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Champions League Last 16 Draw: Actual Fixtures Exactly Same As Rehearsal, Odds Of That Happening About ’1 In 2 Million’

By Chris Wright

What you’re looking at there is actually a Sky Sports News graphic from yesterday when UEFA conducted their rehearsal for the Champions League last 16 draw though, by some bizarre twist of fate/rigging, today’s actual draw has produced exactly the same ties!


The actual draw is as follows…

Galatasaray vs Schalke
Celtic vs Juventus
Arsenal vs Bayern Munich
Shakthar Donetsk vs Borussia Dortmund
AC Milan vs Barcelona
Real Madrid vs Manchester United
Valencia vs PSG

Porto vs Malaga

Amazing. According to an ESPN statistician, the raw odds of getting exactly the same Last 16 draw twice in a row are about 0.00005%, which equates roughly as a ’1 in 2 million’ chance.

As for the draw itself, there are some choice ties but the British teams haven’t fared well – though, to be fair, there wouldn’t have been any particularly kind draws whichever way the balls had dropped (so to speak). That’s the nature of the beast after all: Europe’s creme de la creme.

In all honesty, we can see all three going out with Real Madrid, Bayern Munich and Juventus all looking that bit stronger than Man Utd, Arsenal and Celtic respectively.

Any thoughts Pies fans?

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By Chris on December 20th, 2012 in Champions League. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

30 Responses to “Champions League Last 16 Draw: Actual Fixtures Exactly Same As Rehearsal, Odds Of That Happening About ’1 In 2 Million’”

  1. Bangolero says:

    I don’t understand the odds calculations at all. I’ll concede I’m no ‘ESPN statistician’ but here’s my thinking …
    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?

  2. Hendo says:

    You are forgetting that we know the 8 home teams and 8 away teams already plus teams are kept apart from teams from the same country or from their same qualification group. So a team like Real Madrid could only ever be drawn against 5 different teams.

  3. Bangolero says:

    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!

  4. Norwich fan says:


    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!

  5. samio says:

    @Bangolero: Well, one thing that immediately comes to mind is that your calculations ignore seedings and other restrictions on possible match pairs. Any one team has at most 7 possible opponents (#1s or #2s from all the other groups) and in many cases even less, because teams from the same country can not be drawn together at this stage.

  6. DIF Sthlm says:

    @Bangolero: Without having seen the actual draw, I think I can correct your calculations in some areas. What you have calculated is the number of possible orders to draw 16 different teams (or marbles or whatever).

    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.

  7. David says:

    The way you have presented the odds is disingenuous. It is the odds of getting the draw that came out twice in a row and not the odds of getting any draw the same twice in a row. It is like shooting a gun and then drawing a target around the bullet-hole and saying what a good shot you are. The real odds of getting two of the same draws in a row is in the order of 1 in thousands.

  8. Garnuzzi says:

    The ESPN guy would be right if the draw was totally random: 15 x 13 x 11 x 9 x 7 x 5 x 3 (x 1) = 2,027,025. Bangolero, you’re calculation is way off. If the draw was totally random the first team could be paired with 15 different teams, the next team could only be paired with 13 different teams, the next team with 11 teams and so on.

    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.

  9. FeintZebra says:

    Its 14,000,000/1 to win the lottery by picking 6 random numbers out of 49 random numbers – random being the stand out word here.

    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.

  10. Jock says:

    This is one of the more high brow conversation on Pies.

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

  11. Guy says:

    Looks like nobody reads the comments above them. Beat the dead horse some more.
    “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!

  12. Bangolero says:

    Thank you one and all for the input!
    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.

  13. [...] The Champions League draw was identical to the rehearsal! Weird [...]

  14. Independent Adjudicator John from Bedforshire says:

    The most important thing about this draw is that it implies UEFA is corrupt, and that, personally, is all I care about.

  15. Tihanyi László says:

    The exact number is 5463 counted with those special constraints.

  16. Ian says:

    @independent adjudicator john…
    Any football related organisation whose acronym contains the letters FA is definitely corrupt.
    Shame really.

  17. Vasko says:

    I saw Madrid play Espanyol last weekend and they looked sloppy all over the pitch and far too reliant on Ronaldo. If United can shore up their back line and Van Persie and Rooney produce the displays they have been domestically, I beleive United can go through.

  18. Tim says:

    The “test” draw combination was not in the same order as the “official” draw combination. So this is not a restriction. But there are other restrictions:
    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….

  19. D says:

    This is fake

  20. hurricane says:

    Bangolero: I would say your calculated odds are about the same as Arsenals chances of winning the cup! only joking!

  21. Applesauce says:

    Too much math, not enough puns.

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

  22. Dannyboy says:

    Bangolero, please stop talking maths

  23. Bangolero says:

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

  24. Philipp says:

    Sorry to say so, but your calculations are all wrong. There are 5405 possible draws. How does one know? Have a computer program evaluate all combinations in less than 1 second!

  25. seb says:

    odds are 1 to 528. around 0.2%. Tim is almost right for the specific combination, not for whatever combination repeated. also, the way draw goes makes things not that easy, as some pairs are more likely to be drawn than others. probabilities for pairs are. so it is pretty hard to calculate odds for the repeat.
    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

  26. Richard says:

    Tihanyi László is the only poster to get the right answer so far – there are 5463 possible draws so the odds of repeating the practise draw is 5463.

    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.

  27. danny says:

    that means that the champion was meant to be

  28. ronaldo just kick shot at wolf man let he block them and run through with the follow true

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