I'm reading Theory of Poker and I'm having a little trouble understanding expectation. ANY help would be appreciated tons.

If there's $750 in the pot (villain's bet included in pot size) and I'm facing a $300 bet that I only want to call, how many times do I need to win for this to be profitable? And how do I write it (express it in the proper form)?

In the example above, I'm getting 2.5:1 on my money. So if the first time I call I lose, I'm out $300.

The second time I call and lose I'm out $300 again, for a total of a $600 loss.

The third time I call I win and am up $150 ($750-$600).

I need to be good 1 time out of 3 for my call to have +EV, correct? Would I express this 1:3 or 1:2?

Again, thanks a sh|t ton.

do you mean how good you have to be on the river to make it a profitable call? if so:

(300(what you have to call))/(750(whats in the pot) + 300(villans bet) + 300(if you call)) = you have to be good 22% of the time.

if you are talking about pot odds:

750 + 300 = 1050. you need to call 300 into 1050. 1050/300 = 3.5 to 1 odds.

Yeah, I meant how often I need to be good to make a call +EV. Although, Number1hater, I think I explained my example wrong. The $750 pot in the example comes from his $300 bet into a $450 pot. In that case does it become

(300[what I have to call])/(450[what's in the pot])+300[villain's bet]+300[if I call]) = ~29%?

I just wanted to clarify that the total pot after villain's bet is $750. Since that's the case I was right about needing to be good 1:3 or ~29% of the time. Number1hater, you have helped TONS. Grazie, amigo.

yup, thats correct. its always (what you have to call)/(pot size + villans bet on river + what you have to call) = how often u have to be good to make the call profitable.

If you are getting 2.5 to 1 pot odds, you need to be good 1 in every 3.5 times, for this to be 0 EV (dead even money). If your good more often than that, this is +EV (you expect to gain money every time you make this play).

And if you are good less often than that, this is -EV (You expect to lose money every time you make this play).

In your original example you gave 2.5:1 odds, but then when you worked the math out in the example you gave 2:1 and said you won some money. 2:1 is more often than 2.5:1, this is why it looked like you net gained.

Obviously you can't win "half a pot" for the 2.5, so multiply both sides by 2 or by a number until they become whole, if you wish to work it out that way. 2.5:1 is the same as 5:2 etc.

Or just work it out by percentage. 300/1050 = 28.57%, so you have to be good 28.57% of the time.

You can check the math by 28.57% of the time you win 750. 71.43% of the time you lose 300.

750 * .2857 = $214.29
-300 * .7143 = -$214.29
Difference = 0

So everytime you have exactly 28.57% chance of being good, it's 0EV, you won't gain or lose anything. Do the same calculation above if you think you are good 33% of the time, and you'll see you will be +EV.

750 * .33 = $247.50
-300 * .67 = -$201.00
Difference = $46.50

And technically it will on average take you 3 times to average this out (.33 is 1 in 3), you will expect to make $15.50 everytime you make this call. It is +EV

says it best

Guys, this help is great. Sadly, all these poker sesh's I've had for the past year or so have been done w/o EV in mind. I've been breakeven for the year, but since profit is the sum of all these edges combined, I might be able to make some this year. Grazie.