QQ utg check raised by 2

[Nfinity]

I see a lot of people wanting this shown down, even UI, so I wanted to extend it to a precarious positition.

Lets say you call, CO folds and SB calls.

River is a complete Blank.

SB checks, BB bets, Hero????

SB has shown the ability to CR here, and he's going to at least call. The Man in the Middle situation really gets to me sometimes, so I'd like to see what others have to say.

[Dave G.]

[ QUOTE ]
SB has shown the ability to CR here, and he's going to at least call. The Man in the Middle situation really gets to me sometimes, so I'd like to see what others have to say.

[/ QUOTE ]

Call. He's not guaranteed to check-raise, and if BBs turn 3-bet doesn't scare him off, then we can safely fold to it if he does. There is however no reason for preemptively folding such a strong hand.

[benkath1]

River: (19.50 BB) A[img]/images/graemlins/heart.gif[/img] <font color="#0000FF">(4 players)</font>
SB checks, <font color="#CC3333">BB bets</font>, Hero calls, CO calls, SB calls.

Final Pot: 23.50 BB

SB has K [img]/images/graemlins/diamond.gif[/img]K [img]/images/graemlins/spade.gif[/img] [img]/images/graemlins/smirk.gif[/img]

BB has A [img]/images/graemlins/club.gif[/img]6 [img]/images/graemlins/club.gif[/img]

Thanks for all the responses. I wasn't sure if this was a showdown hand or not. But like they say, 'If you're not losing a lot with your sets, you're not playing them right'.

I considered folding the river, but the pot was just a bit too big for that.

[xenthebrain]

You are getting 2:13.5 or 1:6.5 here. Where is that close? You have a IMO clean ten outer.
If I miss I would also call for one bet with top freakin set here on the river.

[benkath1]

[ QUOTE ]
Why are you discounting your outs in case someone has a set?

[/ QUOTE ]

Because I'm trying to improve my hand. If someone else flopped a set of 3's then they have 2 of my outs to improve. I know I have top set, but I also know I am behind a flush (95%) and need a boat to win.

[Dave G.]

[ QUOTE ]
Because I'm trying to improve my hand. If someone else flopped a set of 3's then they have 2 of my outs to improve. I know I have top set, but I also know I am behind a flush (95%) and need a boat to win.

[/ QUOTE ]

This is an incorrect way to view your outs. You don't know what your opponents have. In fact, it can be proven that this doesn't even matter; your odds don't change at all.

It's 4am and I'm in a mathematical mood, so I don't know if this post is going to be useful for anyone, but I feel like doing it anyway. So here goes.

I'll use the more simple example of a flush draw. The flop comes with 2 [img]/images/graemlins/diamond.gif[/img]'s. You hold 2 [img]/images/graemlins/diamond.gif[/img]'s, giving you a flush draw. There are 2 people in the pot against you (for simplicity I'll limit it to two; the calculations get too long for more than that).

There are 47 unseen cards, and 9 unaccounted for [img]/images/graemlins/diamond.gif[/img]'s. We say you have 9 outs. 9/47 = 0.191489 which is your probability of hitting a diamond on the turn.

With 2 other people, there's a decent chance that someone has one of your [img]/images/graemlins/diamond.gif[/img] outs. Lets say we take this route, and assume that one of our opponents does indeed have a [img]/images/graemlins/diamond.gif[/img], and we discount our outs to 8. Once you do this, you are no longer dealing with a deck that has 47, or 46 unseen cards. That is, we don't just calculate 8/46 and that's our chance of hitting our diamond on the turn. We are considering the odds of all our opponents cards when we say that someone must have a diamond. So the new number of unseen cards, that is, the cards we are free to draw at, is 43. (2 opponents, 2 cards each).

Here's the correct way to do the new calculation. The probability that an opponent holds 0 diamonds is 38/47*37/46 = 0.6503. We'll call this p0. p2 (someone holds 2 diamonds) = 9/47*8/46 = .0333 and p1 = 1 - p0 - p2 = 0.3163.

Call m the probability of there being a certain number of total diamonds between our two opponents. If there are 0 diamonds between our opponents, there are still 9 diamonds left in our new 43 card deck, and the probability that neither opponent has one is equal to p0*p0. So, m0 = 9/43*p0*p0. (Neither opponent has a diamond).

Similarly,

m1 = 8/43*p0*p1 + 8/43*p1*p0. (Either player has a diamond, but the other does not).
m2 = 7/43*p0*p2 + 7/43*p1*p1 + 7/43*p2*p0 (Either one player can have two diamonds, or both players have one).
m3 = 6/43*p1*p2 + 6/43*p2*p1 (one player has one and the other has 2)
m4 = 5/43*p2*p2 (both players have 2 diamonds)

Now we have a list of probabilities. If we calculate m0+m1+m2+m3+m4, which is the total probability of all of these events, we get a probability of 0.191489. Which, as it turns out, is exactly the same as the initial probability with 47 unseen cards - 9/47.

It doesn't matter if you think your opponent has X card in his hand. Your odds of drawing do not change, because he will only have X card in his hand with some probability. And by considering that someone has that card in their hand, you have to take your opponents cards out of the deck when calculating odds.

Edit: Had p1 and p2 probabilities around the wrong way.

[arsixsixwy]

Fantastic post. Analytical players like me eat this stuff up.

[nomadtla]

I think you've got the odds for a call (esspescially implied). Though I'm fairly certain your up against a flush. I may pay one off UI on the river but deffinately not 2.

[benkath1]

OK. After 3 reads I think I got it. Thanks for the math. I hadn't looked at it that way.

[McGahee]

Call and spike a T.