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Author: molosztash
Date: 12-03-2008, 07:53

I came up with a difficult math problem just starting my day

What is the probability of at least one all-English clash in CL??

First of all, the number of the draw possibilities is 8!/4! = 1680

Second, the number of the draw possibilities that English clubs never play with each other is 4.4.3.3.2.2.1.1/4! = 4! =24

Then, the probability of at least one all-English clash is (1680-24)/1680= 1656/1680 = 104/105

What do you think, is this true??

Author: Ricardo
Date: 12-03-2008, 09:04

Let's say you have 4 red(English) and 4 blue balls(not English). You pick them 1 by 1 and what are the chances that you pick them in pairs red-blue (or blue-red) ?

1st ball: 8/8 (don't matter if you start with red or blue)
2nd ball: 4/7
3rd ball: 6/6 (doesn't matter again-new pairing)
4th ball: 3/5
5th ball: 4/4 (doesn't matter again)
6th ball: 2/3
7th ball: 2/2
8th ball: 1/1

This comes at a 8/35 chance on getting no all-english clashes That is almost 25%
Can that be true? It's a completely differentnumber from the 1/105???

Author: Ricardo
Date: 12-03-2008, 09:28

I see 1 spot where you go wrong: number of posibilities is not 8!/4! but 8!/(4!*4!) :
as well the English as the non-English can be randomized -> not 1680, but 70 possibilities

Or is the 4! meant for the 4 pairings? What about the Liverpool-Roma and Roma-Liverpool permutations? How are these taken into account

Well I am a bit usty with these statistics- I did these 20 years ago...

BY the way I just found out that MartinW did some calculations on forum2 and came to the same number as me...

Re: Probability of at least one all-English clash
Author: molosztash Date: 12-03-2008, 10:05	that seems true... thanks Ricardo... btw, what happened to your prediction page. the Link in Bert's page doesn't work...

Re: Probability of at least one all-English clash
Author: molosztash Date: 12-03-2008, 10:06 Edited by: molosztash at: 12-03-2008, 10:10	I put the 4! for the 4 pairings, their order do not matter, I thought... and I did not take into Liverpool-Roma Roma-Liverpool changes ( I think We must divide 1680 to 16 for home-away matters for all for clashes, I mean 2^4=16. So there are 105 draw combinations.

Author: Ricardo
Date: 12-03-2008, 10:18

24/104 = 8/35 that would end up the same
I don't know what you tried. the link on the "Access 2008" page seems to be working main page:Ricardo

I noticed I havent written itin the log, but I did copy latest domestic results already to it.
I also noticed that there is something wrong with Irlands representative in CL. I will have a look at it later today

Regarding the prediction for team- & countryranking '08, I had a quick look at Liverpools qualification influence: As Inter was the groupwinner I had them qualifying. Now with Liverpool through England takes the first spot in countryranking and the fight for spot 1 in teamranking is extremely tight between Liverpool, Barcelona, ManUtd and Milan.... I will publish this maybe tonight after 6 out of 8 UC-results or otherwise tomorrow

Re: Probability of at least one all-English clash
Author: amirbachar Date: 12-03-2008, 11:12	It's funny - yesterday I calaculate in in my head. Ricardo is right, it is much easier to do it that way (Conditional probability) and not by the definition (number of possible "bad" draws / number of possible draws)

Re: Probability of at least one all-English clash inCL
Author: Ricardo Date: 12-03-2008, 13:44	@amirbachar Yes it is true, with these numbers it is easier to do it my way. But how is the formula, I mean with bigger numbers it is calculatable, but which formula is to be used for it.....

Author: Mykalobakus
Date: 12-03-2008, 14:57

The number of significantly different draws (i.e. draws which differ with the home teams listed alphabetically) is 8!/4! = 1680.

The number of significantly different pairings (i.e. draws where the teams are listed alphabetically in each match, then the matches listed alphabetically by home team) is 8!/(4! * (2^4)) = 105.

The number of significantly different pairings where English teams are kept apart is 4! = 24.

The probability of English teams being kept apart is therefore 24/105 = 8/35 as stated above by various routes.

The general formula where there are n English teams is therefore
Probability = (2^n)*((n!)^2)/(2n)!

By the way, I was formerly known as Mikalobakus, but it wouldn't let me log in again (:-(

So I changed an i into a y, and registered again. I guess with I,i,Y,y,1 and O,o,0 there are quite a number of obvious aliases, but it would be convenient to keep the same user name - if anybody knows how...?

Author: amirbachar
Date: 12-03-2008, 19:20
Edited by: amirbachar
at: 12-03-2008, 19:28

Generally, in Ricardo's claculation, if there are m English clubs and n clubs in total (if n>=2m of course):
Assuming that the non-English club drawn first (it doesn't really matter). Now the probability to draw an English team is m/(n-1).
Again, let's assume the same thing and now the probabilty to draw an English team is (m-1)/(n-3).
That way until there is one English team left and then the probabilty is 1/(n-(2m-1)).
so the total probabilty is p = m!/((2m-1)*...*(2m-(2m-1)))=
=m!/((n!/(n-2m)!)/((n/2)!/(2^m*(n/2-m)!)))

if n=2m: p = (m!)^2*2^m/(2m)!

(I used the fact that 1*3*...2n-1 = (2n)!/((2^n)*n!) in the proof)

Re: Probability of at least one all-English clash
Author: amirbachar Date: 12-03-2008, 19:30	Nice - we both got the same expression in different ways.