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Old Dec 15,2009, 12:24 AM   #1
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Why Pot Odds Calculation So Complex?

I am learning currently how to calculate Pot Odds in my head and I am uncertain why there is so much complexity in the calculation.

First up they say to calculate your Outs and multiply by 4 for your percentage to hit on the Turn, and multiply by 2 for your percentage to hit on the River. As an example I will say that I have 6 outs which equals 24% for the Flop betting. Okay, this part isn't complex.

(Insane Complex) But next up they say to calculate Pot Odds, you need to take the amount to bet which - as an example I will make $10 - into the pot $30 and your odds become 3 to 1. Then add the 3+1 to get 4 and divide 100% by 4 and you have 25% Pot Odds. Why so complex?

(Somewhat Complex) Why not just mentally add the $10 you would bet to the $30 and say that $10 is 25% of $40? Sure seems like a lot easier calculation method to me when you have 15 seconds to act. But even this is complex to my method below!

(Not Complex) Why not just take your outs - which in my example are 6 outs for a 24% (25.2%) chance to win, and divide the pot by 25%? So if the pot is $50 and I divide that by 25% I will get $12.5. Since I only have to bet $10, then I should make the bet as my Pot Odds are better than my chance to win.

Please someone explain how my "Not Complex" method would fare differently long term to the more Complex methods.

Thanks,
Licentia

Last edited by Licentia; Dec 15,2009 at 04:01 AM.
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Old Dec 15,2009, 12:42 AM   #2
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Example

Here's an example:

The Flop cards are placed on the table and I calculate that I have 6 outs. That gives me a 25.2% chance to win the hand (so long as I am correct about those 6 outs giving me the winning hand).

We'll assume $60 in the pot. We'll also assume that I am required to Call $20 to proceed. According to my method, $20 equals 33.3% of $60 which is more than the 25% expectation to win the hand, and therefore I wouldn't make the Call.

According to the Pot Odds calculations I have learned today, I would calculate the $60 pot and $20 bet to be 3 to 1 Odds and then divide 100 by 4 which gives me 25% Odds. So according to Pot Odds theory I would be required to make the Call; though only by a slim margin.

Why in the world would I want to call when I only have a 25% chance of winning the hand and I am betting 33.3% of the Pot size? That sounds like a negative EV to me over the long term, at least compared to my method.

Any "Expert" opinions available?

Thanks,
Licentia

Last edited by Licentia; Dec 15,2009 at 04:04 AM.
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Old Dec 15,2009, 06:53 AM   #3
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I'm not a pro but it feels more like a general poker question so I'll give you my two cents anyway.

With 6 outs twice you'll win 1 time out of 4 roughly.

Calling 20 on the flop to win 60 means you'll have to win 1 time out of 3 if someone shot both of you with a tranquiliser gun and made you check turn and river. Calling a 20 dollar bet made into a 60 dollar pot is calling 20 to win 80 giving you 1:4. I think that's a big part of the confusion


More interesting though are the implied odds (pot odds adjusted for future betting) that play a big role in this situation. If you expect your opponent to fire most turns when you don't hit and check-fold when you hit you're in terrible shape. Then you only have 6 outs once (since you can't call turn) and no money to win if you hit. This would mean you take the pot 1 time in 8 and you'd need to win 1 in 3 so you'd snapfold.

If you think your opponent will check turn and pay off most of your bets though you're in way better shape. You'll win 1 time out of 4. And the time you win you'll get another bet in on the river, or two bets on the turn and river. In this case you're calling 20$ in order to win 60 + any bets you make when you make your hand. Best case scenario you make your hand on the turn and get a 60 dollar bet in on the turn, and then a 150 dollar bet on the river giving you 20 to 170 or about 1 to 8,5 which is way better than the 1 time in 4 that you'll win (given the he has no redraws).



Pot odds calculation in itself isn't so complicated but in NL implied odds are more important than ever and when trying to figure those out you'll always deal with incomplete information making it almost impossible to play perfect poker. Feel free to ask if there was something there that didn't make sense


Edit: One out once is worth about 2,2 percent (1 card divided by 45 cards left in the deck) so I don't it's right to multiply your outs by 4


Edit again: I think the problem with your insanely complicated method is that if he bets 10 into 30 then you need to call 10 with win 40 giving you 1:4 so you're probably just confusing yourself (making my long post unnecesary ^^' ) All those methods should work just fine and yield the same result.

1. Calculate number of outs
2. multiply by 2,22% on the flop or 2,27% on the turn (this is the chance of improving on the next street)
3. multiply pot by result from point 2 ( 25% is 0,25 obv)
4. Result from 3 is the amount you can call on this street in a vacuum

Simple enough method to use on the go



Last edited by Richard~; Dec 15,2009 at 07:21 AM.
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Old Dec 15,2009, 12:13 PM   #4
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Quote:
Originally Posted by Licentia View Post
Why in the world would I want to call when I only have a 25% chance of winning the hand and I am betting 33.3% of the Pot size? That sounds like a negative EV to me over the long term, at least compared to my method.
Sounds like you're confusing odds with probability. A 25% probability of winning is equivalent to 3:1 odds against winning. That means if you have a 25% probability to hit your outs and are getting 3:1 pot odds, you're net 0 EV if you call. If you're getting slightly better pot odds (or have a slightly better probability of winning, perhaps by factoring in the probability that your opponent could be bluffing and you're already ahead, etc.), then you have correct pot odds to call.

In short, you should either convert all the numbers (from pot and chance to win) to either odds or probability - whichever you feel most comfortable with.

Relating probability to odds against (odds:1):

probability = 100%/(odds + 1)

odds = 100%/probability - 1

Personally, I find it easiest to figure out my probability (from the rule of 2.1% per out on the turn; 2.2% per out on the river), then convert that to odds (1/probability - 1) and compare to the odds I'm getting from the pot (which I find easier). For example, say I've got 6 outs on the river. I figure that's almost 20% to win. 100%/20% - 1 = 5 - 1 = 4, so I need at least 4:1 pot odds (maybe a little less if there's a chance I could already be ahead, or might have a few extra outs that I wasn't factoring in, like an over-card that might pair to win). Then it's just a matter of counting the pot, including any uncalled bets, and dividing that by the amount I still have to call (or intend to bet, as the case may be).

Last edited by iNano78; Dec 15,2009 at 12:24 PM.
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