Posted Tue Jun 15, 2004 5:00 pm GMT by Ninja
Let's say that you're holding 89 of spades, and the flop is 4 of diamonds, 6 of spades, and 7 of spades. Somebody else goes in holding A 4. You have 6 outs for an 8 or 9, 9 outs for a flush, and 6 outs for a straight (one each of the 5 and 10 are of spades). This makes 21 outs. Using the outs rule (X 2 + 2) You should hit one of those on the turn or river 88 % of the time. That's 7 out of 8. But there are more cards in the deck that you will miss if you hit. How does this work out?
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Posted Tue Jun 15, 2004 11:54 pm GMT by ShoelessJoe
3 8's
3 9's
8 straight outs (4 10's + 4 5's)
9 flush outs (A, 2, 3, 4, 5, T, J, Q, K)
That gives you 23 outs out of a possible 45 cards at this point(52 minus your two, the three on the flop, and your opponents two)
23/45=0.511
Then after the turn if one of your outs doesn't come up then there are 44 unseen cards now.
23/44=0.527
Taking into account that your opponent cannot get any of his outs (2 4's or 3 A's) with 45 outs (7/45=0.155) or after the turn with 44 outs (7/44=0.159).
You take your probabilities (0.511+0.527) minus his (0.155+0.159) which gives you about 0.73 or more than a 3 to 1 favorite to win the hand.
Posted Mon Jun 21, 2004 10:31 am GMT by Adam Marshall
It's a little more complicated than that.
First off, Joe, you counted the Ts and 5s twice. Once in the flush outs and once in the straight outs. You should have 21 outs. That's a common mistake whenever you have a straight flush draw.
3 Eights, 3 Nines, 15 Straight and Flush possibilities. 45 cards left on the turn, 44 on the river.
The math would be...
1 - (45-21)/ 45 (44-21) /44]
1 - 24/45 23/44
1 - 600/1980
1 - 0.303030....
0.696969...
or about 69.7%
That's yer chance of hitting one of your outs. It doesn't take into consideration the fact that if you hit an 8 or 9, you still can lose if the other card is an Ace or 4. That doesn't account for a whole lot, but changes it to an overall chance of around 69.0%.
If a rag hits the turn, you still have around 47.7%.
Posted Mon Jun 21, 2004 10:58 am GMT by Dave B
And that is the chance of hitting. If he has A 4 of spades that takes aways all the spades as outs. Also-if instead of A 4 he has and 8 or 9, then your pair outs can make a str8 for him.
The only gauranteed outs are the 8 that give you a str8 (or str8 flush).
Posted Wed Jun 23, 2004 2:29 pm GMT by CamBam
Just for my education....you all seem to agree the odds are figured taking into account the pocket cards of the opponent....so in figuring the odds post flop, you calculate the remaining cards as 52-2 (you) - 3 (flop) -2 (opponent)
I thought the 2 opponent cards were not figured in the odds, since they are unknown....therefore post flop the remaining cards are 47 to use in the calc and not 45
Am I mis-informed? I've picked up my info from reading some poker books on odds ... but maybe not the right ones
Please clarify.....
Thanks
Posted Wed Jun 23, 2004 3:17 pm GMT by ShoelessJoe
Typically in a game you won't figure in your opponents' two hole cards but since in this case we knew the opponents two hole cards we figured them into the odds.
Usually though, unless you're POSITIVE of your opponents' hand, don't figure in their hole cards into the odds.
Posted Thu Jun 24, 2004 4:16 pm GMT by CamBam
Gotcha....thanks
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