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View Full Version : I hate math- Hold'em hand


J.R.
03-24-2004, 11:01 PM
Not that I could do this at the time, but I called, caught a K and paid off a river bet to be shown a J.

Playing 10-20, I raise with KhKd, got called by the SB. 5 sb in the pot.

Flop was Tc 9s 8h. Check, I bet, SB calls. 7 sb

Turn Qd [Tc 9s 8h]. Check, I bet, SB check-raises.

I am getting 6.5-1 to call here (the rake is 3 bucks, so its what, 6.333-1). I know SB has either 2 pair or a J.

If he has a J, (he does not have JJ), I have 3 outs to most likely win the whole pot.

If he has 2 pair, my clean outs are 2 kings and 4 jacks. I also have 6 other outs to make Kings up, but if I make kings up I won't know whether my opponent is full or not. So I have 12 outs but don't know what they all are.

Just trying to figure out:

1) what percentage of time my opponent needs to have 2 pair to make a turn call correct assuming I call the river if I make two pair or a set and bet if I catch a J (assuming my opponent check-calls with a 5 straight on board)?

Thanks.

I keep getting lost because of all of the possible outcomes, how/can I (no math genius) do this math?

If I catch a J, I bet and win 7.5, but I may have 3 or 4 jack outs. (This assumes my opponent check-calls the river, which may be optomistic.)

If I make a set, I win against 7.5 versus 2 pair and lose 2 bets against a J, and have 2 outs here.

If I make Kings up, I win 7.5 against a worse 2 pair, and lose 2 bets against either a straight or a full house, and I have 6 outs here.