View Full Version : Less Straightforward Holdem Pot Odds

09-07-2003, 10:01 AM

I'm just trying to evaluate some less clear-cut pot odds situations.

* Firstly I have a query about when you flop a 3 straight and 3 flush on the flop. I assume the way to work this out is 1/5x/15 + 1/5x1/5 = 0.08. What does this number mean? Does that mean i need pot odds of about 12:1 to continue?

* I'm also wondering what to do with those situations where an opponent has a hand but potentially not as strong as yours. Do you factor in their chances of improvement also?
Say for example, I have two overcards (AK) on the flop and they have flopped a small pair holding 65. I have 6 outs at beating their hand, but there is also a chance that they may improve to 2 pair or trips with their 5 outs.

Another example, I have a four flush on the turn versus a set. They have 10 outs to improve to quads or a boat. I have 9 outs to get my flush.

I would appreciate any advice on the above situations.



09-07-2003, 02:36 PM
* Firstly I have a query about when you flop a 3 straight and 3 flush on the flop. I assume the way to work this out is 1/5x/15 + 1/5x1/5 = 0.08.

[/ QUOTE ]

Not exactly. You need a little more information to go on. To get an exact answer one would need to know what cards, since you may pick up an inside draw or outside draw on the fourth card, but an outside draw may be ruled out under certain circumstances.

Likewise, you don't want to double count a card that makes your flush as one tht makes your straight.

You might be able to estimate this as (8/47)*(4/46) + (10/47)*(9/46) - (2/47)*(1/46) =~ 0.056 or 5.6%

I'm sure someone will come correct my mistakes /images/graemlins/crazy.gif

09-08-2003, 07:43 AM
Re the 2d part of your question, absolutely, you have to factor in your opponents chances of improvement. This step is almost always omitted in discussions of "outs" because there is an implicit assumption that if you hit you have the best hand. (The one time it is discussed is when you are drawing to a straight, but one of your straight cards is also a likely flush card for your opponent, so you reduce your outs by one).

To put it a little more mathematically, "pot odds" is just the algebraic equivalent of the "break even" or "premium" calculation. The simplest pot odds calculations are when you are deciding whether to call a bet made by the single remaining opponent. If x is the probability of winning the hand, you break even when


to convert a Probability to odds needed(where the bet is 1), Odds=(1/P)-(1/1) (eg if P=1/3, Odds=3/1-1/1=2/1)

so for the above,

odds = (pot+2*bet)/(bet)-bet/bet=(pot+bet)/bet...the familiar form.

But we got there by looking for "x" the probability of winning. That encompasses not only the fact that we make our hand, but that it will be good when we do make our hand, and (less commonly thought of) that we arent already ahead. Ie x is a special case of

P(we are ahead and stay ahead)+P(we are behind but move ahead)=

P(ahead)*[P(neither improve)+[1-P(he improves to the lead)]+[1-P(ahead)]*P(we improve to the lead)

In the usual pot odds determination, we think we are behind, ie P(ahead)=0 so the above reduces to P(we improve to the lead), which, if I werent so tired, could be expanded to terms for P(make hand)*(1-P(he also makes hand)), ie the straight draw vs flush draw situation.