Folding the nuts?
Today I come across a very interesting situation that made me seriously think.
100BB effective stacks, 6max table
BTN raises 3.5 BB 
SB calls 3.5 BB
Flop (8bb) 
SB checks
BTN checks
Turn (8bb) 
SB checks
BTN bets 6bb
SB goes All in for 96.5bb
BTN ?
Now why is this such an interesting spot, because every time here we are against another ace. Our problem is that some of these aces will have a flush redraw. And whenever we call in this spot against a flush redraw our EV is -10bb against all other aces our expectation here is +9.5 bb
Hence if he makes this move more than 50% with ace with redraw we lose money by calling.
Now an intelligent opponent will understand this concept and often will push directly with his free rolling nuts. Hence we should actually fold the nuts in some of these situations. This sounds like a good line to exploit people that do not want to call huge overbets only to split.
This move is obviously great if he have the suit ace, no matter whether we have flush draw or now. Hence if we are against thinking players and we don’t have the suit ace we still can try this huge overbet push line just to get our opponent to fold his nut hands sometimes.
Lets state the problem in a different way, assume we are the small blind from the above example and we don’t have the suit ace. We know we are against thinking opponent and we have reason to believe he will fold some of his non suited aces. With how deep effective stacks is it profitable to try this move?
The decision whether this is a good push or not is based on what is the pot/stack ratio and how often do we need him to fold the nuts. During this computation we take into account only the situation where our opponent also has the nuts.
Suppose x is the current pot, y is the size of the effective stack and z is the frequency with which our opponent will fold.
Our opponent will hold the suit ace, 1/3 time and this ace will be suited approximately 1/10 times (I try to take into account the people’s general preference for suited aces).
ev = z*x - 0.1*0.2*y +(1-z-0.1)*x/2
z = 0.036*(y/x)
which for our example from above translates to the opponent can safely push in this situation if he believes we can fold an ace 16% of the time. And against a thinking opponent this is very reasonable. Suppose an even more extreme example, on the turn SB direcly open pushes, then his push should actually make you fold? 46% of your aces hands, which is kinda infeasible. Lets say that against thinking opponents 20% is reasonable number . Then the maximum pot/stack ratio with which you should try this play is 5.5.
I guess this is the main result from this computation if the effective stacks are not more than 5.5 times the current pot, then an overbet AI move against thinking opponent will be profitable.
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