Talk:Pot odds

message: We should have something about implied odds... Evercat 19:18, 15 July 2005 (UTC)
replies: text: 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)

text: 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)
thread_title: Old discussion

message: 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)
replies: text: 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)
thread_title: Reverse pot odds

message: 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)
thread_title:  Probability calculation 

message: 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)
replies: text: 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)
thread_title:  Implied pot odds 

message: 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)
replies: text: 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)
replies: text:  Aha! Thanks a lot! --78.0.69.133 12:17, 23 September 2007 (UTC)
thread_title:  Implied pot odds example 

message: Perhaps someone could edit the use of the word "weak". In the section on reverse implied pot odds we have the expression "If the opponent is weak or bluffing" appearing twice.  This is ambiguous. Does this mean that the person is a weak player, or that he has a weak hand? Eclecticology (talk) 00:01, 8 October 2009 (UTC)
thread_title:  weak 

message: A bet of $6.43, resulting in pot odds of 4.11-to-1, would make his opponent mathematically indifferent to calling.
While this may work out to be zero-EV for us in theory, in practice its terrible, since the villain will usually have implied odds to make the call (we might river a set/2pair and be reluctant to lay it down, or we might click the wrong button accidentally). Also, if we make him break even on his call, we aren't going to profit from our bet. We want him to pay too high a price, which will cause him to make a mistake and then we profit. 93.139.67.66 (talk) 16:41, 25 February 2010 (UTC)
replies: text: In practice, when implied odds factor into the equation, it becomes more of a game based upon inutition and instinct, ie... trying to work out what his implied odds are most likely to be and making the appropriate bet size based upon that. With regards to aiming to make an opponent mathematically indifferent to calling, it's in order to allow us to capitalise on our opponent's mistakes. From a game theory standpoint, betting more than what Nash would tell us to do would cause our opponent to fold more often than necessary. Of course, Nash isn't the be-all and end-all of game theory but I hope I've helped with your understanding of the reason behind why we want our opponents to be mathematically indifferent to calling. JaeDyWolf ~ Baka-San (talk) 16:11, 7 May 2012 (UTC)
thread_title:  Manipulating pot odds 

message: I am trying to write an example for using pot odds with card suit symbols, but Wikipedia rejects them as inappropriate since I am a new user. What can be done? Editor794 (talk) 18:35, 17 September 2021 (UTC)
thread_title:  Card suit symbols rejected 

message: The given application of the Addition Law of Probability is incomplete as it doesn't subtract the probability of both cards being the straight card.
The addition law of probability combines the chances of making the straight on the turn (4/47 = 8.5%) and on the river (4/46 = 8.7%) to give the player an equity of 17.2%, assuming no other cards will give them a winning hand.
I believe it should be 4/47 + 4/46 - (4/47 * 3/46) = 8.5% + 8.7% - 0.6% = 16.7%
I didn't want to edit because I'm not entirely sure I've done the math right. Cjgeist (talk) 17:18, 19 September 2022 (UTC)
replies: text: I think your result is much closer to the correct one than the one in the article, unless there's something I'm not understanding about the situation described (I'm not a poker player). I would do 4/47 + 43/47 * 4/46, or alternatively, 4/47 + 4/47 - (4/47 * 3/46). W. P. Uzer (talk) 19:59, 19 September 2022 (UTC)
thread_title:  Addition Law of Probability