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FA Cup probability

Discussion in 'Mathematics' started by adamcreen, May 14, 2011.

  1. adamcreen

    adamcreen Occasional commenter

    The match this afternoon (as you may know) is between Man City and Stoke, who are both in the Premiership (20 teams) and were due to play each other today anyway.


    So what is the probability that if both Cup Final teams are in the Premiership, that they would have been playing each other anyway?


    I think it's easy, but as with many of the probability questions on here, I'm sure there's a good discussion to be had!
     
  2. I do not think it is easy

    The matches in the premiership are set prior to the first round of the FA cup


    So you would have to look at the probability that it is those specific teams

    So you have probabilities of them being drawn against each other ... then do you have to work out the probabilities that
    they win each match

    If they play a non-premiership team do you assume they win (since you have a stated premise of "given they are both premiership teams" ... or is that made more complicated by the fact that the team they are playing just has to lose to a prem team at some point

    Do you work backwards from the final considering which each would have played ... is so do you assume that they had the same chance of playing team C in round 5 as the team that actually beat them in round 4 did
     
  3. adamcreen

    adamcreen Occasional commenter

    I was trying to set the easier problem of: GIVEN that it is teams A and B in the final, what is the chance that the Premiership fixtures that day featured the same draw? I didn't really want to think about the previous rounds of the FA Cup at all. To be honest, I don't think about football anyway!!
     
  4. I think you also need to turn the clock back several years and assume that all league matches are played at the same time (or, at the very least, on the same day). In actual fact, there are normally games on a Saturday, Sunday Monday and often Tuesday, Wednesday or Thursday which are all counted as being on the same "matchday" (or part of the same "matchweek"). Which day a game is played on is far from random (TV rights, desirability of a team, dates of games in a nearby ground, police requests and cup clashes to name just a few), and the rules of probability don't really apply. Relative frequency could help, but then what would you do for a newly promoted team?
    So ... there are 380 games played in the premier league. So, I think, given the various assumptions, the probability must be 2/380 = 1/190 (since the matches A v B and B v A both satisfy the desired outcome).

     
  5. Who cares?
    Stoke lost - and that certainly makes me happy!! [​IMG]
    Why?
    The clue is in the name mates!
    Valed
     
  6. ahhhhh, I see ... i've often wondered what your name was about. Perhaps a corruption of (and pronounced the same as) "veiled" ...
    Then sometimes I think it's "val-ed" (value education? or your name is val and you work in education?)
    But now it makes sense!
     
  7. DM

    DM New commenter

    That's what they mean by valed reasoning.
     
  8. harderfaster

    harderfaster New commenter

    Given that every team is playing this weekend and already given the fact that the two clubs are in the final, surely there's a 1/19 chance that these two teams were playing each other this weekend? So given that City had a match this weekend already, P(they a playing Stoke) / P(they are playing any team) = 1/19.

    As mentioned, as soon as one of these conditions (both teams are already in the final, time of game is not important - rather just the weekend) is dropped it becomes much more complicated, and one boiling down to subjective probabilities (otherwise we have to assume all teams have an equal chance of reaching the final, which is silly) and so there would be no one answer...
     
  9. harderfaster

    harderfaster New commenter

    You neglect the fact that there are 10 matches this weekend!
     
  10. Oh yes, of course! How stupid of me! If it's any excuse, I wrote that post after about 7 hours of mind-numbing work on my portfolios of evidence for GTP ... If I never see the letter "Q" again for a few years I'll be a happy (and, hopefully, qualified) teacher!
     
  11. harderfaster

    harderfaster New commenter

    Good luck with that! In all honesty I don't know whether then simply dividing it by 10 is the end of the problem but I have 7 hours of revision ahead of me so I'm not going to think too hard about it!
     
  12. Yep, divide by 10 and you're done.
    The easiest way to look at it is (given the various assumptions) that there are 38 matchdays (or gameweeks), 2 of which will have teams A and B playing each other. 2/38 = 1/19.
     
  13. Yes !!
    Used to be Valed Threat long time ago, the irony being that Vale offered not even a 'veiled threat' to other teams .......
    I am exposed .......
     
  14. Piranha

    Piranha Lead commenter

    I think the easiest way of looking at is that if (say) Stoke would have been playing this weekend, there is a 1/19 probability that their opponent would have been Manchester City. If we assume that all clubs would have been playing this weekend, then the problem seems pretty easy.
     
  15. I am really struggling with the basic concept here.....
    Not that of the probability but the conjunction of the two words 'Stoke' and 'play' in the same sentence.....
    Up the Vale! [​IMG]
     

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