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Old 10-28-2004, 02:39 PM
Vanquish Vanquish is offline
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Default Hold\'em Odds

Hey people, how's it going...

So I've heard and read (just on this board) conflicting information on hold'em odds and how to calculate them. I'm looking for some solid information or at least a discussion of which way is probably the right way. What I'm mainly concerned with is decision making post-flop in NL games. Ya know, do I have the odds to call an X sized bet and chase my flush draw, etc. etc.

Here's how I learned it, and this may be wrong... you can tell me. Btw, I'm most comfortable with percentages rather than odds since I don't gamble outside of poker and percentages is what I learned in math class.

Post-Flop betting:
Chances you'll make your hand on the turn: outs/47
Chances you'll make your hand on the turn or river: This is the one that is confusing, so I'll discuss it below.

Post-Turn betting:
Chances you'll make your hand on the river: outs/46

Okay... Chances you'll make your hand on the turn OR river. I have read a variety of poker odds web sites and looked at a number of odds calculators, and there are definitely different ways people are calculating the odds for this. My instinct is to go with math, and not to trust sites that base odds on however many simulated hands. I mean, why go with a simulation when I can calculate precisely how often that third Ace I need is going to come up?

This is my understanding of the statistics behind poker odds:
If you want to know the chances of events X AND Y happening, like catching a runner-runner flush, you multiply the chances of each happening individually. Here, it would be (10/47)*(9/46), which is 4.17%.

What I don't get exactly is how to calculate the chances of event X OR event Y happening. For example, what are the odds of completing an inside straight draw by the river? The chances of catching it on the turn by itself is (4/47), and the chances of catching it on the river if you missed on the turn is (4/46). How do you put those together? You don't multiply. Adding doesn't seem to make sense. I saw one site that calculated the odds of you NOT getting the cards you need, since that can be an "AND" situation and you can multiply. It was like this:

Odds you'll complete your inside straight draw:

Start with odds you won't hit on turn AND won't hit on river: (43/47)*(42/46) = 83.53%

Subtract this number from 100% to find out chances that this WON'T happen (i.e. chances you'll have you straight by the river): 100% - 83.53% = 16.47%

Is this correct? Is there another way? I thought this was the right way, but I'll show one last example where someone else's numbers contradicted mine.

In the MTT forum, there was an all-in situation where it was basically

A[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/club.gif[/img] against 3[img]/images/graemlins/spade.gif[/img] 3[img]/images/graemlins/club.gif[/img]

The flop was: 2[img]/images/graemlins/diamond.gif[/img] 4[img]/images/graemlins/heart.gif[/img] 5[img]/images/graemlins/diamond.gif[/img]

In the thread, the guy with As said he had about a 75% chance to win and backed it up with numbers from http://www.twodimes.net/poker/. But the guy with 3s had only four outs (the sixes, since the other two threes would give As a straight) or he would lose. According to the statistics as above, that means he only had a 16.47% chance to catch a winning card, making the guy with As about a 83% favorite. I say about because I get that if both a 3 and a 6 came then they would tie with a straight to 6.

So... okay, I've laid out my thought process and some examples. Tell me why I'm right or why I'm wrong, and thanks very much for doing so.
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  #2  
Old 10-28-2004, 02:46 PM
Cleveland Guy Cleveland Guy is offline
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Default Re: Hold\'em Odds

In the thread, the guy with As said he had about a 75% chance to win and backed it up with numbers from http://www.twodimes.net/poker/. But the guy with 3s had only four outs (the sixes, since the other two threes would give As a straight) or he would lose. According to the statistics as above, that means he only had a 16.47% chance to catch a winning card, making the guy with As about a 83% favorite. I say about because I get that if both a 3 and a 6 came then they would tie with a straight to 6.

Just to clearify - he does have 6 outs, because the 2 As also give him a straight. So he now does have 6 outs.

6 outs is about 23%, plus he could hit runner runner3, or 3-5 or 3-4, giving him 4 of a kind or a full house.


As for the rest of your question, I am not a math wiz, so I go to the ultra basics. You have almost a 2% chance to hit any card in the deck. So give yourself 2% per out, per card remaining.

6 outs - with 2 cards remaing is about 24%.

That's pretty close to the 75/25 the guy with AA said he was.
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  #3  
Old 10-28-2004, 03:05 PM
jakethebake jakethebake is offline
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Default Re: Hold\'em Odds

There's an article called "Pot Odds Made Easy" by Krieger on cardplayer.com. I don't have the link, but I think it came out in March.
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  #4  
Old 10-28-2004, 04:51 PM
BarkingMad BarkingMad is offline
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Default Re: Hold\'em Odds

[ QUOTE ]
I saw one site that calculated the odds of you NOT getting the cards you need, since that can be an "AND" situation and you can multiply

[/ QUOTE ]

Thats it. You can also use combinations to solve this problem (see pg 103 of Mike Petriv's Hold Em's Odds Book).

Here's why you have to use the odds of not completing the draw when using probabilities.

Addition Rule: To find the chances that at least one of two things will happen, check to see if they are mutually exclusive. If they are, add the chances.

Multiplication Rule: The chance that two things will both happen equals the chance that the first will happen, multiplied by the chance the second will happen given that the first has happened.

If you're looking to fill a draw on the turn or river, you can tell by it's definition that the addition rule doesn't apply because the draw could be filled on each street (the events are not mutually exclusive). You can also tell by it's definition that multiplication rule doesn't apply if you're using the odds of filling the draw in each case. By using the odds of not filling the draw, the requirements of the rule are satisfied. Then you can subtract the result from 1 since the chance of something (filling the draw) equals 100% minus the chance of the opposite thing (not filling the draw).

The method of using the odds of something not happening is the correct way to solve alot of probability problems.

Here's the same thing you did, but for flush draws.

Correct way:
1-(38/47)*(36/46) = 35% or 1.86 to 1

Note: The above result includes the chances of getting runner-runner flush cards. The chances of flopping a four flush and then getting only one of your needed suit on the turn or river is 31.6% or 2.16 to 1. I just learned this from Petriv's book, he uses combinations to calculate it.

Incorrect way:
(9/47)+(9/46) = 38.7% or 1.58 to 1

-Lance
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  #5  
Old 10-28-2004, 04:56 PM
jtr jtr is offline
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Default Re: Hold\'em Odds

Hi, Vanquish.

Good questions.

[ QUOTE ]
My instinct is to go with math, and not to trust sites that base odds on however many simulated hands. I mean, why go with a simulation when I can calculate precisely how often that third Ace I need is going to come up?

[/ QUOTE ]

As someone who basically does simulations for a living, I can appreciate your scepticism here. But rest assured that simulating deals (as opposed to realistic play) in poker is pretty easy and a tool like pokenum is effectively doing the same thing as the enumeration you do in a pen-and-paper probability calculation.

[ QUOTE ]
I saw one site that calculated the odds of you NOT getting the cards you need, since that can be an "AND" situation and you can multiply. It was like this:

Odds you'll complete your inside straight draw:

Start with odds you won't hit on turn AND won't hit on river: (43/47)*(42/46) = 83.53%

Subtract this number from 100% to find out chances that this WON'T happen (i.e. chances you'll have you straight by the river): 100% - 83.53% = 16.47%

Is this correct? Is there another way?

[/ QUOTE ]

This is exactly the right way to go about it. Don't let anyone convince you otherwise. One small caution: don't say "odds" if you're going to work with stuff like 43/47 in your working above. This is a probability (in this case the prob. of not hitting your gutshot on the turn) -- it's not an odds ratio. Odds ratios are the probability of an event happening, divided by the probability of its not happening. So heads on a fair coin is 50% as a probability, but 1 (or 1:1) as an odds ratio. I don't mean to be a geek about this, but there's a real danger of confusion if aren't clear about which one you're using.

Also, the popularity of odds ratios over probabilities in poker reasoning is not an accident. Odds ratios make it very easy to compare the return the pot is potentially offering you on your bet, and the likelihood of your draw hitting.

Finally, I think the disagreement you had in your example was a disagreement about counting outs, not a disagreement about how to calculate probabilities.
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  #6  
Old 10-28-2004, 05:22 PM
BarkingMad BarkingMad is offline
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Default Re: Hold\'em Odds

[ QUOTE ]
A A against 3 3

The flop was: 2 4 5

In the thread, the guy with As said he had about a 75% chance to win and backed it up with numbers from http://www.twodimes.net/poker/. But the guy with 3s had only four outs (the sixes, since the other two threes would give As a straight) or he would lose. According to the statistics as above, that means he only had a 16.47% chance to catch a winning card, making the guy with As about a 83% favorite. I say about because I get that if both a 3 and a 6 came then they would tie with a straight to 6.

[/ QUOTE ]

Poker Stove says aces are a 75.445% favorite. Calculating showdown %'s can be tricky. At first blush it looks like the 3's have 6 outs to win, making the aces a 75.9% favorite to win.

The reason Poker Stove's answer slightly differs from a simple "outs" answer is because it's considering all of the more complex instances, like where both a 3 and a board pair card will come, giving the aces a straight but a full house to the 3's.

-L
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  #7  
Old 10-29-2004, 12:45 AM
Vanquish Vanquish is offline
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Default Re: Hold\'em Odds

You're absolutely right Cleveland about another Ace hitting giving him the straight. And after reading through these other posts I can see that I was over simplifying things a little bit. Yes I overlooked that the turn card can create additional outs for each player on the river, and from there I can see the value of a hand simulator. I'm sure I could sit and calculate the precise probability (yeah, saying "odds" only confuses things) that each would win that hand, but that's the problem. To get the exact number you would have to calculate the probability for each unique hand/board combination, and that's just way too much work. So I recant... I see the value of the hand simulators, but also their limitations, and you can correct me if I'm wrong jtr. They're good for after the fact analysis, but not so much for checking pot odds during play since they are necessarily card dependent?

So thanks to barking for confirming my math, and that seems to be what I'll need to stick with for my pot odds/calling guidelines. I also see the value of the odds ratio format for this rather than using probability, but I'm having a hell of a time mentally looking at a figure like 13.5:1 and quickly figuring out how large I bet I can call given the size of the pot. Anyone else have trouble with this? How do you approach it mentally? Is it: Look at the bet size, multiply it by 13.5. If the pot is larger than what you just calculated, then you have pot odds to call? It might just be me, but that's not as natural a thing for my brain to do... I might have to start carrying an odds sheet around with me. [img]/images/graemlins/tongue.gif[/img]

Thanks again.
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  #8  
Old 10-29-2004, 11:08 AM
spatch spatch is offline
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Default Re: Hold\'em Odds

The important thing is what type of hand you put your opponent on, and if there were other people int the pot.

Here's a good example of a similar ( but more complicated ) hand I had last night and re-ran on PokerTracker.

I have AA on the button, and raise 3 limpers, who all call.

Flop in 7,8,9 in 3 suits. There is a bet by a early position limper, 1 fold and 1 call. I raise, both only call. My assumption that they were on straight draws was correct. One had 66, the other Q,10 suited.

The turn was not a scare card for the straight card, but as it turns out gave the Q,10 suited a flush draw.

Little did I know it, but I was down to about a 45% of winning the pot v.the field. Any 5(4),10(3),J(4),6(2) or flush card had me beat.

The river was a 10.

Just an example of when it gets comlicated.
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  #9  
Old 10-31-2004, 07:34 AM
jtr jtr is offline
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Default Re: Hold\'em Odds

Hey, Vanquish.

Glad you got some useful answers to your question. As for how to decide, in the heat of battle, whether you have the odds to call, you should start thinking of the pot in terms of its size in bets, not dollars. (This is especially important in live play, where there is no handy pot size measure displayed on the screen!) So, for example, let's say you're playing 2/4. You note that there are $20 in the pot at the end of the flop betting. That's 10 small bets. Now we move to the big bet streets, so divide by 2. We have 5 big bets in that pot. One guy bets (+1 BB), he gets a caller (+1 BB again), and then it's your turn. If you've been thinking about the number of bets going in, it's easy to see that the pot is offering you 7:1 on your call. If you needed, I don't know, call it 10.5:1 for a gutshot draw, you could happily fold knowing you weren't getting the odds.

So in my opinion, there's no need to multiply the odds you require by the bet size to get a minimum pot size for calling, although you could certainly do it this way if you want. If you train yourself to always think "that's a 7BB pot", "that's a 14.5BB pot", etc., then thinking about whether you have odds to call at X:1 becomes very easy.
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  #10  
Old 11-01-2004, 11:53 AM
LALDAAS LALDAAS is offline
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Default Re: Hold\'em Odds

Hello all first time poster here.

I am still as green as a frog here, so for my question. All these calculations with pot odds, implied pot odds and reverse pot odds when to decide to call is purely base on the fact that on a gutshot striaght draw with a 10.5:1 is to break even in the long run. Is that corret?

so when looking at pot odds of 10.5:1 in essence what this is saying is that I am giving my self 10.5 oppertunities to hit the striaght, hence in the long run I will at best break even.
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