Pot Odds Calculation - Help! - Need trips or two pair
 

So I'm trying to calculate the pot odds for the following situation: I have two low cards in the hole and I'm against one caller (suppose I was BB or on an ill-advised bluff). I suspect they have a higher pocket pair, and on the flop I hit one pair, which still can't beat his pocket pair on its own. I either need to make trips, or get two pair.

I can calculate the chances of hitting trips or two pair easily enough: I have 5 outs, and 100% - (42/47 x 42/47) = 20% that I'll hit one of my savior cards. (Please correct my math if I'm wrong.)

I have three questions from this starting point:

(1) How do I account for the risk that he'll hit HIS third card?

(2) Are the risks of hitting quads, full house, straight, flush, etc. significant enough to warrant taking that into consideration from this point?

(3) Is this really an appropriate pot odds calculation in the first place, since you may have to bet/call/raise after the turn as well?

Thanks...

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

If it's given that he has pocket aces and you have 7-5 off-suit , and you wish to avoid an ace on the flop . Then the probability you hit two pair or trips is :

6c2*40/48c3 ~ 3.4690%

2) This is low enough to indicate that it's usually a bad idea to play for a 3.5% chance of hitting two pair or trips .

3) You can say something like , 3.5% of the time I will win more money post-flop then I will lose on average . In general , it's not a good idea to call raises pre-flop when you're only hope is to hit a low percentage play like two pair or trips .

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

don't try to make it too complicated. if u have 5 outs just multiply that number by 4 and that's ur chance of winnin by the river (5 outs times 4 = 20%). forget about whether your opponent hits or not. the 2-4 rule is more than enough for any practical calculation in hold em.

so now that u know ur chance of winnin (1 out of every 5) it means u need 4 to 1 on ur money to make a breakeven call.

EXAMPLE:

hero: 8s7s
villain: KcKh
flop: 8c Th 2d
pot: $1000
villain bets $250 and is all in

the pot now is $1250 and u have to call $250. 1250/250 = 5 so u r gettin 5:1. u only need 4:1 to breakeven (from previous calculation) so makin a call here is profitable in the long run.

hope this helps.

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

[ QUOTE ]
(3) Is this really an appropriate pot odds calculation in the first place, since you may have to bet/call/raise after the turn as well?

[/ QUOTE ]

In such a case you would calculate Implied Odds which can get complicated.

For example:

You have $750 and 7 [img]/images/graemlins/diamond.gif[/img]8 [img]/images/graemlins/diamond.gif[/img]
Villan has $750 and A [img]/images/graemlins/spade.gif[/img]A [img]/images/graemlins/heart.gif[/img]

Flop: 7 [img]/images/graemlins/club.gif[/img]3 [img]/images/graemlins/club.gif[/img]J [img]/images/graemlins/spade.gif[/img]
Pot = $300

Villan bets $200
Pot Odds = 2.5 to 1

You have 5 outs to make your hand (2 pair or trips) which is 4 to 1. So you are not getting correct pot odds to call. However, you must take implied odds into account.

Implied Odds = [($ in the pot + $ villian will put in if you hit you hand) / ($ you have to call now + $ you will have to call in the futre if you don't hit on the turn)]

2 situations to consider...

Implied Odds = [($500 + $550) / ($200 + $0*)] = 5.25 to 1

*Assumes if you don't hit on the turn AND villan bets you will fold OR you don't hit on the thrn AND villan checks so you can see the river for free. Also assumes villan will call your all in bet if you hit.

Implied Odds = [($500 + $550) / ($200 + $100*)] = 3.5 to 1

*Assumes if you don't hit on the turn AND villan bets $100 AND you call. Also assumes villan will call your all in bet if you hit.

SO........the 2 situations you consider average to 4.38 to 1. So you decide to make the marginal call...

Turn: 2 [img]/images/graemlins/heart.gif[/img]
Pot = $700

Villan bets $100 (like you assumed?)
Pot Odds = 7 to 1

You have still 5 outs to make your hand (2 pair or trips) which is now 9 to 1. So you are not getting correct pot odds to call. However, you must still take implied odds into account.

Implied Odds = [($700 + $450) / ($100 + $0*)] = 11.5 to 1

*Assumes if you don't hit on the river AND villan bets anything you will fold. Also assumes villan will call your all in bet if you hit.

Therefore, you are getting the right odds to call the bet on the turn.

NOTE: This is a very specific example used to prove a point. Please be aware that the ammount that villan bets will GREATLY affect your odds most likely in a bad way (i.e. he bets more than $100 on the turn, which would be more likely)


Hope this helps...

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

[ QUOTE ]
(1) How do I account for the risk that he'll hit HIS third card?

(2) Are the risks of hitting quads, full house, straight, flush, etc. significant enough to warrant taking that into consideration from this point?

(3) Is this really an appropriate pot odds calculation in the first place, since you may have to bet/call/raise after the turn as well?

Thanks...

[/ QUOTE ]
(1)He has two outs for his third card. But he only has one chance at it since the other card would be your out. One card is about 4.5%

(2)Again, one of the two remaining cards to be dealt has to be yours, so it can make him quads and you will have to determine in one of your outs could help his hand. We need a specific example for this. It can range from not at all to several of your outs help.

(3)Implied odds are much more important than pot odds in NL holdem (unless you are in an all-in scenario). However, if you need the pot odds to make a decision, most decisions lie either well above or well below the odds.

(4)You missed question (4) which is a scenario where villain can also get two pair which are higher than your two pair. As well, the board can pair and counterfeit you sometimes.

There just aren't that many times where a decision is so close that you need to consider what you are asking. And on top of that, you don't know for sure what villain is holding so that introduces another variable which may be much greater than somebody hitting after you hit.

You also need to look closely at your own hand. You may have a backdoor draw. You also need to look at counterfeiting when you have two pair that are not top two pair.

In general, when you are in live play and faced with scenarios like this, you are usually thinking about the big picture like "I have one pair and a backdoor draw. Against an overpair, I have 5 outs. I am ahead of a flush draw. Against the same pair with a better kicker, I have 3 outs". At the table, that is usually all you have time for.

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

Yes he just has 2 outs to make trips but he has quite a few outs to make a better 2 pair than the OP can draw to. With AA on a flop like 792 with OP having 57, OP has 5 outs to improve to 2 pair or trips, villain has 6 outs to make better 2 pair, 2 outs for trips. If both brick 4th, villain now has 9 outs to make better 2 pair, 2 to make trips.

Spend some time and work out a tree of possibilities and the percentages of each. You'll see it's actually pretty bad for you... the tree is something like...

turn:
* OP makes 2 pair - how often does this happen, how often does it hold up
* OP makes trips - how often does this happen, how often does it hold up
* OP bricks, opponent improves (OP folds, I would hope)

river:
* OP improves - how often does this happen (we'll always win, ignoring straight/flush)
* villain improves - we lose

Sorry, I'm kind of in a hurry, so I can't flesh this out.

Re: Pot Odds Calculation - Help! - Need trips or two pair
 

I'll respond to (1) only.

First off, if you are crediting him with a pocket pair, then there are 45 unseen cards rather than 47. The approximation for the probability you will catch one of your n outs with 45 unseen cards is:

P = 4.44*n - nC2/10 = 4.44*5 - 5C2/10 = 21.2%

To adjust for r redraws, you use the following simple approximation:

Q = n*r/10

This works for 46 unseen cards too. For 47 unseen cards, the divisor is 11.

However there is a complication. The number of redraws varies depending on whether you make trips or not. Your opponent can make a higher 2 pair by pairing one of the 2 flop cards you are not paired with. The calc for Q must be split in 2:

Q = 2*2/10 + 3*8/10 = 2.8%

The 1st half says that for 2 of your outs that make trips, your opponent has 2 redraws that make higher trips. The 2nd half says that for 3 of your outs that make 2 pair, your opponent has 2 + 3 + 3 = 8 redraws making trips or 2 pair.

With this adjustment we can now correct your winning chances to be:

P' = P - Q = 21.2 - 2.8 = 18.4

Using poker stove with a hand like 7s6s vs AsAh and a flop of Kh7c2d gives 18.384%.

If you had blindly used the rule of 4 and ignored the redraws e.g. with 5 outs 4*5 = 20%, you would have been close enough for most purposes. The smaller number of unseen cards partly offsets the effect of the redraws. However, with a larger number of outs and/or redraws the difference can be more significant.

Note that my calcs and example hand sidestep the further complications of backdoor draws.