Steven K.

[drkitten]

From the most recent challenge applicant:

My precognitive dreamt pair of numbers for Tattslotto draw 2551 on 12/31/05 are 6 & 11.


KRAMER :


May I respectfully draw your attention to the date given there? This claim is well within the
stipulated 12-month limit, and is quite testable. Although it's not nearly improbable enough to qualify as a "preliminary test" under the standard ruling, it's certainly something worth investigating, and probably worth keeping his file open.

[petre]

My precognitive dreamt pair of numbers for Tattslotto draw 2551 on 12/31/05 are 6 & 11. The odds of picking two from 6 numbers drawn from a total of 45 numbers are not astronomical but they are significant (perhaps you can work them out for me and let me know), and after all, I only consider this as a supplementary to my main prediction, the California Big One Quake-Date.

Should the numbers in the New Years Eve games of chance conducted by Tattersall's be completely different from the ones I have nominated (6 & 11),
I would not consider it a knockdown to my claim of foreknowledge as to the date of the California Big One Earthquake because the latter event is predicted on a precognitively more reliable (and different) state of consciousness for me - hypnogogia rather than symbolic dreaming.

The applicant identifies that this prediction is seperate from his claim, even specifying that failure of the first is not evidence enough that he has no abilities. I suggest he submit an application of the same nature on April 4, 2006.

On a side note, I believe the odds of picking 2 out of 6 numbers drawn from a pool of 45 comes to about 1:132. A source I looked at (which I may post later when I hit a post count of 15) indicated the San Andreas Fault may be in a period of increased activity where shocks of 6 to 7 magnitude occur every 10 to 15 years. Stretching that to a guess that means it will average a 7 magnitude earthquake every 30 years, that puts the odds of predicting a correct 1-month period in which it occurs at around 1:360. I'll leave the subjective judgement of whether or not any of this is 'significant' to others.

[Edit: to try and fix my quotes]

[eri]

Is there a set level of significance that applicants must be able to attain? It seems like most of the tests that were performed had an extremely low probability of being won by chance. I know if I'm writing a paper, I need my data to have a significance of 96% or above to be able to claim something is there (i.e., a relationship between two variables), but these people have to be much more accuarate to pass the test. Now I can understand why; personally, if I were Kramer, I would think that 1:132 odds are far too much in the applicants' favor if I were only doing one test. I dismissed the 'girl with x-ray eyes' even though she only had a two percent chance of doing what she did by chance - I suppose, to show they have the power, the test would need to be reproducable. They'd have to get it right with the same odds several times, to prove they didn't just get lucky. But since this person went ahead and said that even if he were wrong, it didn't mean that he DIDN'T have these powers, I'd say he's on the verge of figuring out for himself that he's just been guessing.

[LostAngeles]

If you wish to earn one million dollars for having accurately predicted the immeasurable human tragedy that only events such as a 7.0 earthquake in densely populated urban locales can truly bring about, you must submit your application within 12 months of said death and destruction, give or take a minute.

WTFPWNED. Kramer kicks the ass.

[DevilsAdvocate]

Is there a set level of significance that applicants must be able to attain?Not really. There must be a mutually agreed upon protocol that defines what "success" would be. The probability and significance could depend on the test and how the data will be compiled and analyzed. The Flame Thrower challenge got into this in abundance.

I’ve seen numbers thrown around on the forum that preliminary should be 1:1000 and final should be 1:1000000. Randi has offered challenge protocols that call for 1:1000 to pass a preliminary. But I can’t find anywhere that JREF has published these as a standard.

Kramer has stated that the protocol for the preliminary will be identical to the protocol for the final. I’m not sure if this means that the test procedures remain the same, but the definition of success is changed, or if the definition of success remains the same and there has been some confusion about the final being more difficult than the preliminary.

Anyone got answers?

[petre]

Four hours to go, and I don't see A big one, much less THE big one:

http://quake.usgs.gov/recenteqs/Quakes/quakes.big.html

Perhaps it was a gracious savings of time that he did not reapply.

[Dan O.]

For completeness, I looked up the Tattslotto (http://www.tattersalls.com.au/lotteries2/home/index.html) results for Draw date: 31/12/2005; Draw num: 2551; Numbers: 9, 39, 20, 12, 15, 43; Supplementaries: 25, 14;

[Thabiguy]

My precognitive dreamt pair of numbers for Tattslotto draw 2551 on 12/31/05 are 6 & 11.
For completeness, I looked up the Tattslotto (http://www.tattersalls.com.au/lotteries2/home/index.html) results for Draw date: 31/12/2005; Draw num: 2551; Numbers: 9, 39, 20, 12, 15, 43; Supplementaries: 25, 14;

Wow! I'm amazed. It may seem like a failure at the first sight, but consider this:
There are 6 main drawn numbers. The difference between the supplementaries is 11.

Neat, huh? But wait!

The first drawn number was 9. 9 is just like 6 upside down. And if you add together the main numbers and subtract the supplementaries, you get 99, which is 11 multiplied by 9 - again, the dreaded nine, actually the inverted 6!

Whoa, this is getting scary. But there's more: if you add all the numbers together, you get 177. 17 is 6+11 and 77 is again a multiply of 11.

There are 11 digits in the main drawn numbers. If you add together the digits of all numbers, you get 51. Add the digits again and you get 6.

Draw number 2551 is a prime number. And so is the draw date: 20051231. (Coincidence?) But if you add both together, the result is divisible by 6, and if you then substract all the drawn numbers, the result is divisible by 11! What are the odds?!

I think it is beyond doubt that there is definitely something going on here.

[Kilgore Trout]

You've considered the difference of the supplementaries, but there's more. Added together they are 39. 9 minus 3 is 6. 9 multiplied by 3 is 27 and 2 plus 7 is 9. 39 divided by 3 is 13, clearly an unlucky number. And to cap it off, 39 modulo 11 is 6.

9s are upside down 6s, yes, but I would go further to say they are wrong, or evil, 6s. There is also the occurrence of the unlucky 13. The failure of the prediction is self-evident, caused by the supplementaries.

[Thabiguy]

Oh my God! As of this moment, 3 Apr 2007, 21:20 UTC, both Kilgore Trout and I have the exact same number of posts: 77 !! And look,

77 = 11 * 7, the lucky number!

But as soon as I finish this post, my number of posts will be 78.

78 = 6 * 13, the unlucky number!

AAARGH!! WHAT'S GOING ON HERE?!!!!

[Kilgore Trout]

I am afraid to post this observation because of the implications it may have.

As I write this, it is now my 78th post (and possibly last, if this does indeed cause the vortex of psychic energy I suspect it may). 78+78=156. 156 is also 1+5=(6) and 6. 156 divided by 6 is 26, 26 is also written two-six, clearly alluding to the prior duality of sixes just mentioned. 7+8+7+8=30 and 30 divided by 5 is...yes, indeed, 6.

Be afraid. Be very afraid. The lack of elevens is truly alarming.

[John Jackson]

I believe the odds of picking 2 out of 6 numbers drawn from a pool of 45 comes to about 1:132.


I make it 1 in 6.6 :eek:

6.6 - contains two sixes. The same number of digits as in 11 - a number that looks the same back to front, upside down, in a mirror, even if you're standing on your head.

This can't be meaningless.