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#320 |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22×3×17×23 Posts |
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#321 |
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Aug 2002
North San Diego County
68510 Posts |
Yeah, I gave up on P-1 testing because of all of the crashes after finding factors. Too much babysitting.
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#322 |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22×3×17×23 Posts |
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#323 |
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6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
981810 Posts |
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#324 | |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22·3·17·23 Posts |
Quote:
4.4 ponits over 2 days for a E6550 = 2.2 per day per core Pretty much exactly what it gets for DC or LL. |
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#325 | |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
22·3·641 Posts |
Quote:
http://mathworld.wolfram.com/StrongL...llNumbers.html |
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#326 | |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts |
Quote:
The Strong Law of Small Numbers is talking about small numbers (or sequences of them) appearing in more places than you might expect because there's so few of them. I don't see how that applies...unless you meant some sort of related "law" regarding the statistics of a small sample size and how that can produce "unexpected" results at a relatively high rate.
Last fiddled with by Mini-Geek on 2010-03-13 at 01:28 |
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#327 |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22·3·17·23 Posts |
For what stats are worth....
I've had 10 PC's that have done 12 P-1 tests in the last 365 days (the lowest is 12 tests). The success rates are: (I understand that the smaller the sample size the greater the chance of being away from the expected value) 2/25 = 8.0% 1/25 = 4.0% 3/49 = 6.1% 1/27 = 3.7% 4/58 = 6.9% 2/12 = 16.7% 8/88 = 9.1% (this is the one that is 6/31 recently) 4/60 = 6.7% 1/34 = 2.9% 4/61 = 6.6% =========== 30/439 = 6.8% 34/471 = 7.2% LIFETIME Guess I'm within the statistical norms... |
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#328 | |
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"Lucan"
Dec 2006
England
2×3×13×83 Posts |
Quote:
used the binomial distribution with n=30 and p=0.066. To get a "feel" for the likelihood of the actual number of successes when the expected number is 2, I resort to the Poisson distribution whenever it is approximately applicable: Probability of 0,1,2,3 or 4 successes is (1 + 2 + 2 + 4/3 + 2/3) / e2 = 0.95 David Noting also that the standard deviation is sqrt(2). Last fiddled with by davieddy on 2010-03-13 at 08:03 |
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#329 |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22×3×17×23 Posts |
My P-1 assignments jumped from 51.8M to 53.0M recently.
At the same time about 30,000 TF Available exponents appeared in the 51-52M ranges. I'm going to take a guess that the 51-52M range was released to do the final bit of TF. I'm going to further guess that these assignments will then be assigned the old (v4 way) letting the tester do P-1 and then LL. Finally, I'm going to guess this is to give the P-1 workers a better chance to stay ahead of the LL testers. Anyone know the real facts? |
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#330 |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
170148 Posts |
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