# steal from BTN with any two calculation

• Bronze
Joined: 04.04.2011
Hello everyone i have search and found somewhere on the internet a formula that you can calculate based on the foldtosteal stat if its profitable to steal from the BTN with any...

the formula is this :

EV=(1.5 * F) - (XBB * (1-F))

F= SB fold to steal * BB fold to steal
XBB= how much BB you raise from the BTN

so if we say that SB folds to steal 80% and BB folds 90% and we raise lets say 3BB from the BTN with any two

EV=(1.5 * 0.72) - (3 * (1-0.72)) = +0,24 so the formula says its profitable to steal in this situation with any two

can anyone confirm if this is correct ?
• 4 replies
• Coach
Coach
Joined: 19.11.2010
Moving to No Limit
• Bronze
Joined: 29.01.2005
Hey,

just compare it with a standard calculation.

x:=foldequity SB
y:=foldequity BB

x*y>=66% for +EV in case you are raising 3bb from the buttom.

0,8*0,9=0,72 -> two persons with 2 foldequities are one person with one foldequity

EV=1,5*0,72-3*0,28=24

Sounds good for me
• Bronze
Joined: 10.10.2010
The formula is correct if the sb is 1/2 bb. On NL5 you need to adjust it slightly.

However, be careful when stealing any two cards, as an observant villain can easily take all your stealing profits by playing aggressively a couple of times with pretty much any hand as well.
• Bronze
Joined: 04.04.2011
thank you

yea the formula is just a formula if villian adjust you must readjust to