Overpair on a scary board

Lucky_River · #1 10-12-2006, 02:56 PM

PokerStars 1/2 Hold'em (10 handed) Hand History Converter Tool from FlopTurnRiver.com (Format: 2+2 Forums)

Preflop: Hero is MP1 with T[img]/images/graemlins/diamond.gif[/img], T[img]/images/graemlins/heart.gif[/img].
2 folds, UTG+2 raises, Hero 3-bets, 6 folds, UTG+2 calls.

Flop: (7.50 SB) 8[img]/images/graemlins/club.gif[/img], 9[img]/images/graemlins/club.gif[/img], 2[img]/images/graemlins/club.gif[/img] (2 players)
UTG+2 checks, Hero bets, UTG+2 calls.

Turn: (4.75 BB) 4[img]/images/graemlins/club.gif[/img] (2 players)
UTG+2 checks, Hero

what is the plan? Bet to protect the Overpair (against Overkards) and after a raise call down or bet/fold
or is the best way to play it c behind and call a bet on the river?

mrcunningham · #2 10-12-2006, 03:01 PM

I'd love to hear from the more math minded what the probability is that an individual holding two cards has at least a single suit when the board shows 4 of them, because it seems that information is necessary to know how to proceed.

Anybody?

Xhad · #3 10-12-2006, 03:03 PM

If you're checking behind here, it's to fold the river UI.

If you're betting here, you should insta-fold to a raise.

I bet here but really, every option makes me want to throw up.

bungyrocks · #4 10-12-2006, 03:15 PM

i concur. bet and puke when raised

Lucky_River · #5 10-12-2006, 05:18 PM

and after puke. Easy fold or call down?

bungyrocks · #6 10-12-2006, 05:38 PM

IMO fold.
edited to say that reads would help here, but without an extremely solid read on the villian i dont think twice about folding here.

wardnine · #7 10-12-2006, 05:59 PM

[ QUOTE ]
I'd love to hear from the more math minded what the probability is that an individual holding two cards has at least a single suit when the board shows 4 of them, because it seems that information is necessary to know how to proceed.

Anybody?

[/ QUOTE ]

Here's an attempt (pretty easy problem but haven't done a probability problem like this since college so may not be correct):

P(at least 1 card is of a given suit) = 1 - P(neither card is of given suit)

P(neither card is of given suit) = P(1st card isn't of given suit)*P(2nd card isn't of given suit given that the first card wasn't of given suit) = (37/46)*(37/45) = .661

So P(at least 1 card is of a given suit) = 1-.661 = .339

So there is a 33.9% that your opponent has at least one card that is a club given that 4 clubs are on the board and neither of your own two cards are clubs. I think my math is correct, but I'm not sure. Also, I think you have to adjust the 33.9% up a few percentage points given that your opponent made the flop call with 3 clubs on the board...

Little_blue · #8 10-12-2006, 07:05 PM

for the math..I think it is simply 1-(37/46 * 37/45) 1 - it comes to about .34 or 1 out 3 of having a club with 2 random card and 4 clubs on the bored.

Nick Royale · #9 10-12-2006, 07:18 PM

[ QUOTE ]
for the math..I think it is simply 1-(37/46 * 37/45) 1 - it comes to about .34 or 1 out 3 of having a club with 2 random card and 4 clubs on the bored.

[/ QUOTE ]
The correct calc (given that we hold 2 non clubs) is:
1 - (37/46)*(36/45)

But given his flop call the probability is higher than that.

Anyhow, I bet/fold without puking or anything.

JJH3984 · #10 10-12-2006, 11:01 PM

I bet/fold here... I don't think its really that close.