Question on 3 team parlay combination

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  • ThePhenom00
    Restricted User
    • 12-01-08
    • 72

    #1
    Question on 3 team parlay combination
    If I bet a 3 team parlay how many combinations are there? Also, how do I calculate this? Thanks for any hel
  • 20Four7
    SBR Hall of Famer
    • 04-08-07
    • 6703

    #2
    I don't get the question. If you pick 3 teams and bet them as a parlay then there is only one combination that's the 3 teams you bet. Now if your asking how many combinations can be made out of 3 games that's a different question. Or are you picking 4 teams and want to see how many combo's of 3 you can make out of the 4.
    Comment
    • Hybris
      SBR MVP
      • 07-22-09
      • 1023

      #3
      3^2=9
      Comment
      • ThePhenom00
        Restricted User
        • 12-01-08
        • 72

        #4
        20four7 exactly how many combinations can be made out of 3 games.
        Comment
        • LT Profits
          SBR Aristocracy
          • 10-27-06
          • 90963

          #5
          Originally posted by Hybris
          3^2=9
          You mean 2^3=8
          Comment
          • Hybris
            SBR MVP
            • 07-22-09
            • 1023

            #6
            Originally posted by LT Profits
            You mean 2^3=8
            yes
            Comment
            • snipeshow23
              SBR High Roller
              • 10-12-09
              • 153

              #7
              3 combinations of 2 teams can be made with 3 games... TEAM A, TEAM B, TEAM C.
              Combo 1 = A and B
              Combo 2 = B and C
              Combo 3 = A and C
              Comment
              • IrishTim
                SBR Wise Guy
                • 07-23-09
                • 983

                #8
                Yup, 8. You trying to maximize a free play?
                Comment
                • RickySteve
                  Restricted User
                  • 01-31-06
                  • 3415

                  #9
                  Originally posted by IrishTim
                  Yup, 8. You trying to maximize a free play?
                  I'm assuming his answer is yes. While great in theory, this practice screams "I have a brain" and you will quickly be backed off, or even cheated by lesser books.
                  Comment
                  • arwar
                    SBR High Roller
                    • 07-09-09
                    • 208

                    #10
                    if you assume that the chance of winnng any leg is 50% which is .5 which is 1/2,
                    the probablity of a combination of these legs (i.e. a parlay) is simply the product of
                    of the indivdual probs. p(2team)=1/2*1/2 p(3team)=1/2*1/2*1/2
                    p(4team)=1/2*1/2*1/2*1/2 etc. Coincidently the final denominators in these calculations will also tell you the number of combinations in the play. Therefore a 2 teamer has 4 combos, a 3 teamer has 8, a 4 teamer has 16, etc., or 2^n where n is the number of legs and the reciprocal (1/n) will be the probability.
                    Comment
                    • tltaylor89
                      SBR Posting Legend
                      • 06-19-09
                      • 19610

                      #11
                      Heres a parlay for you http://1.bp.blogspot.com/_DqRn9yIIX4...ben+forbes.bmp
                      Comment
                      • shantystar
                        SBR Hall of Famer
                        • 11-13-05
                        • 7299

                        #12
                        Originally posted by 20Four7
                        I don't get the question. If you pick 3 teams and bet them as a parlay then there is only one combination that's the 3 teams you bet. Now if your asking how many combinations can be made out of 3 games that's a different question. Or are you picking 4 teams and want to see how many combo's of 3 you can make out of the 4.
                        yes right
                        Comment
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