Calculating EV

[James Boston]

I hear EV discussed, and I understand the gist of it, but don't know how to calculate it. I've recently started trying to keep good records of my playing, and want to know how to figure EV. Currently, I just take my hourly earnings and convert that into big bets per hour. Is this completely out of the ballpark?

[CrisBrown]

Hi James,

EV isn't about your expected hourly return. It's about the value of a given hand relative to the pot odds.

E.g.: You hold JTs on a flop of 8-9-3 with two of your suit, giving you both a flush and a straight draw. There is $250 in the pot, and your opponent bets $350. You put him on a pair (for whatever reasons having to do with your hand-reading analysis). You have 19 outs (9 for the flush and 8 for the straight) with two cards to come, which makes it around 1:2 for you to draw out. Your call would be at 10:7 pot odds ($600in the pot for a $350 call), so the call would have "negative EV," i.e.: the pot odds don't justify your call.

The correct decision is to fold ... and even if the next card is a Q, 7, or one of your suit ... you still made the correct decision.

Cris

[Nottom]

[ QUOTE ]
E.g.: You hold JTs on a flop of 8-9-3 with two of your suit, giving you both a flush and a straight draw. There is $250 in the pot, and your opponent bets $350. You put him on a pair (for whatever reasons having to do with your hand-reading analysis). You have 19 outs (9 for the flush and 8 for the straight) with two cards to come, which makes it around 1:2 for you to draw out. Your call would be at 10:7 pot odds ($600in the pot for a $350 call), so the call would have "negative EV," i.e.: the pot odds don't justify your call.

[/ QUOTE ]

If you are folding straight-flush draws with 2 overcards you are making a big -EV mistake. You are often actually a favorite to win by the end of the hand and should be trying to get as much money in the pot as you can on this flop. Against a pair of 9s you could have as many as 21 outs and 2 shots at it. Even against 2 pair you are probably even money here to win by the river.

[bigpooch]

Exactly. EV is simply the fair value for the play in
question and pot odds is something else altogether (to
help a practical player make decisions).

Here are the number of wins for the JhTh for a flop
of 9h 8s 3h versus a number of hands (you were right
about 98 being close to a coin flip!):

99, 88 or 33: 399/990
98: 504/990
AA: 511/990 or 557/990 (no Ah)
KK: 510/990 or 557/990 (no Kh)
QQ: 452/990 or 480/990 (no Qh)
JJ: 537/990
TT: 615/990
Ah 9x: 631/990
As 9s: 652/990
other A9: 682/990
Ah 8h: 349/990
Ah 8other: 631/990
other A8: 683/990

As you can see from the above possibilities, the
Jh Th is often a favorite!

[Bozeman]

Umm, 600:350 is 12:7 and 8+9=17!=19 and str8 + flush only gives you 15 outs (were you counting overcards too?).

[CrisBrown]

Hi Boze,

You're right and I'm too stupid to do simple math today. I should've stayed in bed. *shrugs*

Cris

[dylhead]

To figure out exact odds you can use the new odds calculator at cardplayer.com. Or you can break out your stats book and do it that way. It can get quite tricky to figure out the exact odds if you haven't had any experience with permutations or independent vs. dependent variables.

[Copernicus]

[ QUOTE ]
Exactly. EV is simply the fair value for the play in
question and pot odds is something else altogether (to
help a practical player make decisions).



[/ QUOTE ]

Pot odds is not something altogether different, they are the odds you need for the EV to be exactly 0.

[Louie Landale]

Yup. Show him your hand and raise all in. Or better yet DON'T show him your hand and raise and take it.

- Louie

[Boris]

lol

[bigpooch]

Pot odds are sometimes insufficient and yet EV > 0. This
is because of implied odds; a good example is from pot
limit draw Jacks-or-better:

Suppose you have T987 all in diamonds with the deuce of
clubs and someone opens for the size of the pot and now are
you saying because you have insufficient pot odds (you have
9 flush outs and 6 other straight outs for 15 out of 47
possible combinations but even this is an approximation!),
you should fold? No, because of the implied odds on the
last betting round!

There are implied odds for future betting rounds and pot
odds only gives a good approximation for whether you
should continue. It is the EV that is the true indicator
of whether you should make one decision over the other. Of
course, if you are drawing and almost have pot odds in a
limit game, you should call because even against a lone
opponent, he must pay you off most of the time when you
get there. You also sometimes must discount factors such
as getting there but you weren't drawing to as many outs
as you thought!

A practical example comes from PL Omaha high: even if your
opponent bets the size of the pot when it is heads up, you
only need ten clean outs on the turn (out of 44 cards) to
have a profitable call even if your sole opponent plays
theoretically correctly. It is because on your ten cards,
you can bet the size of the pot and on some bluff cards
with optimal frequency and your opponent must call you 1/2
of the time to minimize the amount you make on any bluffs.
It is important that none of your outs are "tainted"; i.e.,
they are truly all outs.

Suppose you are in a fixed limit game with one last betting
round after you call to draw to your hand with the pot size
as P before each of you put in one bet apiece (so now the
pot size is P+2 in bets). Your pot odds were 1/(P+2). On
the other hand, if you hit and bet, your opponent's
theoretical calling frequency is (P+2)/(P+3) so you only
need a probability of 1/(P+2 + (P+2)/(P+3)) of completing
your hand or (P+3)/((P+2)(P+4)). As you can see, this
probability is smaller than the required "pot odds"
frequency by a factor of (P+3)/(P+4).

Look at an exact example: Suppose the pot has 3.5 BBs on
the turn when you have Js Th and the board reads

Ks Qh 7d 6c

Your opponent bets the turn and now you can call even though
you don't have immediate pot odds (1/5.5 = 2/11 = 8/44
> 8/46, this last number being the chance you will make your
straight although admittedly your opponent most likely has
at least one card in his hand that you don't need to show up
on the table!) Also, we are disregarding the extra rake but
even that can also be included in the calculations! You
actually only need a 26/165 chance or higher of completing
your hand to have an EV>=0 here. The EV in this example is
38/46(-1) + 8/46(+4.5+11/13) = 31/299 = 0.1037 BBs. Folding
here would be an error of greater than 0.1 BB and would be a
"whopper" using backgammon lingo.

When you have tainted outs, e.g., when your opponent could
have a set and you have nut flush draw in LHE, you must
consider how those situations will reflect on your EV. If
it is not unlikely your opponent has two pair or a set, you
must modify all your EV calculations appropriately, but
don't forget it will depend on your opponent and the
previous betting action.

A good way to think of how pot odds can be applied is to
pretend that you will be all-in if the pot you are
contesting were to have a fraction (P+2)/(P+3) of a bet
(which is almost but not quite 1 in typical pots) added to
it. By the way, don't forget to bet when you hit!

[Copernicus]

I thought it was obvious that I was talking about the equivalence of EV=0 and pot odds in a single bet situation where you are a definite winner if you hit and a definite loser if you dont.

Your EV formulas are quite simplified as well, since a true EV calculation needs to consider probabilities of folds if you raise, the additional chips at risk if you arent a definite winner (a more refined "tainted outs" calculation), as well as the additional bets you may be able to induce from opponents using alternative strategies.

For a beginner like the poster, stick with current pot odds, since even the literature does a poor job on what a correct "implied odds" or EV calculation entails. If your hand reading skills arent well developed the latter two are so error prone that considering them will likely hurt your win rate compared to a simple pot odds calculation, using one draw odds unless it is virtually certain that you will be able to stick around for a second draw if you miss.

[bigpooch]

There is no equivalence if there is a possibility of
future action on the hand; if a player currently has
exactly pot odds to draw, his EV is strictly positive.
As my previous post has stated, if a player almost
(but does not quite) has pot odds in a LHE game, he
often will have an EV>0 by calling.

As you had pointed out raising may have an even
higher EV if you know your opponent folds more than
optimally (game theoretically) or if on the end,
without position, the alternative strategy of
check-raising may have even a higher EV against
particular opponents. Nevertheless, the analysis
was quite simplified to exhibit the clear reason
that for EV>=0, immediate pot odds are not quite
necessary.

To adjust to the pot odds mindset, you need to
adjust the pot by imagining that it is juiced up
by (P+2)/(P+3) BBs.