Durban Skeptics in the Pub December 2012
Location: Pizzetta Pizzaria
139 Davenport Road
Glenwood, Durban
Thursday 20 December
18h00 - whenever!
We will be doing a Secret Santa for this, the last SITP of 2012.
Buy anything up to the value of R30 (it can be anything, but try and make it gender neutral, I have not found much use for the home douche kit I got last time), wrap this item/s in an envelope / box and include an impressive skeptical quote with your gift.
Now, and this is important, make sure that only an even number of you come to the event. The space time continuum will be seriously out of whack if an odd number of guests arrive, and we d
on’t want to happen what happened the last time that happened… erm, yes.
“I don’t believe in astrology; I’m a Sagittarius and we’re skeptical.”
Arthur C. Clarke
There will be the usual skeptical discussion - snarky reparte and heinous blasphemy are both encouraged as heavily weighted ingredients in the proceedings.
Must admit I’m also a bit stumped by the insistence on an even number of attendees. Unless only one person shows up, you obviously don’t need an even number of people to ensure that each leaves with a present other than the one he brought himself. So I assume there is a team debate or a dance afterwards?
Rigil
Well, I knew this was going to eventually bite me in the ass - blame it on my Liberal Arts education, my math is dodgy - thank goodness I stick to organising shit around here, and leave the serious logic up to others.
A prompt reply I got from a Philosophy professor friend of mine shortly after posting the event on FB:
David Spurrett
Dude, you don’t need an even number.
Like · · Unfollow post · 5 December 2012 at 22:01
Clinton Armitage What do you mean?
5 December 2012 at 22:03 · Like
Clinton Armitage Relax man, I was kidding about the space time continuum.
5 December 2012 at 22:04 · Like
Clinton Armitage Um, if it is an odd number, then someone will have a gift to give but will not receive one in return? Or is my math dodgy
5 December 2012 at 22:04 · Like
David Spurrett For any integer number of people greater than one there's at least one way of everyone getting a gift they didn't arrive with. (Put them in a ring, and get everyone to pass one to the left to find the unique solution for two people, and one of the options for three. The number of viable allocations rises as you add gift-carrying people, but doesn't require an even number unless you impose additional constraints. And if folks randomly get gifts one at a time, then you can end up with someone faced with the one they bought irrespective of whether the starting number is odd or even.)
5 December 2012 at 22:34 · Like · 2
Clinton Armitage Blinded by science! Thanks for the mind bender David. This may need some processing to digest.
6 December 2012 at 07:50 · Like · 1
Shereen Mann Dodgy math strikes again
6 December 2012 at 09:21 · Unlike · 1
David Spurrett Just think of folks sitting in a ring and all passing a gift to the left. That shows there's always at least one solution for two or more people.
6 December 2012 at 21:34 · Unlike · 2
Clinton Armitage Holy crap I get it. The ring idea is the key - I was stuck thinking about multiples of pairs of people. Thanks David, I appreciate the info. Logic, it works.
8 December 2012 at 08:12 via mobile · Like
David Spurrett That's why we philosophers get paid so much ... er ... um.
9 December 2012 at 23:10 via · Unlike · 1
Hey, when you don’t know, you don’t know…
;D