Calculating pot odds "backwards"

I've never seen anyone mention this, but when I'm deciding to call or not and calculating pot odds, I do it "backwards". What I mean is that I figure out the odds of my hand improving and then look at the bet I'm faced with and then multiply that bet by the odds (for instance, I'm faced with a $70 bet and my odds are improving are 4 to 1, so the pot has to be $280 or more for me to call).

This is backwards from every way I've ever seen it described, but does it matter? Am I doing it wrong and it's going to hurt me?

Thanks!

Thats how ive always done it.

Glad to hear I'm not the only one! Does anyone else see a problem with doing it this way?

Yes, you have deviated from the norm, and now must be struck with some sort of striking instrument.

If you reach the proper solution, it doesn't matter how you did it, you need to be able to calculate odds quickly, if this way is easier for you, more power to you.

I'm used to counting bets from casino play, so I usually multiply the number of bets in the pot by the number of outs I have, divide by the number of bets I have to call, and compare the result to 45 (number of cards remaining) minus the number of outs I have. The number of cards remaining is obviously greater than 45, but there are a few reasons I use this number, and it doesn't affect the calculation that greatly.

Example: There are 7 bets in the pot, I know I am behind and have 6 clean outs, and I have to call one bet. 7x6/1 = 42 > 45-6, so I call. For two bets I would fold.

Sounds convoluted, but it's second nature to me.

Originally Posted by koolmoe

I'm used to counting bets from casino play, so I usually multiply the number of bets in the pot by the number of outs I have, divide by the number of bets I have to call, and compare the result to 45 (number of cards remaining) minus the number of outs I have. The number of cards remaining is obviously greater than 45, but there are a few reasons I use this number, and it doesn't affect the calculation that greatly.

Example: There are 7 bets in the pot, I know I am behind and have 6 clean outs, and I have to call one bet. 7x6/1 = 42 > 45-6, so I call. For two bets I would fold.

Sounds convoluted, but it's second nature to me.

interesting formula, that only works for limit though. The normal math works for any type of game.

Originally Posted by SteveDonel

interesting formula, that only works for limit though. The normal math works for any type of game.

I learned holdem eight years ago playing limit in home games and casinos, where there is no convenient display of the pot size.

It's pretty damn hard to keep track of the pot in B&M NL, if you ask me.

Originally Posted by koolmoe

I learned holdem eight years ago playing limit in home games and casinos, where there is no convenient display of the pot size.

It's pretty damn hard to keep track of the pot in B&M NL, if you ask me.

Yeah, I haven't really picked up this skill. I generally have a good enough feel for about how big the pot is to make a pretty good decision and not worry about making a thin call here and there. The opposition offline is so horrible anyway...

Yeah, live games require more of an estimate, which is ok with me, as I find pot odds to be slightly flexible to begin with.

I'm a former blackjack dealer, so figuring the pot even in a NL game isn't too difficult for me, thankfully (even scattered about it's not too difficult once you're used to it). Being a former dealer has the added benefit of being able to know exactly how many chips everyone has, which can be useful.

The odd thing about pot odds for me is that I actually have more trouble doing it online than in B&M because I tend to run out of time online. Of course, I keep getting hammered in B&M, but I have a feeling that doesn't have anything to do with my pot odds calculations....

Originally Posted by Brodie

I've never seen anyone mention this, but when I'm deciding to call or not and calculating pot odds, I do it "backwards". What I mean is that I figure out the odds of my hand improving and then look at the bet I'm faced with and then multiply that bet by the odds (for instance, I'm faced with a $70 bet and my odds are improving are 4 to 1, so the pot has to be $280 or more for me to call).

This is backwards from every way I've ever seen it described, but does it matter? Am I doing it wrong and it's going to hurt me?

Thanks!

Have you tried to use texas hold'em calculators for this task?
There are some calculators, which show your optimal bets depending on the pot size, cards, number of players in the pot, etc.

take the number of outs, multiply by 2 and add 1

For example, if you hold [Ah] [Kh]

and the flop is

[9h] [10h] [Js]

you can catch any heart (9), or any queen (4) to have a hand worth betting... 13 outs, X 2 + 1 = 27% odds of improving

this means, roughly, 3.7 to 1 so, the pot must be 3.7 times whatever bet you face (round it to 4 if you wish for simplicity)..

Hope this helps

Originally Posted by DocDann

you can catch any heart (9), or any queen (4) to have a hand worth betting... 13 outs, X 2 + 1 = 27% odds of improving

Remember not to count the same out more than once (Qh). So, there are actually 12 outs.

I do it the easy way...

Say we're on classic flush draw, 35%

If my bet is greater than 1/3 of the pot I fold.

If you want it described mathematically:
If Odds (.35) * Pot < Bet, fold.

You can also do:
If Pot < 3 * Bet, fold
in this situation for a faster estimate.

Also, you can do:
If Pot/Bet < 1/Odds, fold
or
If (Odds * Pot)/Bet < 1, fold.

Just remember that the pot will be bigger once you make the call. If there's 500 in the pot, and you have a 30% to win it, and the bet is 200. Your pot odds is not 200/500 (40%), it's 200/700 (28.5%).

It seems this method gives a higher odds than texas hold'em calculators
Here the results for cards ah kh 9h th js, pot=40, expected contribution by each player=70

N of pl.| Odds Call/Fold/Raise
_______________________________
2 | 70.9% or 2.44:1 raise
3 | 56.8% or 1.31:1 raise
4 | 48.6% or 1: 1.06 raise
5 | 42.6% or 1: 1.35 raise
6 | 38.7% or 1: 1.58 raise
7 | 35.7% or 1: 1.80 raise
8 | 33.3% or 1: 2.01 raise
9 | 31.2% or 1: 2.20 raise
10 | 29.5% or 1: 2.39 raise

Just remember that the pot will be bigger once you make the call. If there's 500 in the pot, and you have a 37% to win it, and the bet is 200. Your pot odds is not 200/500 (40%), it's 200/700 (35%).

I don't beleive your supposed to include your call to the pot in your calculations.

Originally Posted by DrumzCT

Just remember that the pot will be bigger once you make the call. If there's 500 in the pot, and you have a 37% to win it, and the bet is 200. Your pot odds is not 200/500 (40%), it's 200/700 (35%).

I don't beleive your supposed to include your call to the pot in your calculations.

Yes, you are. You win whatever money you put into the pot as well.

mj

No you do not count your call. The last bet you count is the opponent's right before you.

Lets say pot is $50, and opponent bets $10 then the pot size you use for the calculations is $60. After you call it will be a $70 dollar pot but think of it as paying $10 for a chance to win $60.

I'll try to clear this up using a standard draw - a flush draw with 9 outs:

There's a 19.6% chance (on the turn) that your flush draw is going to come in (calculate it yourself or check the odds listing on the main page). You need about 4.1-to-1 to call a flush draw profitably (from any ratio quotation...the most recent place I read this was in Sklansky's SSH, but the ratio will be the same anywhere). So then I'll show you where that 4.1 comes from:

With a 19.6% chance of making your draw, you need the pot to be enough that if you win it 19.6% of the time, it makes up for losing it the other 80.4% of the time. So out of the 100% of the times you'd be playing this, you'll be winning it 19.6% of the time. Simply divide them (100% / 19.6%) and you see that you need to win 5.1 times your bet in order to make the call profitable. Since the bet you make is included in what you make if you win, the pot only needs to be 4.1 times your bet (counting everything that you expect to be in it OTHER than your bet) in order for the call to be profitable.

Example: Pot is $20. The Bet is $4.5. That's a ratio of 4.44-to-1 (less the the 5.1 if your bet isn't counted, but more than the 4.1 since your bet is counted). You call the bet. The pot you will win is now $24.5, for a profit of $20 when you win it. When you do not win it, you will lose $4.5. Say you've made this call 1000 times. You've thus won 1000*0.196*$20=$3920. You've lost 1000*0.804*$4.5=$3618. Making this call instead of folding made you $3920 - $3618 = $302.

- Jeffrey

Originally Posted by rdqlus

Originally Posted by DrumzCT

Just remember that the pot will be bigger once you make the call. If there's 500 in the pot, and you have a 37% to win it, and the bet is 200. Your pot odds is not 200/500 (40%), it's 200/700 (35%).

I don't beleive your supposed to include your call to the pot in your calculations.

Yes, you are. You win whatever money you put into the pot as well.

mj

You should not use your own money to build pot odds.

Here is some good advice on another site I visit regularly: http://www.learn-texas-holdem.com/qu...hands-down.htm
The relevant part reads as follows:

There is another point too here and that is you can't build "pot odds" if the money in the pot is your own. In other words you can't raise a lot on the flop and then use pot odds as a reason for calling later. Pot odds only applies when the return on your investment isn't your own money. That's a key point that many people miss even if they are good players. It is true that once the money leaves your stack, it isn't considered yours anymore but you can't use that as a reason for a pot odds call. Notice how you may for one move be able to justify a call based on odds even if it is your money in the pot, but if you looked back at the entire hand and what you put in for what you could get back, you screwed up. Individual moves don't mean anything if overall you are playing badly and going against math.

You absolutely MUST include your bets from a previous round of betting into the pot odd calculations for the current bet. To remove that from the pot is giong to cause you to lose money by making you fold when the odds give you a call-worthy chance of winning. Referring to the example I made above, there's not much chance you hadn't put in at least $.50 of the $20 pot. $5 doesn't fit within the 4.1-to-1 range of what you should call. But if you fold you're absolutely losing everything you put into the pot previously. Further, you're losing the net gain from calling that bet that I go through above. It's a huge mistake to remove all the money you've bet from the pot when calculating odds.

- Jeffrey

Originally Posted by koolmoe

I'm used to counting bets from casino play, so I usually multiply the number of bets in the pot by the number of outs I have, divide by the number of bets I have to call, and compare the result to 45 (number of cards remaining) minus the number of outs I have. The number of cards remaining is obviously greater than 45, but there are a few reasons I use this number, and it doesn't affect the calculation that greatly.

Example: There are 7 bets in the pot, I know I am behind and have 6 clean outs, and I have to call one bet. 7x6/1 = 42 > 45-6, so I call. For two bets I would fold.

Sounds convoluted, but it's second nature to me.

This is interesting and I have a question. The normal equation for this would be that the pot odds must be better than the odds of winning. So, in terms of math the equation becomes:

P = # of bets in the pot
B = # of bets to call
O = # of outs available.
47: constant, # of cards remaining from which the outs must come.

P/B > (47 - O)/O

which is the same as:

P*O/B > 47 - O

which is your formula except that you use 45 instead of 47.

You hint there are reasons for this substitution. What are the reasons?

Originally Posted by Pyroxene

You hint there are reasons for this substitution. What are the reasons?

52 - 5(board cards) - 2(your cards) = 45

Originally Posted by Pyroxene

Originally Posted by koolmoe

I'm used to counting bets from casino play, so I usually multiply the number of bets in the pot by the number of outs I have, divide by the number of bets I have to call, and compare the result to 45 (number of cards remaining) minus the number of outs I have. The number of cards remaining is obviously greater than 45, but there are a few reasons I use this number, and it doesn't affect the calculation that greatly.

Example: There are 7 bets in the pot, I know I am behind and have 6 clean outs, and I have to call one bet. 7x6/1 = 42 > 45-6, so I call. For two bets I would fold.

Sounds convoluted, but it's second nature to me.

This is interesting and I have a question. The normal equation for this would be that the pot odds must be better than the odds of winning. So, in terms of math the equation becomes:

P = # of bets in the pot
B = # of bets to call
O = # of outs available.
47: constant, # of cards remaining from which the outs must come.

P/B > (47 - O)/O

which is the same as:

P*O/B > 47 - O

which is your formula except that you use 45 instead of 47.

You hint there are reasons for this substitution. What are the reasons?

Yeah, that's the equation, and (in limit) when you consider that you are counting the pot in terms of bets (P = N*B where B is the number of bets) you are actually left with N*O > 47-O. Dividing is not as easy to most brains as multiplication, and this is a way to avoid it (sometimes you might have to divide by two if you are cold calling a raise). I suppose you could add the outs to both sides to get N*O + O > 47, and you would have one number to calculate and a constant to compare it to. It would be a little more difficult to use for cold calling raises, though.

The reason I use 45 is that it's a somewhat round number and makes the math quicker (I have this weird Monthy Pythonesque thing about subtracting from 7 where I'm frequently off by two), and (more importantly) reducing the number of unseen cards from 47 or 46 to 45 adjusts slightly for implied odds.

Originally Posted by rdqlus

Originally Posted by DrumzCT

Just remember that the pot will be bigger once you make the call. If there's 500 in the pot, and you have a 37% to win it, and the bet is 200. Your pot odds is not 200/500 (40%), it's 200/700 (35%).

I don't beleive your supposed to include your call to the pot in your calculations.

Yes, you are. You win whatever money you put into the pot as well.

mj

Counting the bet you may be "about" to make as part of the pot odds seems pretty strange to me.. If you are looking to rationalize your call, then just raise $1000 and then you'll have pot odds for even 1 outers (lol).

You definitely do not count the call you are going to make when you calculate pot odds. If the pot (including the bet you are calling) is $500 dollars and you have to put in $200 to call, you are getting 2.5-1, not 3.5-1. You don't "win" what you have to put in. It's in your stack while you're making this decision so it's your money at this point. You are Betting $200 to win $500.

For example, if you bet $200.00 on the Patriots to win the superbowl, to win $200.00, the odds are 1-1. By including your bet, you'd be saying you get 2-1 odds on this type of bet.


Also, that quote about not being able to build pot odds doesn't make sense. Of course you can take each individual bet and evaluate it on its own when figuring pot odds. If you "raise alot on the flop" and that causes you to have pot odds to call on the turn, then you have pot odds to call. It doesn't matter how that money got in the pot. Maybe your raise on the flop was a bad move, but the money's in the pot now, and you can win it, so it is used in calculating your odds.

-Fishmagician