This is a discussion on Poker odds.. am I right that this is wrong? within the online poker forums, in the Learning Poker section; http://www.pokerology.com/lessons/potodds/ Scroll to Figure 3 just after half way down. Previously on the page it said to calculate pot odds percentage you divide the call size by 


#1




Poker odds.. am I right that this is wrong?
http://www.pokerology.com/lessons/potodds/
Scroll to Figure 3 just after half way down. Previously on the page it said to calculate pot odds percentage you divide the call size by the pot size, so in this case it would be 60/108, which is 55.5%. But it says it's 35.7%? Am I right? 
#2




you are confusing pot odds with percentage they are different
60 to win 108 is 108/60 or 1.8/1 Odds of 1.8/1 as a percentage is 100/2.8 or 35.71% you did it the other way 60/108 is indeed 55.55% 
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I said 60 to win 108 is 108/60 or 1.8/1 when i say 1.8/1 that is 1.8 to 1 Like odds on a dice is 5/1 or 16.6666% You said Those are odds against (1.8:1). 
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re: Poker & Poker odds.. am I right that this is wrong?
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Simple numbers make the concept easier to understand. If he bets $10 into a pot of $10 making it $10 for you to call, your pot odds are 33%... 10 divided by (20+10) or 10 divided by 30... and that's the percentage of the time you need to win the pot to break even on the call. Also, the fold that it advocates in that example is sort of bad advice. While it's only a 32.6% chance of making the nuts or near nuts, there is still more money behind and another round of betting to come. The direct odds are already close enough that he only needs to call a bet that's less than 1/10th of the size of the pot (should you hit your hand) for you to make up the difference. He'll likely call a larger bet than that of course, although it's too bad you don't have position in the example. 
#8




The page previously says
"If we want to know the percentage then we add the bet (call amount) to the pot, to give us a total pot figure. In this example it would be: 150 + 600 = 750. Once we have this figure then we would have to perform the following formula: call amount / the total pot size. In our example this would be 150 / 750 = 0.2, or 20%." Apply this to Figure 3 The pot size is 108 not 168 (look at the figure!) The bet is 60. 60/108 = 0.556. I've done exactly as he says in that quote and it's given me 55%. EDIT: Maybe the writer added the call size twice to the pot by mistake? Because yes 60/168 is 35.7% but the pot is 48, the opp call is 60, making a pot size of 108, as he states: "The pot odds are now 1.8to1 (108 / 60) " 
#9




Pot odds are 108:60. You can reduce it to: 1.8:1 (those are odds against). To convert it to percentage (against): 1.8/(1.8+1) = 64.2% but you can also do 'the other way': 1/(1.8+1) = 35.7% (or 100%  64.2%) which gives percentage for you to make a breakeven call.
(converting odds to percentages: x:y => x/(x+y) [%]) 
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So does this mean then that the advice he previously gives in the quotation about 150/750 is incorrect? EDIT: but it isn't.. maybe i'm missing something here 
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re: Poker & Poker odds.. am I right that this is wrong?
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#13




Thank you.
the remaining difficulty for me and my nonmathematical brain is trying to visualise the correlation between pot odds and draw odds and why you should attribute either to each other. I think perhaps I keep misunderstanding that pot/draw odds (draw odds, is that even a suitable term?) is a longterm game strategy, am I right? To ensure that theoretically speaking you at least break even? Is that even realistic? Or have I missed the point? But if I think about it, say my stack is $200, the pot is $100, the bet is $50 so the total pot is now $150, so the pot odds are 3:1... even if I had 3:1 drawing odds, 25%... I don't think I'd risk a quarter of my stack for that!! EDIT: or maybe I would... maybe they're pretty good odds to worth risking... 
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If you are always 'right' the exact percentage of times given your pot odds, you are theoretically breaking even over the long run. Thats why you ideally want to be 'right more often' to turn a profit > thats why you hear all the time that: draw % > pot odds %. 
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That (entirely correct) explanation you quoted refers to an instance where the pot was 450 and the other player bet 150 into it. How did you think the author came up with the figure of 750 you pasted?? When the player puts 150 into the pot, there is now 600 chips in the middle. If you call that bet (this is where the bolded part of your explanation quoted for me comes in), the pot would become 750. When using ratios, you put the amount of chips currently in the middle on one side and the amount you need to call on the other. When using percentages, the difference is that you divide the size of the bet you're faced with by the total size of the pot after you call, which you'd add both sides of the ratio together to get. 
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Game 1: win 40 Game 2: lose 10 = 30 Game 3: lose 10 = 20 Game 4: lose 10 = 10 Game 5: lose 10 = 0 = break even Ergo your point about having better drawing odds than pot odds. Unfortunately i have no idea what the emboldened part means or how you could apply the method to such... Particularly thrown at 'designing optimal 3betting ranges'! 
#20




I've just had a thought though.
It's all well and good knowing your drawing odds, but more importantly what about your winning odds? Isn't this a massive factor that will affect bet size? Is this of any use... at all... : http://www.pokerprobability.net/calculator/texasholdem.html 
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re: Poker & Poker odds.. am I right that this is wrong?
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