EV Question

This isn’t something that you can calculate by hand. What about the times you river a pair, and check it down and win? Or the times you hit your flush and lose to a bigger flush?

Or even if you were willing to work it out by hand, where are you getting these frequencies from? In practice There’s just no way to know how often villain is folding vs your jam, the exact composition of his turn betting range, and how this range will play the river on various runouts - all of which are required to calculate EV.

Not sure I understand nor agree with above. Probably can't make every single calculation off hand on the table, but putting enough work away from the table to be able to inplement the strategies on the table is definitely doable...If you put your villain on a range by the river then you absolutely can make calculations or at the very least estimated calculations. Example he'll fold x amount (2nd pairs maybe weak pairs) and call y amounts (call tpsk+).

Calling: we win 50 (10 in pot, 10 on turn bet, 30 for river shove) with probability 0.18. We lose 10 with probability 0.82 (using your assumptions, ie that we win if and only if we hit our flush, and further assuming we will always fold when we miss and wonÂ’t ever try to bluff.) This makes our calling EV
50*0.18 - 10*0.82 = 0.8 so calling is better than folding. By the same assumptions with villain having 80bb stack, our EV would be 80*0.18 - 10*0.82 = 6.2, which makes calling even better. In general if villain will pay off x otr our EV is (20+x)*0.18 - 8.2. Set that equal to zero and solve for x and we get x+20 = 45.56 or x = 25.56bb. If villain will pay of at least that much we have a call.

Shoving: a bit more tedious and complicated but similar idea. We win 10 if villain folds, let x be the probability of that. We win 50 if villain calls and we hit our draw. The probability of that is 0.18(1-x). We lose 40 when villain calls and we miss, probability 0.82(1-x).

EV is then 10x + 50*0.18(1-x) - 40*0.82(1-x). This expression simplifies to 10x - 23.8(1-x). Set equal to zero and solve. We get 33.8x = 23.8 or x = 23.8/33.8, about 70.4%. If villain folds at least that often, a shove will beat folding (although calling still could be better yet).

Thanks but I think the shoving fold %age may be incorrect no? We win 20 when he folds (his bet+pot) and will win 90 18% of the time (his stack+pot+our stack)

Think we need him to fold 55% of the time to BE?

EV when he calls = (90*0.18) - 40 = -23.8

.55*20(villain folds and we win 20 in the middle) - .45*23.8 = 11 - 10.71 = 0.29 so just slightly better than BE.

YouÂ’re right on the win 20 part. I somehow forgot that we are shoving over villains turn bet. However our profit when villain calls and we win is villainÂ’s stack plus the money in the pot from prior streets, ie 50. We are focused on the decision point of our turn shove. We should not be counting the money we shove as profit if we win.

With the correction for the profit from villain folding we get x= 23.8/43.8 or 54.3%.

Just as an illustration: suppose (mathematically equivalently) that your stack is 40bb when you reach your decision point on the turn and that you shove for 40. LetÂ’s consider how it plays out if we repeated the spot 2190 times (this number selected to make future math easier). If villain folds your new stack will be 60 (your 40 plus 10 from prior streets and 10 from villains bet). This is a profit of 20. If villain calls and you win, you win 10 from prior streets plus 40x2 from the turn bets, so your stack will be 90 - a profit of 50. Obviously if villain calls and you lose your stack is zero - a loss of 40.

Now letÂ’s assume villain folds 1190 times out of 2190 (54.3%). We win a total of 23800 from his folds. He calls 1000 times. We win 180 of these, for a total win of 50x180 = 9000. Our total wins are thus 32800. We would lose a total of 820 times, at 50 per that would make our total losses also 32800, exactly equal to our wins - breakeven at 54.3%, confirming the above calculation.

I think there's still value in learning to estimate these calculations on the fly.

You run into spots where you know someone is never bluffing, and if you're drawing to the nuts it just becomes a matter of calculating whether you are getting the right pot odds to call.

When a nit bets pot on the turn you can just round off the size of the bet, pot and how much of their remaining stack you think you can win on average when you hit your draw.

Then you might think, "I've got to call 10 to win around 70, and since I'm going to get there around one out of five times it should be a profitable call."

10×5=$50 invested in five of these spots, and if you win $70 one time you're coming out ahead in the long run.

This might not be the most sophisticated way of thinking about the spot but it's often effective, especially in low stakes games where people aren't thinking about concepts like balance, minimum defense frequency, etc.