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probability question

  • 09-03-2005 12:27AM
    #1
    poker_face
    Closed Accounts Posts: 68 ✭✭


    this qn has just been bugging me all day so maybe one of you maths whizz's will let me know how to solve it

    There are 2 football teams playing each other. For simplicity sake let's say that they have to either win or lose the game (i.e. a draw isn't a possible outcome).

    The probability of the two teams of winning any game is as follows

    Team A: P(Win)=.8, P(Lose)=.2
    Team B: P(Win)=.4, P(Lose)=.4
    The question is, whether there's a way of working out the probability of Team A winning that particular match. A method I've tried which seems to give sensible enough answers is to multiply the probability of team A winning by team B losing. Then I'd multiply the probability of team B winning by team A losing. The sum of these two answers is less than 1, so surely these aren't the true probabilities. I then deduced that the factor of the true probability is to 1, would be the same as the calculated probability (team a winning * team B losing) is to the sum of the two possible outcomes. Hence the true probability of Team A winning would be (team A winning * team B losing) devided by the sum of the two possible outcomes. I hope whoever is reading that could follow that. The problem is I just kind of stumbled across this idea and I can't think of the logic behind. Could someone verify that this is the correct method and if it's not could you show me the correct method.
    Cheers

    Calculations for the above: Team A win = .8 * .6 =.48
    Team B win = .2 * .4 = .08
    Sum = .56
    True probability of a Win for Team A = (.48/.56)=.85714286


Comments

  • #2
    Healio
    Registered Users, Registered Users 2 Posts: 2,831 ✭✭✭Healio


    Kind of relates to bookies overrounds:
    This mite help a bit


  • #3
    Pherekydes
    Registered Users, Registered Users 2, Paid Member Posts: 16,274 ✭✭✭✭Pherekydes


    poker_face wrote:
    There are 2 football teams playing each other. For simplicity sake let's say that they have to either win or lose the game (i.e. a draw isn't a possible outcome).

    The probability of the two teams of winning any game is as follows

    Team A: P(Win)=.8, P(Lose)=.2
    Team B: P(Win)=.4, P(Lose)=.4

    This is impossible. The sum of the probabilities must equal 1.

    i.e. If P(B wins) = .4, then P(B loses) = .6

    The probabilities will change according to their opponent.

    So if P(A wins)= .8, then P(B wins) = .2, and so on.


  • #4
    poker_face
    Closed Accounts Posts: 68 ✭✭poker_face


    This is impossible. The sum of the probabilities must equal 1.
    i.e. If P(B wins) = .4, then P(B loses) = .6

    Yeah that's what I meant.

    P(A Wins)=.8

    This is the probability of A winnning any match. Take for example in a league, this value would be got from multiplying the probability for each particular game by each other.


  • #5
    silverside
    Registered Users, Registered Users 2 Posts: 1,372 ✭✭✭silverside


    the way you pose the question doesn't make sense (no offence)

    if A has a probability of winning of 0.8, then that holds regardless how good B is. What you are looking for is the probability of winning when A plays B. You can't derive this just from raw statistics, you have to make a subjective guess. If you want, you can do some sort of model, but the answer will still be subjective.


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