Implied value and draws

[Jordan Olsommer]

[ QUOTE ]
if he knew you had the set, calling is incorrect in this situation, stacks weren't deep enough. but he probably thought you had top pair and if he felt you would pay him off then the call is marginally ok.

[/ QUOTE ]

Just for kicks, I decided to fool around with the EV equation for this play just to see how deep a stack you'd need to have for this play to work out

Assuming the person with the set has the gutshot covered in chips (and we'll use x to be the amount of chips the gutshot has going into the turn), 82.98% of the time gutshot would lose his $15 flop bet, 13.32% of the time he'd win 2x+47 (double up through the set plus the 47 that was in the pot to begin with), and 3.7% of the time he'd lose all his chips (-x).

so we'd have
(.8298)(-15) + (.1332)(2x+47) + (.037)(-x)

and we can find out what the break-even point is for this play by setting the equation to 0.

solving for x, we get x ~= 26.97

in other words, if
a) set has gutshot covered in chips
b) set is going all-in on turn no matter what
and
c) gutshot is going all-in on turn only if he hits his draw

gutshot would need to have a stack of only $26.97 to make this a break-even play (and of course any stack more than that makes it that much more profitable)

so you might ask, why did my calculations above show that the play has negative expectation when clearly gutshot had much more than $26.97 in front of him on the turn? Well, apparently whatever hand converter the orig. poster used got the pot size incorrect. Going into the turn, the pot contains $47. Then set goes all-in for $68.50, and gutshot calls, putting in his own $68.50, making for a total of $137 added to the pot on this round, and a total pot of $184. However, the hand history says the pot only contains $115.50, which either means that something went awry in the calculations, or that UTG+1 got to call Hero's bet for free.

In fact, even the pot size going into the flop appears to be incorrect (based on the action, I get $13 in the pot preflop). Orig. poster, do you know what's up with that? There couldn't have been an ante because aside from the fact that one isn't listed, it would've been 0.40 per player (a bit of an arbitrary figure for an ante, not to mention a pretty damn high one relative to the blinds).

In any event, if the pot contains $47 going into the turn and if Set has Gutshot covered, then Gutshot only needs a stack of $26.97 to make calling the pot-sized bet on the flop correct.

[bkholdem]

[ QUOTE ]
Does implied value make it ok to call a pot sized bet on a draw?

[/ QUOTE ]

Yes, sometimes.

Not only do you have the gutshot cards as outs you can also have flush cards as outs (the third one falls and you bet like you have it).

You also may have overcards that are outs if paired (AK on a J10x board).

The more likely villian is to fire an automatic continuation bet on the flop just because he was the preflop raiser the more likely it will be that he will bail if met with resistance.

The harder it is for villian to let go of an overpair the more likely it is to get his stack when you hit.



That is off the top of my head.

[-Skeme-]

[ QUOTE ]
You are 5:1 to hit your OESD. You are getting 2:1 calling a pot-sized bet. That means, if you hit, over the course of the hand, he needs to pay you off more than 3x the bet you just called, to make your call correct.

[/ QUOTE ]

Then Villain made a good play.

[BZ_Zorro]

But I believe your calculations are incorrect again [img]/images/graemlins/tongue.gif[/img]

The numbers looked extremely odd, he just wasn't making enough money with such a short stack on those odds, so I had a look at the equation.

I think the mistake is counting the money he puts in pot as a gain, which it isn't. What we have to calculate is the net money he wins - the net money he loses. The equation is:

The net money he loses by calling the $15 and missing the draw. PLUS
The net money he wins by hitting his draw and holding up until the river PLUS
The net money he loses by hitting his draw and losing on the river.

So, using your numbers, we have:

(.892)(-15) + (.1332)(x+32) +(.037)(-x)

which solves for x to give:
x = $94.8

which is an amount I didn't have in my stack on the flop(x has to be matched).

So the play is negative ev unless we both have flop stacks greater than $94.8, even assuming I would go all in every single time on the turn.

As for the hand history converter, it's got me stumped...I'll have to double check it from now on.

<edit> and I made a mistake as well, he doesn't win x+32, he wins x+17, which makes x even higher <edit>

[BZ_Zorro]

note that x is the size his stack on the flop, and it has to be matched in order to get all of x. i.e. x is the minimum of our two stacks on the flop.

But anyway, what makes this play awful, besides the fact that it's completely -ev anyway, is that you can't assume you'll get a person's whole stack if you hit your straight. That's just absurd. I may have bet tptk or two pair exactly the same way on the flop, it which case I would definitely have checked the turn after a reasonable player smooth called a pot sized bet. I (and most other average players) would have laid down one pair to action, no question, and pondered long and hard on two pair.

This play is only marginally correct with much deeper stacks, and the absolute assumption that he'll get my entire stack if he hits on the turn. That's a fishy assumption to me, barring some incredible read.

[bkholdem]

[ QUOTE ]
note that x is the size his stack on the flop, and it has to be matched in order to get all of x. i.e. x is the minimum of our two stacks on the flop.

But anyway, what makes this play awful, besides the fact that it's completely -ev anyway, is that you can't assume you'll get a person's whole stack if you hit your straight. That's just absurd. I may have bet tptk or two pair exactly the same way on the flop, it which case I would definitely have checked the turn after a reasonable player smooth called a pot sized bet. I (and most other average players) would have laid down one pair to action, no question, and pondered long and hard on two pair.

This play is only marginally correct with much deeper stacks, and the absolute assumption that he'll get my entire stack if he hits on the turn. That's a fishy assumption to me, barring some incredible read.

[/ QUOTE ]

Go post the hand on the mid-high stakes forum (and change all of the stack sizes, bet sizes, etc X10 their acutal sizes) and see what type of feedback you get. Go see if they say "bad call".

P.S. Sticking to strict by the book odd's play in no limit (never bluffing, semi bluffing, etc) against good players is a terrible idea. Do you agree or disagree?

[Jordan Olsommer]

[ QUOTE ]
What we have to calculate is the net money he wins - the net money he loses.

[/ QUOTE ]

That's not correct - when calculating expectation, you use in your calculations the total amount of money you get versus the total amount of money you give.

To prove it to you via a simple example, lets say that Gutshot Boy is tired of chasing crazy draws at the poker tables and foolishly offers you 2-to-1 odds on a coin flip. If we calculated the net money won versus net lost as you are suggesting, the EV would look like this:

(50%)(1) + (50%)(-1)

for an expected value of $0 per wager! Clearly that's absurd, since you are obviously getting the best of it in this case.

So you might say "but if we're talking about net values here, shouldnt the loss amount be 0 and not -1, since his net gain is 0?" - this is also incorrect and can be proven by another absurd conclusion.

Let's reverse the scenario with the coinflip now, and say that you have for some reason offered Gutshot Boy 2-to-1 on a coinflip instead of the other way around. What's your expectation?

.50*1 + .50*0 = .50!

Because using this method, half the time you'll have a net gain of $1 (Gutshot boy gives you $1), and the other half of the time you'll have a net gain of zero, so either this method is incorrect, or your friend is somehow getting the worst of it on both sides of a 2-to-1 coinflip wager [img]/images/graemlins/smile.gif[/img]

So of course the correct way to do EV for this simple wager is

(% of time you win)(amt you win) + (% of time you lose)(- amt you lose)

or

.50*2 + .50*(-1) = a positive EV of .50 cents per wager

so I believe the calculations in my previous post are correct.

[BZ_Zorro]

[ QUOTE ]
Go post the hand on the mid-high stakes forum (and change all of the stack sizes, bet sizes, etc X10 their acutal sizes) and see what type of feedback you get. Go see if they say "bad call".

[/ QUOTE ]

The stack sizes in relation to the call make this play always negative ev, even on the crazy assumption he gets all of my stack every time. The maths show that clearly. That's the end of story as far as I'm concerned, at any limit. But I'll post and see if they know their stuff.

If we both had $200 stacks and my opponent only potted monsters and/or was very aggressive, I would call without hesitation in that situation.

[ QUOTE ]
P.S. Sticking to strict by the book odd's play in no limit (never bluffing, semi bluffing, etc) against good players is a terrible idea. Do you agree or disagree?

[/ QUOTE ] Completely agree. Keeping player notes on predictable players is where most of my money comes from, not fish. I'm not advocating by the book play by any means.

[Jordan Olsommer]

The key in this example isnt that he was a huge dog to make his draw though - it is the assumption that you have a monster hand and will go all-in on the turn no matter what, whereas he will only go all-in if he has you beat.

In other words, when he wins, he wins the maximum. When he loses, he loses the minimum. The only time you win the maximum is when both scenarios coincide (he hits his draw but you outdraw that on the river)

Its pretty much the same reason Doyle evangelizes mid-sized suited connectors in "Super System" (and AK, come to think of it) - because you know exactly where you are with these hands after the flop. So when you win with them, you win big because they're covert and can make some huge hands. On the other hand, when you lose with them, you lose only your limp -in bet before the flop, since you're not going to put an extra dime into the pot if you don't hit the flop pretty hard.

[Jordan Olsommer]

There's another example I just remembered that might help us understand the situation:

In Matt Matros's excellent new book, he talks about how Jesus Ferguson spoke at a rec.gambling.poker banquet/conference/somethingorother and outlined a theoretical problem: you and I are in a full ring game with me in the Sb and you in the Bb, and everybody folds to us. I make a sizeable raise, and in doing so accidentally flip over my hand, which you see (and I am aware that you see) are the two black Aces.

The question is, given that you know that I have A[img]/images/graemlins/spade.gif[/img]A[img]/images/graemlins/club.gif[/img], and that I know that you know it, what hands would you call that raise with?

It's a tough problem, but the answer turns out to be any hand at all. The reason being because you know precisely where I'm at in the hand at all times, since you know what I'm holding, and thus you will be able to bluff me out all the time when the flop misses me, and you'll never play a big pot when I make a monster hand. This is another scenario where you win because my only outcomes are to win the minimum (I get lucky and flop quad aces....only to have you of course fold at the first bet) or to lose the maximum (I catch a big hand but you catch a bigger one). You, on the other hand, can only lose the minimum or win the maximum. So yours is a very profitable position to be in in this scenario.

See what I mean? If we stipulate that you'll go all in on the turn no matter what comes up to protect your hand, then the gutshot draw becomes an extremely powerful weapon, since youll never know when he's made his hand until either a) he folds, in which case you've won only his $15 from the flop, or b) he says "I call" after you go all-in and you lose all your chips.