re:the draw
Re: re:the draw
No just using up the station names for a bit of fun first person to guess the station name drawn out of the hat on june 1st out of 57 gets 10 points for example.the same for each day until 30th.
- hwolge
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Re: re:the draw
Does this imply that you won't reveal the content of the 56 left over eggs on the day? I smell potential manipulation here...
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Re: re:the draw
I could empty all of the 57 eggs that were not not used to ensure a fair draw was carried afterwards i suppose.
- hwolge
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Re: re:the draw
I guess offering it would in reality be good enough.... After all, we'd probably trust you anyway!
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Re: re:the draw
Time for an independent adjudicator methinks ...
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Re: re:the draw
Did I say that?
One thing only do I know, and that is that I know nothing - Socrates.
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Re: re:the draw
Not in words no but implied independant adjudicator means u cant trust me on merit alone.
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Re: re:the draw
It doesn't necessarily mean that. Most independent adjucidators are appointed to ensure that fair play is seen to be done, not because people think that foul play WILL be done.
Anyway, it doesn't really matter what I think because I'm not involved in any of this.
Anyway, it doesn't really matter what I think because I'm not involved in any of this.
One thing only do I know, and that is that I know nothing - Socrates.
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Re: re:the draw
This is a link to a random drawing of the kinder eggs to prove said facts.http://www.youtube.com/watch?v=sNMuowqaa1A
- hwolge
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Re: re:the draw
Great video, Andi!
After watching it, the matematician/statistician/drawer in me makes a few observations:
* As a "proof" of all eggs holding different stations, I think the video falls short. Obviously, lots of manipulation can take place afterwards...
* Anyone dare to guess the maximum number of eggs one could draw on random (putting them back afterwards) before the accumulated risk of hitting the same one twice goes above 50%? (Not putting them back immediately would of course had been a better method anyway). Maybe there was a retake even?
* Do I have to be blindfolded since I now know the colour of 3 of the eggs of three of the stations? Or will there be some other obfuscation method?
Regardless, I'm happily looking forward to executing my randomising duty next Friday!
After watching it, the matematician/statistician/drawer in me makes a few observations:
* As a "proof" of all eggs holding different stations, I think the video falls short. Obviously, lots of manipulation can take place afterwards...
* Anyone dare to guess the maximum number of eggs one could draw on random (putting them back afterwards) before the accumulated risk of hitting the same one twice goes above 50%? (Not putting them back immediately would of course had been a better method anyway). Maybe there was a retake even?
* Do I have to be blindfolded since I now know the colour of 3 of the eggs of three of the stations? Or will there be some other obfuscation method?
Regardless, I'm happily looking forward to executing my randomising duty next Friday!
Three times Zone 1 Challenge winner
Official record holder in the 2008 Guinness Book of Records, pg 199
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Re: re:the draw
Andi, you are going to have to paint all the egg shells the same colour now......
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Re: re:the draw
It was just to prove that there were different names in each capsule and my father did change them afterwards does anybody have a velvet bag.?
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Re: re:the draw
42 draws I believe for 57 eggs.hwolge wrote:* Anyone dare to guess the maximum number of eggs one could draw on random (putting them back afterwards) before the accumulated risk of hitting the same one twice goes above 50%? (Not putting them back immediately would of course had been a better method anyway). Maybe there was a retake even?
Re: re:the draw
I'm sure it's much quicker than that. I have the probability falling below 50% on the 10th egg.42 draws I believe for 57 eggs.
p(all unique after n draws) = (57/57)*(56/57)*(55/57)*(54/57) ..... *((58-n)/57)
p(1) = 1.000
p(2) = 0.982
P(3) = 0.948
p(4) = 0.898
p(5) = 0.835
p(6) = 0.762
p(7) = 0.682
p(8) = 0.598
p(9) = 0.514
p(10) = 0.433
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