(11.5)^2, the odds of three are 1/(11.5)^3 and so forth. If you run the math, all this really does is increase the chances of a negative result by an order of magnitude as compared to the original math I provided. Your odds of experiencing four NOPs, even given these very charitable assumptions, are therefore worse than your odds of winning a four ball daily lottery in a Northeastern state. Moreover, the average number of additional timers you wait under Lemming's proposal are, as advertised, just under 0.1 when you solve the infinite series with N=1.
So there you have it: rigorous deductive proof of everything I have claimed. Empirical evidence is irrelevant here; all it will do is confirm the assumptions and fill in the proper value of N, which is probably > 1. Increases in the value of N will decrease the risks of long timers very rapidly, since the denominator is exponential.
Conclusion: the risks of long waits under Lemming's proposal are minimal.