Math question about a knockout tourney

    • Luupainaja
      Joined: 16.03.2010 Posts: 1,148

      Lets say that a certain play gives me -x% (eg -1%) $EqDiff in Wizard. But if i win a hand, i win a bounty that's y% (eg 2%) of the pricepool. Is it correct to assume that i have to win the hand x/y*100% of the times to BE (eg 1% loss is 50% from the 2% bounty)?
  • 2 replies
    • kurrkabin
      Joined: 12.10.2010 Posts: 5,976
      Hi, luupainaja!

      To be completely honest, I am not sure how the calculation is done. Following the logic, if we have somebody/ies covered, we can take the bounty for their knock out, therefore we can loosen up our range. Unfortunately, I am not quite sure how to break down the math.

      Maybe you can forward this question to one of our knockout players such as superuorme? ClickHe has a blog here in the english community. He also has a solid background playing these games and I think he can be more helpful than me in this area.

    • CollinMoshman
      Joined: 20.09.2011 Posts: 451
      The usual way is to convert the bounty to an approximate # of chips and use that to adjust, e.g. bounty is 20% of buyin so approximately worth 300 chips (changes though based on stack distribution).

      Off the top of my head I'd think you could add the term W * y to the diff% to modify for the bounty.