Exploiting Short-Stacks who Steal a Lot
I'm going to discuss our expectation vs a short-stack's pre-flop raising and folding to 3-bet ranges in late position. It so transpires that the best way to exploit an aggro short-stack is to shove with a lot of air. What follows doesn't take into account the fact that a short-stack who plays well is likely to adjust his opening range and calling 3bet range when you play back. Also, i'm only speculating as to a short-stacks opening and folding to 3bet ranges: bare in mind this is largely a mathematical exercise.
Let's say we have plenty of information on our short-stack. We deduce that his late position opening range is 17.2% of hands, something that might look like:
22+,A7s+,K9s+,QTs+,JTs,ATo+,KTo+,QJo
We're also pretty sure that he folds all of these hands to a 3-bet shove except these 4.2% of hands:
TT+,AQs+,AKo and AQo 50% of the time.
This means he calls with 24.4% of the hands he opens, i.e. probability he calls, p(call) = 0.244; probability he folds, p(fold) = 1 - p(call) = 0.756
The game is $1/$2. Pre-flop: short-stack is $40 deep. He open raises for $6 in the CO and the Button and SB fold. You are in the BB and must decide how to act. The pot is $9.
The equity of a random hand versus his calling range is:
Code:
117,464,054,400 games 142.802 secs 822,565,891 games/sec
equity win tie pots won pots tied
Hand 0: 73.276% 72.71% 00.57% 85408646638 664832731.00 { TT+, AQs+, AKo, AQo half the time }
Hand 1: 26.724% 26.16% 00.57% 30725742300 664832731.00 { random }
i.e. 26.16% - the probability that our random hand wins is 0.2616.
Our expectation for shoving then is:
<FTR> = {expectation when he folds} + {expectation when he calls}
<FTR> = p(he folds)*(pot) + p(call)*[p(we win)*(size of pot when he calls shove) - cost of shoving]
<FTR> = 0.756*($9) + 0.244*[0.2616*($81) - $38]
<FTR> = $6.80 + (-$4.10)
<FTR> = +$2.70
What about if EVERY TIME WE DO THIS we are (by some miracle) holding 72o. Against his range we now have just 21.07% equity, i.e. p(we win) = 0.2107.
<FTR> = 0.756*($9) + 0.244*[0.2107*($81) - $38]
<FTR> = $6.80 + (-$5.11)
<FTR> = +$1.69
Let's say that he instead calls our shove with 9.2% of hands, i.e. with 53% of the hands he raises with (p(call) = 0.53, p(fold) = 0.47). Against the range,
22+,ATs+,KQs,AQo+
a random hand has 32.6% equity; p(win) = 0.326.
<FTR> = 0.47*($9) + 0.53*[0.326*($81) - $38]
<FTR> = $4.23 + (-$6.14)
<FTR> = -$1.91
In fact, to make this play against this range, we need to find a range of hands that whose probability is greater than or equal to p(win) where,
<FTR> = 0.47*($9) + 0.53*[p(win)*($81) - $38] >= 0
i.e. p(win) >= 0.37 or 37%
