View Full Version : Odds question...another method?

10-22-2003, 04:20 PM
I read an old post on rgp by Abdul where he uses a different (I say different because I've never seen it in any of the books I've read) of calulating odds (to tell you the truth I'm not sure if its odds or what).

if C(1+P) > D, where C is the number of
cards that are outs for you, P is the size of the pot, and D is the number of cards unseen in the deck. If the number computed is less than the number of cards in the deck (D) then fold, if it's greater then call.

My questions are:

Is this a valid method of computing "odds"?

If it is a valid method what makes it so?

What exactly does the number represent?

I will admit I'm a total math idiot when it comes to these things. I've memorized the most basic odds tables in order to help when at the table. I was just wondering if this method had any merit.


10-22-2003, 05:01 PM
It's just another way of expressing pot odds vs odds of hititng your hand if you only have to call one bet.

Example: You are drawing for a gutshot straight on the turn.

Number of cards left in the deck (D): 46
Number of cards that help you (C): 4
Number of cards that don't help you (D-C): 42
Number of bets in the pot (P): 11

Do you call?

Using the most common method:

Pot odds are (P:1) 11:1 (11 bets in the pot vs the one bet you must call)
Odds of hitting your hand are ((D-C):C) 42:4, or 10.5-1 (number of bad cards vs number of good cards).

Since 11:1 > 10.5:1 you call.

If we take the equation P:1 > (D-C):C, multiply both sides by C, we get P*C > D-C. Then add C to both sides, we get C + P*C > D. Factor out the C on the left side and we get Abdul's equation:

C*(1+P) > D

10-22-2003, 05:51 PM
It's valid, and it's the same as memorizing how large the pot needs to see the next card for 1 bet when you have C outs. What you memorize is D/C - 1, and P has to be larger than this.

P > D/C - 1

For example, for a flush draw, C = 9, and the odds against making it on the turn are 47/9 - 1 = 4.2-1. The minus 1 just converts it to odds. If P > 4.2 (or 5) then we can call 1 bet. This is the same as:

P + 1 > D/C


C(1 + P) > D

It's only useful if you have some number of outs that you didn't memorize. It works on both the flop and the turn. On the turn D = 46.

10-22-2003, 06:33 PM
If we take the equation P:1 > (D-C):C, multiply both sides by C, we get P*C > D-C. Then add C to both sides, we get C + P*C > D. Factor out the C on the left side and we get Abdul's equation:

C*(1+P) > D

[/ QUOTE ]

Strictly speaking "C*(1+P) > D" is not an equation. It is an inequality. An equation has an "=" in it.

10-23-2003, 08:45 AM
Thanks Pudley.

How do you adjust if you have to call two bets?

What would the mathematical expression look like?


10-23-2003, 10:14 AM
How do you adjust if you have to call two bets?

Even on the flop, if you can make your hand in one card, you should still be looking at the odds with 1 card to come. If you don't have the pot odds to see 1 card for 1 small bet, then you won't have the odds to see 2 cards for 3 small bets (since the turn bet is double). This is true as long as we are just talking pot odds and not including other bets that you will collect if you make your hand. If you have a draw that requires hitting two cards, such as a backdoor flush draw, then you usually just add 1 out for the backdoor draw and still look at the pot odds for 1 card to come.

Lots of people have various tricks for calculating pot odds, and there is much confusion about how to handle 2 cards to come. Personally, I'd rather memorize a few numbers than do arithmetic during a hand, especially considering all the other stuff you need to think about, pot odds being the least important. They are really just a starting point anyway. The pot odd numbers never change, so there is no sense in calculating the same ones over and over again hundreds of thousands of times for the rest of your life. All I do is remember how big the pot needs to be for me to see the next card when I have from 4 to 12 outs:

<font color="blue">4,11</font>
<font color="red">6,7
<font color="blue">11,4</font>

These numbers are exactly the same for both the flop and the turn since 47 vs. 46 cards remaining doesn't make a difference. Remember, the formula is 47/outs - 1 = pot or 46/outs - 1 = pot, and you round up to the next whole number.

You may want to add a few more numbers, but if you have less than 4 outs you generally aren't going to play unless the pot is huge, and if you have more than 12 outs you generally will play unless you know you're drawing dead, so you will know what to do without knowing the exact numbers. Of course if you are contemplating calling 1 or more raises, the procedure is the same except that you divide the pot by the number of bets you need to call.

Be careful though, because except for 5,9, and 11 outs, all of these pot values are only marginally correct, that is, they don't give you any overlay. You will be making adjustments for them anyway based on many other factors such as the possibility of making your hand and still losing, implied odds, the possibility of a raise behind you, etc. You want to be conservative and leave yourself some overlay, so all other things being equal, you may want to add 1 to most of these numbers, or simply realize that they are very close. Leaving an overlay will reduce your fluctuations without costing you much EV, and if your judgment is less than perfect, it will save you a lot of EV.

Notice that these numbers are easy to memorize because some of them occur in pairs as I've indicated by the colors above. There are really only 6 pairings you have to memorize.

People worry about the separate set of probabilities for 2 cards to come, but those really aren't used very often unless you are taking effective odds or implied odds into account. They aren't used for pot odds. For example, it's useful to know that it's 2-1 to make a flush or straight draw in 2 cards so that if you are last to act on the flop with more than 3 callers, you should raise since you will be putting in 1 bet for 3 of your opponent's bets.

10-24-2003, 08:25 AM
Bruce, I think he's asking what it looks like if you have to call 2 bets in the current round (since in my answer I stated this inequality is correct only if you have to call one bet).

In this case, divide the left side of the inequality by 2 (or 3, or however many bets you have to currently call).

10-24-2003, 01:06 PM

10-24-2003, 01:28 PM

I have another question.

"I'd rather memorize a few numbers than do arithmetic during a hand"

From reading all the Sklansky/Malmuth books I get the impression alot of calulations are going on (at least when they play).

I curious as to how much math is "required" in the heat of the battle?

You've commented on doing "arithmetic" during a hand. Is it common for the "new breed" of player to snap off a number of odds calculations, or is this usually something reserved for post analysis of a hand?

I'm sure there isn't a hard rule on this, but what other calculations (besides comparing the odds of making your hand and pot odds) are worthwhile (if any)?

I realize that "math" may be more important to some than others. I also realize that it is just one facet of the game. Because of my inexperience I'm not sure how I should approach this part of the game. In other words is it enough to compute the odds of making your hand and compare it to the pot odds, etc.

Up to this point I have neglected the mathematics of the game because I am mathematically challenged. I would like to improve in this area, any suggestions would be greatly appreciated.


10-24-2003, 01:41 PM
Pot-odds comparison isn't exactly rocket science, its basically simple arithmetic and is easy enough to do on the fly even for someone without a strong math background. Many players memorize the common odds situations (flush draws, gutshots, 2nd pair) and thats more than enough to make a call based on the size of the pot, especially in limit holdem. I'm sure situations can come up in NL games where a player might need to do some on-the fly calculations, but these are pretty rare.