-EV with the best hand??

[jba]

[ QUOTE ]
If both have pot odds, then betting is +EV for both.

[/ QUOTE ]

you are wrong.

say there are 10 bets in the pot on the turn, you have a made hand and your opponent is on a draw that will come in 1/4 times.

if you bet, your opponents EV is:

EV=.75*-1+.25*11=2

if you don't bet, your opponent's EV is:

EV=.25*10=2.5

clearly it is EV- for your opponent to have to call a turn bet.

your first post makes no sense to me.

[ QUOTE ]
[ QUOTE ]
If both have pot odds, then betting is +EV for both.

[/ QUOTE ]

you are wrong.

say there are 10 bets in the pot on the turn, you have a made hand and your opponent is on a draw that will come in 1/4 times.

if you bet, your opponents EV is:

EV=.75*-1+.25*11=2

if you don't bet, your opponent's EV is:

EV=.25*10=2.5

clearly it is EV- for your opponent to have to call a turn bet.

your first post makes no sense to me.

[/ QUOTE ]

you're a bit confused about my posts.

[jba]

[ QUOTE ]
you're a bit confused about my posts.

[/ QUOTE ]

that's what I said.

[ QUOTE ]
[ QUOTE ]
you're a bit confused about my posts.

[/ QUOTE ]

that's what I said.

[/ QUOTE ]

I'm gonna go though them in a few minutes when I get the chance. Maybe I made a mistake (woudn't be the first time).

[Obliky]

[ QUOTE ]
say there are 10 bets in the pot on the turn, you have a made hand and your opponent is on a draw that will come in 1/4 times.

if you bet, your opponents EV is:

EV=.75*-1+.25*11=2

if you don't bet, your opponent's EV is:

EV=.25*10=2.5

clearly it is EV- for your opponent to have to call a turn bet.

[/ QUOTE ]

The way i have always understood it is that both calling a bet AND checking are +EV for your opponent..but checking is more +EV so he would prefer to check. Is this not right?

[bobbyi]

[ QUOTE ]
your bet is EV+ for you and EV- for him. the sum is zero.

his call is EV- for you and EV+ for him. the sum is zero.

[/ QUOTE ]
Before cards are dealt, it is a zero sum games. At the point of view where a decision is taking place during a hand, it is a constant sum game where everyone's equity will sum to the pot size because money already in the pot is now "dead money" even if you originally contributed it.

The EV's do not sum to zero; they sum to the pot size (disregarding details like rake still to be taken).

[jba]

[ QUOTE ]
[ QUOTE ]
your bet is EV+ for you and EV- for him. the sum is zero.

his call is EV- for you and EV+ for him. the sum is zero.

[/ QUOTE ]
Before cards are dealt, it is a zero sum games. At the point of view where a decision is taking place during a hand, it is a constant sum game where everyone's equity will sum to the pot size because money already in the pot is now "dead money" even if you originally contributed it.

The EV's do not sum to zero; they sum to the pot size (disregarding details like rake still to be taken).

[/ QUOTE ]

the ev of the decision will sum to the pot size, but the ev of the *difference* of two decisions from the perspective of each player, will sum to zero -- it is a zero sum game after all (disregarding details like rake).

edit: what i mean is when I said "your bet is EV+ for you" i did not strictly mean the bet is EV+; that is obvious as he is a favorite with dead money in the pot. what I specifically mean is "a decision to bet has a higher expected value than a decision to check" (this is fairly common shorthand on these forums, no?). you are correct that either decision, bet or check, are EV+ and each of those values sum to the pot size with the opponents respective EV

[sethypooh21]

On a betting round after the flop, there is money in the pot. Everyone who still has a chance of winning the hand 'owns' a piece of that money, this is your pot equity. If I have the best hand currently (which means my PE is greatest), then I not only own the lion's share of the money already in the pot, but I get even more from each additional bet that goes in.

Conversely, if I don't have the best hand, but have a decent shot of winning the hand, it is +EV to call a bet, as between the options of folding (and 'surrendering' all of my equity in the pot) and calling one bet, calling one bet is a better option. It would be *best* for me not to have to call a bet, but that option isn't up to me...

[Obliky]

Thanks for the help, i think i get it now [img]/images/graemlins/smile.gif[/img]

I have a feeling everyone is saying pretty much the same thing but in different ways...would anyone like to make it painfully clear with a little example showing the EV calculations in a hand?

Thanks

Obliky

I've been putting a lot of thought into this lately, as I'm reading TOP, and immediately could think of situations in which even if both players flipped their cards up, they would play out the hand just as they would if their hands were concealed. The key, as many have pointed out, is the issue of when you turn your cards up. And even in the most extreme situations, the "Your EV + Their EV = 0" rule holds up.

Imagine a hold 'em game in which there are no antes or blinds, and the action is folded to the last two players. The first holds A [img]/images/graemlins/spade.gif[/img] A [img]/images/graemlins/club.gif[/img], the second A [img]/images/graemlins/heart.gif[/img] K [img]/images/graemlins/spade.gif[/img].

The second player doesn't know it, but he's pretty much screwed. Factoring in the rare chops, the second player only has about a 5.94% equity stake in any money going into the middle, with the first player having the other 94.06%. Still, AKo is a big hand, especially against a random opponent, so the second player is ready for action. The FTOP holds up here, because if the hands were turned up, the second player would fold without calling a single bet.

Before looking at the EV, just imagine that there are four rounds of betting pre-flop, and no raising. Unrealistic, of course, but it clarifies the point. So here are the EV calculations of each round:

First Round:

AA: (-1*.0594)+(1*.9406) = $0.88
AK: (-1*.9406)+(1*.0594) = -$0.88

Total EV = $0

Second Round:

AA: (-1*.0594)+((1+2)*.9406) = $2.76
AK: (-1*.9406)+((1+2)*.0594) = -$0.76

Total EV = $2

Third Round:

AA: (-1*.0594)+((1+4)*.9406) = $4.64
AK: (-1*.9406)+((1+4)*.0594) = -$0.64

Total EV = $4

Fourth Round:

AA: (-1*.0594)+((1+6)*.9406) = $6.52
AK: (-1*.9406)+((1+6)*.0594) = -$0.52

Total EV = $6

Clearly, the total EV each time is the pot. So each decision (independent of the pot) is neutral EV, even though the EV balance may be positive. Furthermore, imagine the flop comes down:

2 [img]/images/graemlins/heart.gif[/img] 6 [img]/images/graemlins/heart.gif[/img] 10 [img]/images/graemlins/heart.gif[/img]

Nice board for player two. All redraws and chops have been eliminated: if a heart comes, player two wins, and if it doesn't, player one wins. So even if the players flip their cards up, player one will still bet, and player two will still call. The FTOP applies to the hand, but not to this particular point in the hand. EV of just the flop call:

AA: (-1*.0594)+((1+8)*.9406) = $7.06
AK: (-1*.9406)+((1+8)*.0594) = $0.92

Total EV = $8

Player two can finally make a +EV decision, but that doesn't mean that the total EV is positive. Rather, as before, the EV is simply the size of the pot.

I'm really just elaborating on what others have said in this thread: the pot skews the EV calculations. The $8 pot on the flop creates a +EV decision for both players, but if we ignore the pot, the bet/call decision itself is neutral EV, just like the bet/call decision before it. The pot is the sum of a series of neutral EV decisions, and is therefore itself neutral EV. A neutral EV pot combined with a neutral EV flop decision is a neutral EV hand. The truth of Sklansky is, of course, revealed, and the universe is in order.