NLHETP EV Question

[Dante1126]

I've just started reading NLHETP and I have a disagreement I'd like to get thoughts on. More than likely its probably just my simplistic thinking, but I'd enjoy some constructive thoughts on it.

On pg.21-24 Sklansky and Miller talk about thinking in terms of expection, which I understand.

The problem posed is how much to raise at showdown when you hold the pure nuts when your opponent has bet $50 in a pot of $100 with both your stack sizes equal at $500 left.

Small=$50
Medium=$150
Large=$450(all-in)

With the formula EV=(Pcall)(S), Pcall=chance of call, S=Size of the bet.

My first instinct while reading this was a $150 bet for the reasons they apply, and becuase it has a good chance of being called (they estimate 40%). Thus $60=(0.40)($150) for the medium sized bet; and $90=(0.20)($450) for the large sized bet. Making the large sized bet the best EV choice of the three.

My thinking on it though is why isn't the medium sized bet actually a better choice? I understand the formula and the higher EV of the large bet; but in my mind if the situation presents itself 10 times the profit made from a call 40% of the time is $240 on the medium bet. While the profit made from a call only 20% of the time with an EV of $90 in only $180 on the larger bet.

While lower EV, wouldn't the higher chance of being called make the medium sized bet be more profitable and the best play over the long run? Am I just not running the math through my head right? Admittedly I'm not a heavy math player. I understand to a degree, but I'm no GT player.

I'd appreciate any thoughts.

[Ed Miller]

You're doing the math wrong. EV represents the AVERAGE win over a large number of trials. So the play with the higher EV will be the more profitable one.

To do it your way, you need to multiply $150 by 4 and $450 by 2.

[Dante1126]

I figured I was missing a simple point, I understand the problem now. I appreciate you taking time to respond.