Probabilities Inquiry

Collapse
X
 
  • Time
  • Show
Clear All
new posts
  • Mcducky
    SBR Rookie
    • 05-02-11
    • 40

    #1
    Probabilities Inquiry
    This may be a stupid question, but can someone explain the below to me?

    Place bets on the nine at craps illustrate a situation where the house has an edge. These bets have 40 percent chance of winning and pay 7-to-5, giving the bosses 4 percent edge.

    If the bet has a 40 percent chance of winning and it pays 7 to 5 (7/5 =1.4), how does the house have a 4 percent edge?
  • LT Profits
    SBR Aristocracy
    • 10-27-06
    • 90963

    #2
    7/5 odds has break-even of 5/12 = 41.7%
    True odds are 40.0%
    .417/.400 = 1.04 = 4% edge
    Comment
    • Mcducky
      SBR Rookie
      • 05-02-11
      • 40

      #3
      So you are basically saying 2 things:

      When the house gives you 7/5 odds on a bet, you need to win at a 41.7% frequency to break even?

      The difference between the break-even percentage and the payout for winning percentage is the house's edge?

      Thanks for the quick reply.
      Comment
      • LT Profits
        SBR Aristocracy
        • 10-27-06
        • 90963

        #4
        Originally posted by Mcducky
        So you are basically saying 2 things:

        When the house gives you 7/5 odds on a bet, you need to win at a 41.7% frequency to break even?

        The difference between the break-even percentage and the payout for winning percentage is the house's edge?

        Thanks for the quick reply.
        Correct.
        Comment
        • indio
          SBR Wise Guy
          • 06-03-11
          • 751

          #5
          Originally posted by Mcducky
          So you are basically saying 2 things:

          When the house gives you 7/5 odds on a bet, you need to win at a 41.7% frequency to break even?

          The difference between the break-even percentage and the payout for winning percentage is the house's edge?

          Thanks for the quick reply.
          While all the %'s you have been given are correct, sometimes it's good to visualize the basics to get an understanding of an edge.

          If I roll two dice, there are 36 possible ways for it to land, six of those will be 7's, and four of those will be 9's. So obviously I should roll six 7's for every four 9's when I roll dice (thanks to variance, that only has a 25% chance of happening on the first 10 results, but that's for another thread).

          Let's say I roll dice and bet $10 on every roll and receive $24 back (7/5 odds) when I roll a 9, and
          all rolls other than 9 or 7 don't count. If I bet $10 on ten rolls that count, I would bet $100 total. If four of those ten rolls are 9's, I would get back $96 ($24 x 4). So out of my $100 in bets, the house has paid back $96 and kept $4, hence, 4% hold.
          Comment
          SBR Contests
          Collapse
          Top-Rated US Sportsbooks
          Collapse
          Working...