Maths (sports) question....

[Dauphin]

6 is nonsense. I was working with 3/8 * 2/7 and my mental faculties got harangued.

[b etor]

yea that's what i did the first time.

[Wernazuma III]

Originally posted by Dauphin


You made a mistake.

There are 28 possible finals, and 3 of them have all English clubs.

Yeah, you're right. Adding 6+1+1 gives 7, when you're me.

[MRT144]

Originally posted by KrazyHorse
Eat the peanuts out of my sheeeittt, pilgrim.

So this is the wisdom you were talking about?

[Solly]

It's still what I said earlier, 19/42

[Drogue]

Originally posted by Krill
I wonder if we can get 3 English teams into the semis...

Originally posted by reds4ever
...or even the final!

Am I the only one who burst out laughing at this?

[KrazyHorse]

Originally posted by Doddler
It's still what I said earlier, 19/42

What, the answer to the first question?

No, it's not...

List the teams in some permuted order. Pair up the teams in positions 1&2, 3&4 etc
The number of permutations of 8 teams is 8! = 40320

However, the pairings have some symmetries associated with them. Flipping team 1 with team 2 does not have an effect on the actual pairings. Therefore you first divide by 2^4 as a symmetry factor

The next symmetry is on any permutation of the 4 pairs. i.e. 12345678 is the same as 56347812. This next symmetry gives a factor of 4! = 24

So the number of distinct sets of pairings is 8!/(2^4 * 4!) = 105

Your answer is thus ruled out (there does not exist n an integer s.t. n/105 = 19/42)

[Krill]

nope Drogue

I just wasn't going to alert reds to the mistake so more could get a laugn out of it...

[KrazyHorse]

Originally posted by Wernazuma III


A quick calculation, taking that all teams have equal chances to go up, gives 0,125 (or 7/56) as chance for an all English final.

Who needs KH?

(And now let's hope I didn't make a mistake)

You did, however.

Assuming all teams have an equal chance of winning any game against any other team, the probability of an all-England final given 3 English teams in a field of 8 is 3/28

The probability of having a semifinal round with all 3 English teams in it is 1/14

The probability of having a semifinal round with exactly 2 English teams in it is 3/7

[Wernazuma III]

Originally posted by KrazyHorse
You did, however.

Which has already been pointed out. I forgot adding a 1/56 I had on my paper, which gave me 6+1+1=7
My calculation was a bit too quick, then, it seems.

[MRT144]

KH, show your work so we can end the idiocy once and for all.

[LordShiva]

Liverpool has the easiest game

But once Milan beats Bayern, they'll have a smooth ride to their 7th championship

[Solly]

Bascially it is more likely that two English teams played each other than didn't. KH is just playing the fool and spouting nonsense.

[Solly]

Originally posted by LordShiva
Liverpool has the easiest game

It will turn, for them, into the most difficult Can the Kop get revved up for PSV, I doubt it. Only Barcelona and John O'Shea get it going these days

[KrazyHorse]

Originally posted by Doddler
Bascially it is more likely that two English teams played each other than didn't. KH is just playing the fool and spouting nonsense.

Sure, Stewie. Whatever you say...