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amtannn
02-12-2003, 02:21 PM
If you are playing hold em, and after the flop you have an open ended flush draw (giving you 9 outs). But your opponent has two pair (giving him 4 outs to beat your flush), how do you calculate the odds that you win? I know there is a 35% chance to hit the flush, but how do I factor in the fact that there is a 16% chance my opponent will beat that???

BruceZ
02-12-2003, 09:16 PM
Assume there are 2 suited cards on the flop. If he has 2 pair with the 2 suited cards, then you have 9 outs. If he has 2 pair with the unsuited card, then you only have 8 outs since one will fill him up or he has it in his hand. Assuming you have the full 9 outs, then the probability that you hit your hand by the river is actually 36.4% rather than 35% since there are only 45 unknown cards instead of 47 since you know he doesn't have one of your suit. 1-(36/45)(35/44) = 36.4%. Now the probability that you hit your hand on the turn and he hits his hand on the river is the product of these two probabilities which is (9/45)*(4/44) = 3.3%. Then the probability that he hits on the turn and you hit on the river is also 1.8% This probability is only significant if you won't be able to get away from your hand if he hits on the turn. 36.4% -1.8% - 1.8% = 32.8% for the chance of hitting your flush and winning.

You can also compute this with poker software. Using www.twodimes.net/poker (http://www.twodimes.net/poker) gives 33.13%. The small difference is due to the possibility of the flush draw making trips.

<pre><font class="small">code:</font><hr>
Running: pokenum -h kc 6c - ah 9h -- ac 9c 2d :
Holdem Hi: 990 enumerated boards containing Ac 9c 2d
cards win %win lose %lose tie %tie EV
Kc 6c 328 33.13 662 66.87 0 0.00 0.331
Ah 9h 662 66.87 328 33.13 0 0.00 0.669
</pre><hr>

amtannn
02-13-2003, 08:28 PM
What still doesnt make sense, is that the other player's 16% chance of making his hand only lowers my odds by about 3%? Can you explain that? I thought that it would make more sense to calculate the odds by multiplying 35% (or 36.4% which represent the odds of making my hand) by 84% (the odds that the other player doesnt make his hand) for a total chance of 29% to win. Why is my calcualation wrong? What logic am I missing?

BruceZ
02-13-2003, 08:51 PM
I thought that it would make more sense to calculate the odds by multiplying 35% (or 36.4% which represent the odds of making my hand) by 84% (the odds that the other player doesnt make his hand) for a total chance of 29% to win.

You can't multiply these numbers because they are not independent. If you make your hand on one of the two cards, it cuts his odds of making his hand since now he has to make it on just one card. The odds that he doesn't make his hand when you make your hand on one of the cards increases to 91%. Now you can multiply 36.4% times 91% get 33.1%.

the other player's 16% chance of making his hand only lowers my odds by about 3%? Can you explain that?

It's only 16% of the 36.4% of the time that you make your hand that matters (which is 6%), and then it's only half of that or 3% because when you make your hand it cuts the chance of him making his hand by half to 8%. That's why the concern about about making your hand and getting beaten by a bigger hand is sometimes (but not always) overstated, especially heads-up.

amtannn
02-13-2003, 10:25 PM

02-16-2003, 09:27 PM
What if instead of Ac 9c 2d, the board is Ac 2c 9d? The difference being that the 9c can come, hitting you both. I've always found that this is where it gets really expensive. Can you run the numbers on that? Thanks.

BruceZ
02-17-2003, 12:30 AM
Now the probability that you hit your hand on the turn and he hits his hand on the river is the product of these two probabilities which is (9/45)*(4/44) = 3.3%.

Make that 1.8%. The rest is ok.

BruceZ
02-17-2003, 01:20 AM
That's the other case where he makes 2-pair with the unsuited card giving you 8 outs instead of 9. Calculating as before, your chance of making a flush without the 9c becomes 1-(37/45)(36/44) = 32.7%, and the chance that he makes a full house on the other card when you make a flush without the 9c is
2(8/45)(4/44) or 3.2%. The 2 is there since he can make it on either card. Don't think you can generally multiply by 2 to get your odds for 2 cards to come. It is exact here because we are talking about each player making their hands on different cards, and it is the same if you swap the order of the turn and river. So the odds that you make your flush and he doesn't make his hand is 32.7% - 3.2% = 29.5%. The poker simulator verifies this, the small difference again being due to trips. Of course if you think you will lose more money, then you need to make a bigger adjustment in your required odds than this difference would indicate.
<pre><font class="small">code:</font><hr>
pokenum -h kc 6c - ah 9h -- ac 9d 2c
Holdem Hi: 990 enumerated boards containing Ac 2c 9d
cards win %win lose %lose tie %tie EV
Kc 6c 298 30.10 692 69.90 0 0.00 0.301
Ah 9h 692 69.90 298 30.10 0 0.00 0.699</pre><hr>