RE: Euro 2013 Group C Permutations

Dear Reader,

I’ve been doing some maths, and I’ve stagnated  myself into an unfavourable position. To any mathematicians out there, I’m requesting assistance!

I have a finite set, S, where Ew=England Win, El=England Loss, Ed=England Draw, Sw=Spain Win, Sl=Spain loss, and Sd=Spain draw.


Ex can only map with Sx, where x=outcome of the match

So far, my 9C2 = 9!/2!(9-2)! = 32 (9 permutations, out of a size of 2, gives rise to 32 possible combinations).

But this is a wrong calculation, since ExSx is being mapped to EySy (ExSx*EySy).

Which would look like this:

England win, Spain win AND England loss, Spain loss. (EwSw*ElSl)

I would even get final outcomes like this, for example:

England win, England lose, AND England draw, Spain draw. (EwEl*EdSd)

Ignoring some ideas of quantum mechanics, this is impossible outcome in reality.

ExSx is being treated as an independent event, when in fact, it is a final outcome.

So my question is: How on earth do I make it, so Ex can only map with Sx, where Ex and Sx are independent events that occur simultaneously into a final outcome (Ex*Sx)?


OK. So I’ve worked it out.

To map Ex with Sx, the formula would be indeed p(n,k)+n=n^2 just as I speculated. I’ve tested it on large and small sample sets, and it all works.

Screen Shot 2013-07-16 at 14.00.25

BUT! This only works if k=2

For larger/smaller k, I’m trying to think of another formula… or a general formula. But I’m working on it.

p(n,k)+n=n^2 probably exists in some form already… my ignorance of statistics and mathematics prevents me from claiming this as my own. But if so, I’m taking it. <INSERT LAST NAME>’s law, anyone? I jest. This is laughably easy, and dear Maths friends, do feel free to laugh. I know I would too. But I find this incredibly fun.

You should too.

Yours truly,

A Combinatorialist