PDA

View Full Version : What is the relation between pot odds and probability here?

10-28-2002, 06:48 AM
Hi

I'm having a conceptual problem here I was hoping one of you sharp guys can help me with. I understand the relationship between odds and probability, that they are 2 ways of saying the same thing. For example if the odds are 4 to 1 I know the probability can be got by adding them for the denominator and the 1 is the num so 1/5 chance of success.

My problem is when I use this same logic for pot odds. Let's say there is \$40 in the pot and \$10 to call. I believe this is correct to say 40:10, or 4 to 1 pot odds. I'm having a problem however because this doesn't seem analagous to the prior example as I'm uncomfortable making a fraction out of this 1/5 or a percentage 20%. I think my problem is that this is no longer a probability problem because it's just a ratio between two numbers, I'm not trying to determine the chance of an event occurring. Yet I'm still using probability notation. I want to figure this out without comparing the pot odds to the odds of making my hand, I want to use the pot odds alone to come up with the theoretic breaking-even point, if I can.

Then, saying the pot odds are 4:1 as above, how do I use this information to determine my break-even percentage of how often I must make hand or win the pot to break even. I know it's 1/5 but, again not sure how they get this.

Buzz
10-28-2002, 08:17 AM
Shattered -

You write, "For example if the odds are 4 to 1 I know the probability can be got by adding them for the denominator and the 1 is the num so 1/5 chance of success."

That is correct, but usually one would first calculate the probability of making a hand and then convert that probability to odds of making a hand. The main reason, as I see it, to convert the probability of making a hand to odds, is to make a comparison between the pot odds and the odds of making a hand. When the pot odds are greater than the odds of making a hand, you have favorable odds to call a bet. Although there may be such a thing as "pot probability," I've never heard of it. (A second reason to convert probability to odds is that many people seem to be more familiar with the term "odds" than the term "probability.")

"My problem is when I use this same logic for pot odds. Let's say there is \$40 in the pot and \$10 to call. I believe this is correct to say 40:10, or 4 to 1 pot odds."

Yes.

"I'm having a problem however because this doesn't seem analagous to the prior example as I'm uncomfortable making a fraction out of this 1/5 or a percentage 20%."

Seems there is no particular reason to go in this direction. I suppose you could, but it would seem oblique to most people.

"Then, saying the pot odds are 4:1 as above, how do I use this information to determine my break-even percentage of how often I must make hand or win the pot to break even. I know it's 1/5 but, again not sure how they get this."

Out of every five times, you figure to lose \$10 four of those five times. That's a total loss of \$40.

Out of every five times, you figure to win \$40 one of those five times. That's a total win of \$40.

Over five times, you figure to win as much as you lose, a "break even" situation.

In reality, with odds of four to one, you generally have to do something many more than five times, like hundreds or even millions of times to see the true four to one calculated odds reflected accurately. It would be like taking five cards, A, 2, 3, 4, and 5, shuffling the five cards, then picking a card at random, then repeating this process, over and over again. The probability of drawing a particular card is 0.20, (making the odds four to one against), but you may select the same card more than once and not select another card at all the first five times you draw.

Hope this helps.

Buzz

Ray Zee
10-28-2002, 08:59 PM
pot odds give you the information needed to decide if it is worth playing on. so in your example of the pot offering 4 to 1, you do need any hand that will win 20% of the time or more. pot odds do you no good if you can not figure out what your chances of winning might be. the more exact you are the less mistakes you will make.

10-29-2002, 06:18 PM
Thanks Buzz and Ray but I'm still confused so let me give an example. I can do the probability comparison of pot odds to chance of making a hand but this is just a touch different. I mean, it's the same thing basically but I don't know why and somehow I couldn't infer it from your answers. So please be patient. But this is really driving me crazy.

Let me give just a very easy example and I really hope someone could PLEASE help me figure it out. Don't skip any steps cause I get lost so easy. Here's the example: You bet the Spurs to beat the Lakers at the standard -110. You bet \$110 to win 100. What percentage of the time must you win the bet to break even including the vig? (I know the answer but I don't know how they got that, which is the important part).

Shattered

10-29-2002, 07:43 PM

Ed Miller
10-29-2002, 07:48 PM
Let x be the fraction the of time you must win to break even. Thus, you will win \$100 x% of the time, while you will lose \$110 (1-x)% of the time. Your EV is the sum of these possibilities. To break even, your EV must be &gt;= 0. Thus,

100x - 110(1-x) = 0
210x - 110 = 0

Solving for x,

x = 110/210 = 52.4% of the time

Buzz
10-29-2002, 10:40 PM
Major Kong - I like your answer.

Here's another way for the non-algebra minded.

"You bet \$110 to win 100. What percentage of the time must you win the bet to break even including the vig?"

Shattered - Maybe you can intuitively see that if you play twenty one times, winning eleven and losing ten, you will break even.

That is, \$110*10 = \$100*11. (I don't know if that's obvious to you or not.

If it is, you can see that you need to win 11 times out of 21.

If you want the answer in terms of percentages, 11/21 = 0.5238, so that 52.38% is how much you must win (leaving 47.62% for losing). It's the same answer as Major Kong, but less elegant.

I think you have to either set up an algeraic equation or somehow intuitively see an equivalency relationship, whatever works best for you.

Buzz

Bozeman
10-30-2002, 12:39 AM
Ok, if we were to define "pot probability", it would be the ratio of the amount you need to call to the pot you will win=pot now + amount you call. "Pot probability" is then the fraction you must win to make this a break even bet.

PP=bet/(pot+bet)

I used to do this, but odds seem to work a little bit better.

Craig

For example, suppose the BB is all-in, and it is folded to you in the SB. PP=(1/2)/2=1/4, so you should play any hand that wins at least 25% of the time against a random hand. For holdem, this is all hands (32o has 32.3% pot equity). Now that I think about it, fractional pot equity is the term that we are looking for: the fraction of the pot that you will win in the long run.

11-01-2002, 07:38 AM