EV Error in Article?

[T-Rod]

I'm no math expert, but is there an error in Aisthesis article's EV calculation? In his example, there is a 25% chance of hitting and winning. The pot is at 160 and it costs 60 to call. He computes an EV as (1/4)*160 – (3/4)*60 = 40 – 45 = -5.

If you win, don' t you get 220 back? That's 160 plus your call amount of 60. The way I compute it is if I win, I get .25 of 220 which is 55. However, it costs me 60 to do it so I shouldn't. Are these two formulas the same?

I notice that other Pot Odds articles typically add in your call amount to the EV calculation. Kent's article titled Calling X into a Y pot (all-in) seems to do this.

Am I missing something here? Which is the proper way.

In another example it could be different. Say the pot is $10 and some guy bets the pot and you have a 33.3% chance of sucess. Call? Well if you don't add in your call to the winnings... pot was 10, gut raised it to 20. Okay 1/3 chance of winning 20 which is an EV of about 6.66 and it costs me 10 play. Yuck. negative EV so don't call. OR if you add in your amount... pot was 10, guy raised it to 20, you call bringing pot to 30. So you win a total of 30 when you hit. You have a 1/3 chance of hitting so that's an EV of 10 with it costing you 10 to call. A push.

What am I missing? Grr

Tim

[Nortonesque]

trodgers, you use the net result from your pre-call position for EV. So if you call the 60 and win, you're up 160 from where you were. So if you make this call, you'll be up 160 25% of the time, and down 60 75% of the time, from your previous position. Notice that means folding always has an EV of 0.

Kent's article is essentially the same formula, but rearranged. So if your chance of winning is n, and the pot is Y and the amount you have to call is X, break-even EV would look like:

n*Y - (1-n)*X = 0

in Aisthesis' formula. So if you solve for n you get

n = X/(Y+X)

which is Kent's formula, for the minimum chance to win that you need to call.

In the example you give, it costs you 10 to play, but you only lose that 10 2/3 of the time, so it really costs you 6.66 to play (hence, the call is 0 EV).

Personally I find it easier to just look at the pot odds I'm getting and the odds against my hand (this is just another version of the same formula). Calling 10 to win 20 is 2:1 pot odds, and 33% chance of winning is 2:1 against my hand. As long as the pot odds are greater than (or equal to) the odds against your hand, calling is correct.

[T-Rod]

I knew I was mistaken somewhere. I forgot that 1/3 of the time I don't lose my 10 so cost to play was only 2/3 of 10 or 6.66.

So to stop a flush draw (which is about 35% of hitting) I should be betting at lease slightly better than the pot?

Tim

[Nortonesque]

35% is for the flush to appear by the river, so you don't really need to worry about that stat unless you're putting them all in (or calling all in). If that's not the case, then you're charging them to see 1 card, rather than 2.

It's 4.2:1 against the flush hitting on the next card, so you need to charge a little over 1/3 of the pot for the call to be -EV for them. That assumes that you won't pay off if the flush hits though. Generally it's best to just pick a standard bet, somewhere between 2/3 pot to full pot, so that they won't be able to pick any patterns out of your betting.

[T-Rod]

I completely agree about you making it an negative EV for each street so you only need a 2/3 to full pot bet. However, I'm so used to people being willing to pay just about anything to draw that I just go ahead and assume they are going to draw it out (and thus the 35%). Kinda funny.

Tim