Flopped set on a suited flop: Cap of fold?

[kiemo]

I think the flop play is a pretty easy call

But what happens on the turn if Monkey plays on and a blank or another [img]/images/graemlins/diamond.gif[/img] falls?

Does he bet out if checked to him?
How many bets does he call?
One bet to him, call, capped on the way back, what now?

With the huge pot how many bets can he call on the river if he doesnt fill up? 0,1, more?

These are probably just simple pot odds questions, but its late and I am tired.

[spacemonkey57]

Ouch. Thanks for being honest guys. This isn't much of an excuse, but this was actually my third set in about 200 hands and I dropped big pots with the first two. I wasn't even thinking about my boat outs because I was positive the board wouldn't pair. I didn't have any reads, but through my first few hands at the table it looked typical party loose passive, so I didn't think anybody would jam this much with a single diamond. I guess I should have bit my lip and capped the sucker.

I'll save the results for now so that the morning crew can give me more of the berating that I deserve. Anybody have any guesses on the villan's hands?

[SnakeRat]

You probably had the best hand.
I put CO on a set of deuces, the other guys on 2pair or pair flush combos.

If you never folded a set you would not be way off.

[Nottom]

[ QUOTE ]
I put CO on a set of deuces, the other guys on 2pair or pair flush combos.

[/ QUOTE ]

The old put them all on something I can beat theory ... excellent.

[Nate tha' Great]

[ QUOTE ]
I'll save the results for now so that the morning crew can give me more of the berating that I deserve. Anybody have any guesses on the villan's hands?

[/ QUOTE ]

I'll guess roughly 5 [img]/images/graemlins/diamond.gif[/img] 7 [img]/images/graemlins/diamond.gif[/img], T [img]/images/graemlins/club.gif[/img] J [img]/images/graemlins/spade.gif[/img] and A [img]/images/graemlins/diamond.gif[/img] 9 [img]/images/graemlins/club.gif[/img].

[EdSchurr]

And this should be the math, unless I have no clue what I'm doing...

Assume someone has made the flush and won't fold.
Making the boat by the river: !(!15%*!22%)=34%

Raise? 34% win rate, 25% contribution: Yes. +EV
Eg. You bet 1 unit and three people call it. 34% of the time you win 3 units, and 64% of the time you lose 1 unit.
34%*3-64%*1=.38 units

Call? I don't know. How can you accurately predict the future betting? Anyway, if the raise is correct the call can't possibly be greater EV.

[Jimbobobb]

Let's make the math super simple.

Monkey: AA

Board: A B C

On the turn, you have 3 B's and 3 C's and 1 A that can give you a boat/quads. 7/47 which is approx 14.8%

Turn: A B C D

On the river, you have 10 outs to a boat or quads (3 B's, 3 C's, 3 D's, and an A) whish is 10/46 or approx 21.7%

14.8 + 21.7 let's just say is around one third of the time you'll improve to a boat or quads by the river. That means odds are good that you'll win this pot one third of the time. Well, there are 3 people willing to call any bets you make. Like EdSchurr said above me, the situation is clearly +EV to raise, since you will get back in the long run 1.38 bets per bet that you make in this situation. What you're doing here is pushing your pot equity edge. On that flop you can say 'I'll win this 33% of the time'. That's much more than your fair share against a field of 3.

[SnakeRat]

When I flop a set I tend to put my opponents on something I can beat, maybe you are much smarter than me though.

TT would probably raise preflop.
Flopped flushes are extremely rare, even given the flop action.

[Atropos]

I think this call is rather simple, you look at the pot odds, if they are anywhere near 5.6:1 you call, if not you think about the implied odds which are very very big if someone flopped a flush. And I think there is always a 10% chance at party that these players raise and reraise the flop without a flopped flush, so folding would be a very big error in this situation I think.

[EdSchurr]

The only problem I see is adding the probabilities. I'm not 100% sure I'm right, but here is the reasoning for the way I did it (some discrete math required...though...I'll explain it):

You want the probability of hitting on the turn and/or hitting on the river. The math function you want is "or".

So (x)OR(y)=the percent of hitting on the river from the flop.

You can't just add to do OR. There is, in fact, no really easy way to do it. But AND is easy -- you just multiply the terms together -- and we can make OR look like AND by DeMorgen's rule/law/something.

So we negate the equation twice, because !(!(1)) is still 1. (I'm using the ! symbol to be negation, from comp sci)

"!" by the way is the same as doing 1-x. !(!(1)) is 1-(1-(1)) which is 1.

The reason we negate the equation is because negating an OR produces an AND. So !((x)OR(y)) is (!x)AND(!y).

So, with some rough numbers, !(!(20% OR 30%)) is !(80% AND 70%). To do the AND function, try .8*.7 which is .56. So we have !(56%), which is finally 34%. I'm not sure about the best way to do calculations on the fly. Practice I guess. There are usually approximations that exist but I don't know them.

And so concludes our discrete math for poker lesson, which may or may not be wrong, and which may or may not be applicable. [img]/images/graemlins/crazy.gif[/img]