The calculation is a bit of a pain to do by hand so the easiest way is to write a short computer programs (4 lines will do it) that cycles through each of the 40320 possible ways of matching the runners up to group winners, then eliminating those draws that match teams from the same group of country.

]]>PSG S04 Mal BVB Tur FCB Bar Man

Por – 11,6 19,0 12,4 13,3 12,4 18,3 13,1

Ars 13,1 – 22,2 14,2 15,1 14,2 21,3 –

Mil 14,5 14,7 – 15,5 – 15,5 23,2 16,6

Mad 18,4 18,8 – – 21,8 19,5 – 21,5

Don 11,7 11,8 19,3 12,6 – 12,6 18,7 13,3

Val 18,4 18,8 – 19,5 21,8 – – 21,5

Cel 12,4 12,5 20,1 13,2 14,5 13,2 – 14,1

Gal 11,6 11,8 19,4 12,6 13,5 12,6 18,6 –

if you put question like this: what are the odds pairs would be same as drawn yesterday (say fixed) it would be at maximum 1/7*1/6*1/5*1/4*1/3*1/2 = 1/5040. and much less with the constraints = 1/528. and finally, order makes it easier to repeat. did you know that after first 4 pairs were drawn there was just two possible options, with two of four pairs fixed for sure

]]>Oh, brilliant!! Great idea! You are, sir, a Messi of puns. That is just so neat and elegant. ]]>

Maybe you should rename the website “Who Ate All The 3.14159s?”

]]>1. not same group

2. not same country

3. group winners in same pot, group 2nd in one pot

Let’s say they draw the 2nd team from a group first. There will be these possible combinations:

Valencia 5 (not the other 2 Spanish group winners and not Bayern)

Madrid 5 (not the other 2 Spanish group winners and not Dortmund)

Milan 6 (not Juventus and not Malaga)

Arsenal 6 (not Manchester and not Schalke)

Donetsk 7 (not Juventus)

Porto 7 (not Paris)

Celtic 7 (not Barcelona)

Galatasaray 7 (not Manchester)

There are 50 possible combinations. That means the chance for 1 combination is 1/50 (0.02) which is 2%.

Draw the same combination twice:

0.02×0.02=0.0004 which is 0.04%

That is my guess….

]]>Any football related organisation whose acronym contains the letters FA is definitely corrupt.

Shame really. ]]>

But I’m afraid the only one that gets a coconut is Hendo in post 3 at 12:28pm. At least he got his response in while I was typing that I’d realised what was wrong myself.

I think the joint worst responses have to be David (#8) “The way you have presented the odds is disingenuous.”, and Garnuzzi (#9) “Bangolero, you’re calculation is way off. “. Well, in the first case, I wasn’t being in the least disingenuous, there was no deceipt, I wasn’t trying to trick anyone, I was genuinely puzzled. And in the second case, I think saying I was way off is how I began the whole thread. (BTW: I think it should be “your” calculation, not “you’re” calculation! Or what’s the odds on me being as bad at punctuation as I am at statistics?)

So, I take it no-one had any views on the football and that maths is much more interesting than the Champions League? I bet we’re all really holding our breath for the tie of the round, Porto v. Malaga.

]]>“It’s almost as if it was meant to be that way, and that someone, perhaps a mysterious egomaniacal little French dwarf, had ordained it this way. Such powers are not given to us ordinary mortals”< ha! ]]>

Compared to the usual pseudonymous GUNNAR going “YEAH RVP IS GONE BUT WE GOT SANTIR CARZOLE!”

]]>As guys above have said, you have 16 teams split into two pots, so thats a 1 in 8 chance. Take into account 4 Spanish, 2 Italian and 2 English teams can’t play each other.

Barca couldn’t play Valencia or Real Madrid, so the rest if the teams (Porto, Arsenal, Milan, Shaktar, Celtic and Gala) all had a 1 in 6 chance of being drawn against them. Same with Malaga.

Then, taking into account the odds after each tie has been drawn, they reduce dramatically (ie: By the time Arsenal had been drawn, they couldn’t play Shalke 04 or Juve or indeed Man United, so down to 1 in 5)

My head hurts.

]]>But obviously all that doesn’t matter, because ESPN forgot about all the regulations that are already mentioned here. I read somewhere that the actual odds are about 2100.

]]>1) I believe there are constraints in this draw, that teams from the same country can not be drawn to face each other. If so, we need two buckets where all teams from one country are in the same one (ie 4 ESP, 2 GER, 1 SCO, 1 TUR vs. 3 ENG, 2 ITA, 1 POR, 1 FRA, 1 UKR). This reduces the number of possible combinations.

2) We’re only looking for the chance to copy one specific set of 8 pairings (the one from the rehearsal), regardless of order the pairings or the teams (home/away) are drawn.

3) This means that the probability of drawing the team from the first bucket is always 100% – we will draw one of the teams from that bucket. The probabilities of combining the first team with the same as in the rehearsal are 1/8, for second team 1/7, third 1/6 etc.

4) The combined “odds” for this calculation is 1 in (8*7*6*5*4*3*2*1) which is 1/40320. Not very high, but much more probable than 1 in 10 billion. :)

Please note that I’m not sure about the rules for this CL draw, I just try to explain the math behind.

]]>The problem with your calculation is that you have assumed the draw is completely random. You need to consider that a group winner can only play a group runner up so at first there is a 1 in 8 chance. Also, teams from the same country cannot face each other in this round. Hence, it is quite a complicated calculation!

]]>It’s amazing how a few minutes resting in a dark room can clear your thoughts.

I’ve just realised the error of my ways. Of course, this is UEFA. You never have truly random draws the way I was calculating. It’s all controlled by so many restrictions and constraints. So a team that finished 1st in the league phase can only be drawn against a team who finished 2nd. So that immediately reduces the first odds from 1-in-16 down to 1-in-8. And do they still prevent teams from the same country being drawn against one another? That’ll reduce the odds even further with 4 Spanish, 3 Germans, 2 Italians and 2 English.

Blimey, when you start to take these things in to account, it was almost inevitable the draw would come out the same way twice. In fact, there’s probably on one way it can be drawn. It’s almost as if it was meant to be that way, and that someone, perhaps a mysterious egomaniacal little French dwarf, had ordained it this way. Such powers are not given to us ordinary mortals. We just fumble around with odds of billions to one against anything going right! ]]>

I’ve got 16 balls (no, no, don’t, no, keep quiet at the back there, no, missus, please) so the odds on any one team being picked first is (wait for it) 1-in-16. And the second team is now 1-in-15. So that particular pairing has odds of 1-in-(16×15) or 1-in-240. Now of course, pairing Team A with team B is exactly the same as pairing Team B with Team A – it doesn’t matter which comes out first – so the odds are halved to 1-in-120.

Repeat the exercise for the remaining 14 teams and you have 1-in-(14*13)/2 or 1-in-91 for the second pairing.

And so on with odds of 66, 45, 28, 15 and 6 for the next pairings. By the time you get to 2 teams left, there’s no doubt any longer (i.e. 1-in-(2×1)/2 = 1) so we can exclude them.

Multiply together all these odds of 120 x 91 x 66 x 45 x 28 x 15 x 6 and you get a combined accumulator of nearly 82 billion to 1 against. Even if we don’t care which actual tie Team A and Team B get paired in (i.e. the first at 120-to-1 or the last at 6-to-1) that only reduces the odds by a factor of 7 or 8 (I haven’t got that one quite worked out yet!)

So I still think the odds are somewhere between 10.2 and 11.7 billion to 1.

Where have I gone wrong? ]]>