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Server Time: 2/11/2012 2:41:13 PM PACIFIC |
 Probability / Odds, JasonHoldEm, 4. Dec 2002 22:23 | ||
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| Could someone point me to a tutorial (or book, or whatever) for poker probabilities? This is something I'd really like to learn, but not being a math wiz I can't figure it out on my own. For example: I'm holding Ac Qc Flop is As 2c 5c What are the chances that I could get trips by the river? or two pair? or a full house? What are the chances that I could make a flush? What are the chances some other guy is going to get (or already has) a straight? Thanks in advance, jasonHoldEm | ||
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| Re: Probability / Odds, The Fish, 5. Dec 2002 11:24 | ||
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| "Could someone point me to a tutorial (or book, or whatever) for poker probabilities? This is something I'd really like to learn, but not being a math wiz I can't figure it out on my own." Although I am a bit of a beginner I would suggest Sklansky's Theory of Poker, I feel it does a relatively good job of explaining pot, effective and implied odds. (although I think you are just inquiring about pot odds). "For example: I'm holding Ac Qc Flop is As 2c 5c What are the chances that I could get trips by the river? or two pair? or a full house?" I know you were just asking for books about odds, I figured I would try my hand (no pun intended, at explaining this for you). Bascially, you have to compare the number of cards needed to the number of unseen cards(after the flop this number is 47). So if you are curious what your chances are of getting trip aces on the turn you must ask yourself how many unexposed cards are there (52 - 5) <- the two in ur hand plus the three exposed onthe flop. How many cards make your trips? 2 -- the Ah and the Ad. So: you subtract the 2 cards that will help u from the 47, giving you 45 (that don't help you) and 2 that do. So you get 45 to 2 or reducing this: 22.5 to 1 ( so we have 45 to 2 ... of x to y, x being those cards(unexposed) that don't help you, and y being those cards that do help u) For two pair apply the same process, there are now 3 queens, 3 twos, 3 fives and 2 aces that help you. NOTE: if u said: "what is the probability that I will get exactly two pair you would not include the aces because they give you trips. But I am assuming you want to know the probability that you will get 2 pair OR BETTER. So you have 11 cards that will give you 2 pair or better. So, again, we subtract 11 from 47 to get 36 cards that don't help you. Use the x to y method used above we get 36 to 11, which becomes 3.272727 to 1. From a strategic standpoint, you must be aware that, although, catching another 2 or 5 may increase your hand to two pair, you must be cautioned that it could give one of your opponents trips or better(especially if you are playing with really loose players or the big blind is still in and there were no raises pre-flop), beating your two pair. Also, please note that these probabilities are only for the turn. But you were asking by the river. This gets a little more complicated. You now have two draws(not technically a draw ...but .....meh) to make your two pair, trips, full house etc. I will just use the example of trips, instead of doing them all. So, we know the probability of getting trips on the turn (22.5 - 1 or about 4.4444% ). So clearly if we have two draws to make our hand our chances must increase. I am not 100% on this so be aware this could be completely wrong. That being said, we must add our chances of making trips on the river alone to the chances of making trips by the turn. (When I say "making trips on the river alone" I do not mean BY the river, which we are in the process of calculating). So, assuming we didn't catch trips (or two pair) on the turn there are now 6 exposed cards (2 in your hand plus the 4 on board). So there are 46 unseen cards ( 2 of which help you and 44 that do not). Again using the x to y thing: 44 to 2 = 22 to 1 or 4.54%. So now we take the odds that we will get our trips on the turn (22.5 to 1 or 4.44%) and add the odds of making trips JUST on the river (22 to 1 or 4.54%). So we get 8.98%. That is ur probability of getting trips by the river. I believe the first part, the part about getting trips by the turn is pretty darn accurate or I think at least. I also think this part is correct but I could be wrong, so consider yourself cautioned! "What are the chances that I could make a flush? " - I am not going to answer this one, cause you are probably 'board' with my response already. And, besides, you use the exact same method as the above example. Remember to use the x to y thing (unseen cards that do not make the desired hand (x) to the number of unseen cards that do make your hand (y)). Please also note that this answer is straight up probability, it says nothing about strategy. Note if you can tell, by reading hands that someone probably is holding big slick (ace king) or pocket aces you have 0 chance of getting trip aces in case someone is holding pocket aces and very slim chances if they are holding big slick. "What are the chances some other guy is going to get (or already has) a straight?" Well, this gets pretty tricky. There are two considerations in this question. First straight up probability. Secondly, knowledge fo the table(people playing at the table), well kinda. I am not going to answer the question about the chances of your opponent getting a straight, there are too many variables to control in that question. The question you are asking about him already having a straight is more manageable. So clearly he must have 3 and 4. Now here is where knowledge comes in. You must have a good read on the players you are playing with, would they play 3 4 ? Also, you must be abe to read hands, if the people you are playing with do play 3 4 are they playing it now? I would submit that (most of the time) the only person playing 3 4 would be the big blind if they have not had to call any pre-flop raises. OR if you have a first time player playing with you. As for probability, strictly probability, your opponent needs to hold any 3 AND 4. So there are 4 threes in the unseen cards and there are 4 fours in the unseen cards. So we apply the same concept as before, since you know there are none on board and you don't hold any you have 52 - 5 (seen cards) = 47 unseen cards. So there are 8 cards that would give your opponent this hand, so 47 - 8 = 39. So the probability your opponent holds EITHER a 3 or a 4 would be 39 to 8 or 4.875 to 1. (20.5%) Now for the probability that your opponent holds both the 3 and 4 we take the remaining 46 unseen cards subtract those cards that "help your opponent", 4, which gives 42 to 4 or 10.5 to 1 or 9.52%. But instead of adding the percentages like we did before, we must multiply them. So take the percentages as decimals 0.205(20.5%) and 0.0952(9.52%) and multiply. We get 0.019516 or 1.9516%. Remember, don't multiply the percentages i.e./ 9.52*20.5, you will get 195.16% chance (impossible, unless you are one of those "give 110%" kinda guys). I hope this shed some light on your question about probability in general. Hopefully you understand the idea of what I was attempting to show. I know this is a very long and irritating post but there are just two final cautions: 1) This is pure probability. I have said little to nothing about strategy, which must be used in conjunction with probability. 2) When considering your chances of making a hand by the river (after the flop) you must be sure that you plan on making it all the way to the river if you are counting on those odds. For example, if you say to yourself my chances of getting trips by the river are about 9% and right now I am getting proper pot odds to call but then you end up folding after the turn you were not getting that 9% probability you were in fact only getting the 4.4% (chances of getting trips by the turn). This is a trap I found myself in. Also, you must remember that you will probably have to call a bet on the turn, possibly more bets. Anyway, I think my method is correct, I hope it has helped you. I highly recommend The Theory of Poker by Sklansky. But, being rather new to poker I have done a limited amount of reading. So I am sure there is probably an equal treatment of probability in some other book that I haven't read, but thus far the only book I have come across with a chapter dedicated to pot odds and the like is The Theory of Poker. However, if you are very new to poker (like myself) I would recommend reading Caro's Fundamental Secrets of Winning Poker. It draws attention to many important concepts that sometimes go over looked. The idea of pot odds is closely related to the boarder concept (in Caro's book) of poker being a game of decision making, CORRECT decision making. Anyway, if you have any questions about my posting please e-mail me. And to anyone reading this post besides JasonHoldEm who spot errors in my method(or any concept mentioned) please let me know! Ciao for now, Ben | ||
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| Re: Probability / Odds, JasonHoldEm, 5. Dec 2002 12:51 | ||
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| Ben, Thank you for the wonderful post. It certainly did a great deal to answer my questions. Yes, I am a relative newbie to poker, I will add the theory of poker to my growing list of poker books I need to read. Thanks again, Jason | ||
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| Re: Probability / Odds, mickblueeyes, 5. Dec 2002 14:38 | ||
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| Michael Petriv's Hold'em Odds | ||
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