tubechallenge.com

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.

Does this imply that you won't reveal the content of the 56 left over eggs on the day? I smell potential manipulation here...

I could empty all of the 57 eggs that were not not used to ensure a fair draw was carried afterwards i suppose.

I guess offering it would in reality be good enough.... After all, we'd probably trust you anyway!

Time for an independent adjudicator methinks ...

Sounds like u dont trust me Mr Tubeguru

Did I say that?

Not in words no but implied independant adjudicator means u cant trust me on merit alone.

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.

This is a link to a random drawing of the kinder eggs to prove said facts.http://www.youtube.com/watch?v=sNMuowqaa1A

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!

Andi, you are going to have to paint all the egg shells the same colour now......

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.?

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?

42 draws I believe for 57 eggs.

42 draws I believe for 57 eggs.

I'm sure it's much quicker than that. I have the probability falling below 50% on the 10th egg.

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