[jojobinks]

EV is, simply put, the amount any bet will average winning or losing. in poker and other forms of gambling, we often talk about "good bets" or "bad bets." EV answers the question "how good (or bad) is it?".

Now, i steal some examples from sklansky. (this is from theory of poker, pages 9-10 (paraphrased))
if you flip a coin with a friend, and you win $1 each time you hit heads, and lose $1 each time you hit tails, your EV is 0. here's how you do the math:
.5(1) + .5(-1)=0
in english: 50% of the time you win a dollar, and 50% of the time you lose a dollar. you add up the expectation for your wins, and subtract the expectation for your losses.

now, let's say you have a dumb friend that thinks tails comes up more often than heads. he'll pay you $2 for every heads, and you only owe him $1 for each tails. now the equation looks like this:
.5(2) + .5(-1) = .5
now your EV has moved from 0 to 50 cents. your EV for each flip is 50 cents (or fitty cent, as you wish).

important point: on any given flip, you'll either win $2 or lose $1. but your EV is 50cents b/c it's what you'll average per flip, over the course of many flips. it's what you can mathematically expect.

in poker, EV calculations go from nearly that simple to quite complicated. here's a simple one.
the pot is $100, and you're headsup. the board reads

your hand:

on the turn, your opponent goes all in for $30. you need to call. what's your EV for the call?

you have 8 outs to the absolute nuts and you're sure he's got a K or a J, so pairing the 5 or 6 is no help. you know 6 cards (board plus your 2), so there're 45 cards left in the deck. you have 8/45 to win $130, and 38/45 to lose $30 (what's in the pot is no longer yours to lose).

in math:
(8/45)(130) + (37/45)(-30) = -1.6 (corrected to have numerators add to 45 )
over a million times you make this call, you average a loss of $1.60

if, somehow, you know he's got QT (for a straight draw), then you have an extra 6 outs (all 5s and 6s) and you know two more cards, and the call becomes profitable.
(14/44)(130) + (30/44)(-30) = 20.9 (corrected to have numerators add to 44 ))

okay. that's what i've got for now. it can be much more complicated (and will be, in NLHE:TP).

for now, discuss.

[jldecarlo]

JoJo answer me this:

In your first example, you "know he has a J or a K." Doesn't that increase the number of known cards from 6 to 7. You know your cards, the board, and, effectively, one of his (50/50 K/J). You know that at least one of his cards is not one of your outs. Does this change your calculations.

Similarly, in your second example, you know he has QT. Now you know 8 cards.

This brings me to my biggest problem with calculating EV. Sklanksy always says things like " suppose you know your opponent is on a flush draw 40% of the time, has two pair 40% of the time, and is bluffing 20% of the time . . ." If I knew that, I would be a MUCH better poker player. How the heck are you supposed to make accurate judgments like these? (This leads to a discussion of how to put your opponent on a hand. Something I'd like to discuss if anybody's interested.)


Discuss.

L

[jojobinks]

Originally Posted by jldecarlo

JoJo answer me this:

In your first example, you "know he has a J or a K." Doesn't that increase the number of known cards from 6 to 7. You know your cards, the board, and, effectively, one of his (50/50 K/J). You know that at least one of his cards is not one of your outs. Does this change your calculations.

Similarly, in your second example, you know he has QT. Now you know 8 cards.

This brings me to my biggest problem with calculating EV. Sklanksy always says things like " suppose you know your opponent is on a flush draw 40% of the time, has two pair 40% of the time, and is bluffing 20% of the time . . ." If I knew that, I would be a MUCH better poker player. How the heck are you supposed to make accurate judgments like these? (This leads to a discussion of how to put your opponent on a hand. Something I'd like to discuss if anybody's interested.)


Discuss.

L

you're right, and i'll fix my OP. (after the stud tourny)

as to the estimations...well, you have to guess. the better you know your opponent (and the game you're playing) the closer you'll get. and of course, this is theoretical. b/c in poker information is nearly always incomplete, you have to make guesses and hope they're good.

mostly, this calculations are good when you're away from the table and thinking through situations that you were in or that you might be in later.

[gagners]

Great post JOJO. I enjoy these much more than the "what would you do in this situation" posts. At least there's some theory behind this - as opposed to preferences. These are the types of posts that benefit all. Thanks again.

[coyote]

I think the first thing to discuss is whether or not it is correct to use your assumptions about what your opponent holds to adjust the odds, i.e. reducing the # of unknown cards in the calculation.

neither calculation as written add up to 100%. 8/45 + 38/45 = 46/45

14/44 + 32/44 = 46/44

not a big deal in terms of expectation, but as far as correctly making calculations, I've never seen sklansky/miller et. all reduce unkown cards in a calc based on their assumption of an opponents holdings. The only guarantees are the 2 in your hand and whatever is on the board, and instances where there was an exposed card.

[Wedge Rock]

Also important is to realize that you are putting him on a K or J in the first example and a QT in the second example... You had better hope that assumption is correct and he really doesn't have 33... Because then, you can't win, whether you pair up, or make the straight...

That was one of the hardest things I had to overcome when thinking about theory... how do you *know* what the other person is holding. What I came to realize is that an educated guess, based on experience and betting patterns that leads to a logical conclusion is better than a stab in the dark... But when I'm left with a guess (for example, does he had a K or J or does he have the QT?) and I can't discern which he is holding, I always figure based on a worst case scenario...

[Jambine]

Originally Posted by jojobinks

......okay. that's what i've got for now. it can be much more complicated (and will be, in NLHE:TP)......

LOL. Great post jojo. I’ve been reading those two pages for about two years now.

[Spaceman Spiff]

Let me see if I'm thinking about this correctly, but just from a different angle. Same situation, I would consider that I have to pay $30 to win a pot of $130, so I'm getting 4.33 to 1 on that bet. If I assume that making the straight will win the pot, I'm something like 4.5 to 1 against that happening. So I'm not quite getting the right price to make the call and that's reflected in the slightly negative EV. Is that correct?

So EV, doesn't really take into account implied odds does it? It's just the calculation of the moment at hand. In this case there are no implied odds since it's an all-in, but say your opponent had another $100 in his stack. And you're fairly sure you could get most or all of the rest of that if the straight hits. Does an EV calculation take this into account at all?

[SpaceMonkey]

Originally Posted by Wedge Rock

You had better hope that assumption is correct and he really doesn't have 33... Because then, you can't win, whether you pair up, or make the straight...

Making your straight would mean the board doesn't pair, so a straight still beats a set as far as I remember.

[Wedge Rock]

Originally Posted by SpaceMonkey

Making your straight would mean the board doesn't pair, so a straight still beats a set as far as I remember.

Oops.

I originally considered the possibility that the other player had two higher hearts when I wrote that, but realized there weren't two hearts on the board through the turn, so a flush wasn't a possibility... I editted my response, but neglected to change the straight draw...

My bad.