Talk:Pot odds

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The link to the Missouri gambling site is now out of date and needs to be updated.
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Article rewrite

I rewrote the article (again). I consolidated implied odds and effective implied odds since they are both the same basic thing and Sklanky's convention isn't universal (see Harrington on Hold'em). I tried to improve the examples and make the article read easier from top to bottom. The only potentially controversial statement I made is that the precise calculation of implied pot odds isn't well documented for hands that are not certain winners or certain losers depending on cathing an out or not. I'm pretty well read, but I haven't read everything. If anyone has knows of a verifiable reference that discusses it, please contribute!--Toms2866 00:03, 10 May 2006 (UTC)[reply]

Another bold rewrite. I found an example of implied pot odds for probable winners in NLHE by Sklansky & Miller p.163. Article probably should be better referenced (e.g., references for each assertion), but I don't have time right now. --Toms2866 02:49, 2 April 2007 (UTC)[reply]

Removed this Note in the intro -

Note: Pot odds are not a measure of probability, but a simple win-to-cost ratio to simplify expected value calculations in poker. Thusly, 10:1 pot odds are said to be lower than 20:1 pot odds, whereas a 10:1 probability is said to be higher than a 20:1 probability.

while it's valid information, having it so early on in the article is confusing.

i also added the "Converting Pot Odds to Percentage Values" because later, in one of the examples, that knowledge is taken for granted.

the article might benefit from some drawing odds/pot odds examples since that's the #1 use for pot odds for beginners. possible resource: http://www.turningriver.com/texas-holdem-drawing-odds.html (the bottom chart)

i changed the wording in the example about manipulating pot odds because it seemed peculiar to have a section heading about manipulating odds, then state that the optimal bet would be one that eliminates the difference that manipulating the pot odds makes.

the example in the bluffing section of the article could use some elaboration. added a link to the pot odds and bluffing page. Tgbob 02:25, 22 May 2007 (UTC)Tgbob[reply]

Old discussion

We should have something about implied odds... Evercat 19:18, 15 July 2005 (UTC)[reply]

Yep, I agree, but ... a seperate article.... ? I will try to add somethig.. -Abscissa 11:40, 18 February 2006 (UTC)[reply]

I added a weak discussion of implied odds (and also manipulating pot odds). I figure a weak something is better than nothing. Additional rigor illustrating the mathematics of how to compute (and not compute) pot odds/implied pot odds and how they are used in game situations would benefit the article. --Toms2866 14:46, 24 March 2006 (UTC)[reply]

We should add something about reverse implied odds, where your odds may appear worse than they seem.--Toms2866 13:42, 21 April 2006 (UTC)[reply]

I added effective implied odds (more than one card to come) and reverse implied odds, per Sklansky definitions of those terms.--Toms2866 13:03, 3 May 2006 (UTC)[reply]

Reverse pot odds

The explanation is bizarre. Summarising:

Alice thinks that if her opponent bets on the river, she must be beaten.
Alice intends to call on the river. (she's factored two $10 calls into her betting strategies)

What?? Anyone have a proper source for this? Stevage 07:58, 25 July 2006 (UTC)[reply]

I see no problem with either the explanation or the example. Your paraphrase is a bit odd: Alice doesn't not expect to to beaten if here opponent bets the river--she only recognizes that it's more likely. She calls because it's still quite possible that a river bet will be a bluff or a weak hand, but she has to factor into her pot odds calculation the fact that it's more likely that a bluffer will give up; thus, she needs a higher probability of winning than straight pot odds would imply. This is covered in many of the standard books. --LDC 16:06, 25 July 2006 (UTC)[reply]

?

"With two cards to come, the approximate percentage probability is: (number of outs) x 4 - 1."<-- Did you mean that post flop with the turn and river coming up your odds of hitting after both have occurred is approximately what you said above? You might want to point that out.

Probability calculation

Its pretty strange to express the "win-to-loss" ratio as it is in the beginning of the page. For one, i've never seen anyone express their odds as 1 to anything, its always along the lines of "I'm getting 5 to 1 odds on my money". Second, using the given formula with odds expressed that way, it gives you a lower percentage with higher odds. Ex: 5 to 1 = 1 / (5+1) = ~16% ; 7 to 1 = 1 / (7+1) = 12.5%. It just seems very strange and out of the ordinary to me. Static3d 19:32, 15 August 2006 (UTC)[reply]

Is this right?

The opening paragraph of this article is defining 'Return', rather than 'Pot Odds'. 'Pot Odds' are a consideration of the combination of 'Return' and the odds of improving a hand; Not just getting $100 back from a $10 bet.
This is a common misconception, and I hear it often as a professional dealer, most commonly pre-flop.
Rather than my drawing out a long explanation here, I refer to the URL at the bottom of the definition's page:

How to Calculate Pot Odds

This is an accurate, albeit brief, example.
BarryD9545 (talk) 12:06, 18 September 2008 (UTC)[reply]

The discussion at the introduction of the article seems to be mostly about probabilities of winning, not pot odds. Unless I'm mistaken, "pot odds" refer simply to the ratio of the pot compared to a given call or bet. Pot $50, call $10 = pot odds of 5 to 1. Expected value compares pot odds with actual probability: Pot odds greater than actual probability is positive expected value. Would it be possible to make this intro much more focused on just pot odds, and leave any discussion of probability to a later paragraph? Stevage 03:33, 2 April 2007 (UTC)[reply]

You're right, the first few paragraphs here are terrible, and don't really describe pot odds at all. I'll work on a rewrite today. --LDC 05:54, 2 April 2007 (UTC)[reply]

I rewrote the first part, but I don't have time right now to fix the latter parts--I'll get to them after I get back from work. --LDC

Reads much better now. Thanks!--Toms2866 14:42, 4 April 2007 (UTC)[reply]

Opening paragraph odds

Please stop "fixing" the math in the first paragraph that's already accurate. 1000-to-1 pot odds means that one must win the pot once every 1001 times to break even; once in a 1000 will be profitable, because one will win $1000 once and lose $1 999 times, giving an average profit of a tenth of a cent. Remember that "1000 to 1 odds" means the same as "1 in 1001", not "1 in 1000". --LDC 15:51, 5 April 2007 (UTC)[reply]

LDC is right. Player is contemplating a $1 call to win a $1000 pot. If probability of win = 1/1000, player puts in $1 1000 times, loses 999 times and wins $1001 once ($1000 pot plus his $1 call), profiting $1.--Toms2866 23:55, 5 April 2007 (UTC)[reply]

You're right, LDC. I knew I shouldn't have gotten that lobotomy. Sorry!Ronald King 02:56, 6 April 2007 (UTC)[reply]

Sure, the "math" is correct, it's the definition that is absolutely wrong! The absolutely correct definition of Pot Odds are as described in this link that is used at the bottom of the definition's page:

         "How to calculate Pot Odds"

As shown in this link as well as the other links on the page, Pot Odds are neither the amount of a bet into a pot, nor the number of times a called pot is needs to be won; The first is the typically incorrect understanding held by neophyte poker players, the second is an oversimplified version of a basic premise of Pot Odds, probably derived from improper understanding of these pages referenced at the bottom of the article:

         "Pot Odds & Implied Odds"
         "Using Pot Odds & Calling  Bluff"

I'd suggest reading those articles in full, as well the others listed:

         "Intermediate & Advanced Pot Odds Calculations"
         "Pot Odds vs Pot Equity"

The often misconstrued definition of "Pot Odds" should be corrected. All the more so because the correct definitions are linked within the very pages of the definition itself.

I'm surprised to an extreme this hasn't been corrected out before now.

BarryD9545 (talk) 10:48, 26 October 2008 (UTC)[reply]

Implied pot odds

I think the second example has some problems:

Implied pot = $20 + $50 = $70, not $50.
Alice can lose no more than her opponent on the final betting round, so she can lose no more than $50, not $100.
Alice needs to plan what she will do in the last betting round if she does not make her flush. Presumably, she will fold, so her loss on the last betting round will be $0, not $100 (or $50), in that case. Her gain will also be $0 in that case.
Alice's probability of losing when she makes her flush depends on whether her opponent is drawing to a flush. It should be less when her opponent is not drawing to a flush. I assume that she has estimated that it is 20% when her opponent is drawing to a flush and 0%, not 20%, when her opponent is not drawing to a flush. If this assumption is incorrect, then the example needs to be clarified.
Alice also needs to estimate her probability of losing when she does not make her flush. Presumably, she has estimated that it is 100%, regardless of whether her opponent makes her flush. This would explain why she plans to fold if she does not make her flush.
When computing EV, the call amount should be added to the implied pot before multiplying by the probability of winning, so in the EV calculation, the correct amount to use is $70 + $5 = $75.
EV cannot be computed without Alice estimating the probability that her opponent is drawing to a flush. Since Alice seems to be worried that her opponent is drawing to a flush, she might estimate this to be greater than 50%, say 60%.
EV should be computed as the probability-weighted sum of the amounts she will win from all possible outcomes. There are 8 possible outcomes, based first on whether her opponent will draw to a flush or not, then on whether Alice will make her flush or not, and last on whether Alice will win or not. The number of non-zero terms in this sum can be reduced to 4 by noting that the amount Alice will win or lose if she does not make her flush is $0, because she will fold in that case.
So given all of the above, EV = (60% * 15% * 80% * $75) - (60% * 15% * 20% * $50) + (40% * 19% * 100% * $75) - (40% * 19% * 0% * $50) = $5.40 - $0.90 + $5.70 - $0.00 = $10.20. Since the cost to call is only $5, Alice should call, not fold.

Comments?

--Ronald King 07:40, 12 April 2007 (UTC)[reply]

I removed the second example under "Implied pot odds" for the reasons Ronald cited. The EV of a probable winner is only the product of the implied odds and estimated probability of losing. Adhall 10:43, 9 May 2007 (UTC)[reply]

Implied pot odds example

Now, I'm a beginner, so I'm probably making a mistake somewhere, but... "Second to last betting round" would be turn, right? So 4 cards are on the table. And Alice has 2 in her hand. That's 6 cards. She has 4 outs. So, her pot odds are: total number of cards (52) - cards she sees (2 in her hand and 4 on the table) : 4 outs = (52 - 6) : 4 = 11.5 : 1. And the article says pot odds are 10.5 : 1. What am I missing? --78.0.86.212 20:39, 22 September 2007 (UTC)[reply]

You're confusing "odds" and "probability"; probability is the ratio of favorable outcomes to all outcomes: 4/46 in this case, or 1/11.5. Odds are the ratio of favorable outcomes to unfavorable outcomes, 42 : 4 in this case (subtracting the four winners), or 10.5 : 1. Remember, a "1 in 4" chance or "25%" or "1/4 probability" is 3 to 1 odds, not 4 to 1. --LDC 00:15, 23 September 2007 (UTC)[reply]

Aha! Thanks a lot! --78.0.69.133 12:17, 23 September 2007 (UTC)[reply]