# Math question about a knockout tourney

• Bronze
Joined: 16.03.2010
Hi!

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
• Bronze
Joined: 12.10.2010
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.

Cheers!
• Coach
Coach
Joined: 20.09.2011
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.