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Re:what are the odds of this More than a good approximation, it's the correct probability. Posted in: Poker Discussion	BettinBenny Aug 9th, 2008
Re:the art of bluffing Regarding the 393..9 hand where you put your opponent on an overpair, what are the prerequisites you must recognize to know he's going to fold an overpair there' Do you need to have seen him lay down what you thought were made hands before' Is he a tight player' I've noticed I have a very difficult time determining when/who will lay down this sort of hand when I try to represent trips and have found it better to avoid these types of bluffs as a result. Posted in: Poker Discussion	BettinBenny Aug 1st, 2008
Re:the art of bluffing AQ giving him the 9 clubs and 6 overcard outs' I think in that situation he's ahead of KJ. Posted in: Poker Discussion	BettinBenny Aug 1st, 2008
Re:the art of bluffing Now wait a second. You asked for good and bad criticism. The idea is to have a discussion about these hands - not to mindlessly just believe what you say. I applaud you for writing an article that encourages some thought about when/how to successfully pull off bluffs, but you should be better at accepting the criticism that comes with the intellectual stimulation you're encouraging. Why are you bashing the guy for having an opinion that's different than yours' He didn't attack YOU, he just attacked the way you played the first two hands which you opened up for criticism. Posted in: Poker Discussion	BettinBenny Aug 1st, 2008
Re:the art of bluffing How is focusing on the number of outs he has irrelovant' Your "great" play is putting all your chips at risk when the worst he's going to check, raise, call you with is a coin flip scenario (you're lucky he had the worst hand he could have had and not two overcard clubs or better). Your argument sounds as solid as someone talking about how great they are for pushing 66 preflop and getting someone to call with AK. Posted in: Poker Discussion	BettinBenny Aug 1st, 2008
Re:the art of bluffing He has 12 outs, 9 clubs and 3 queens, and you have a 52% chance of winning the hand - way to get all your chips in when you have such a huge edge. Don't ask for someone to back up his opinion and then berate him when his response makes sense. His reasoning for how you played those first two hands badly is sound. Posted in: Poker Discussion	BettinBenny Aug 1st, 2008
Re(1): re: Ship Happens hand --- Why Check-Raising the River is -EV pariah, In response to mattg's assertion that T9 is "way more likely than Q9" with regards to how the hand played out... Why can you not account for percentages of different villian hands' You seemed to have no problem narrowing down 150 some odd starting hands to a dozen or so based on something. Why not weight those dozen hands based on that same something (the way the villian has been betting during the entire hand). In any case, it may strengthen your argument for why check-raising the river is a negative expected value play. Given the action, we might guestimate, for example, that ... Posted in: Poker Discussion	BettinBenny Jul 27th, 2007
Re(4): A challenging math problem # 2 $$ Under, What if you modify your logic a little. There are 2n-1 total games played in a round. Since players are being eliminated in every round, two players will either meet in one of these games or they won't. -BettinBenny Edit: This isn't working out the way I envisioned - I was hoping to find a mutually exclusive way of looking at the problem since I know jay_shark likes it, but I can't figure out exactly how it works. Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(7): A challenging math problem # 2 $$ The solution above does not work out to 1/2. Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(5): A challenging math problem # 2 $$ I don't think you should just answer "nope;" the solution is correct, yielding the same 1/2n-1 result. Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(3): A challenging math problem # 2 $$ An elegant solution: 21 * 22 * ... * 2n-2 _______________ 21 * 22 * ... * 2n-1 Where the numerator is the number of games in the previous round and the denominator is the number of ways to arrange the games for all rounds. Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(1): A challenging math problem # 2 $$ n/2n-1 1/2n-1 (Proof seen above after several edits.) -BettinBenny Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(1): A challenging math problem # 2 $$ log2 n = Number of Rounds n = Number of Rounds 1/2(x-1) = Chance a player makes it to round x 1/2(2x-2) = Chance two particular players make it to round x. 2n/2x = 2n-x = Number of games in round x 1/2n-x = Chance two particular players play in round x 1/2n-x = Chance two particular players play in round x n�x=11/22x-2 1/2n-x => n�x=11/2x 1/2n-2 => 1/2n-1 Posted in: Off Topic	BettinBenny Oct 13th, 2006
Re(1): a good one for the math wizards Chance to hit an underpair on the flop is around 13%, so she needs to be getting around 9 to 1 on the amount she's investing. When considering implied odds, the amount of your raise here doesn't matter as much as the size of your stack behind. For a correct play, assuming she knows you have an overpair, she needs to be sure she can get 1800 of onaccounted-for chips with her 200 chip call (either by getting all your chips, getting you to make a big bet or gettin you to make a big call). Here, it looks like there is 425 in the pot, so she wants to be able to get around 1400 more from you when ... Posted in: Poker Discussion	BettinBenny Oct 10th, 2006
Re(3): A challenging Math problem Okay, yeah. .52x+1 is the probability that all of the coins land in any particular sequence. 22x is the number of sequences (out of 22x+1 total sequences) in which more heads are flipped in Player 2's group of x+1 coins. So the full probability can be written as: .52x+1 22x. Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(2): A challenging Math problem Perhaps the elegant solution involves a simple proof that the combinations of ways to have greater than n heads tossed in x+1 trials versus x trials is 22x. Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(1): A challenging Math problem The best I can do: P= .52x+1x+1�k=1��x+1��k�k-1�w=0��x��w�� x+1�k=1��x+1��k�k-1�w=0��x��w��=x+1�k=1��x+1��k�2x - x�w=k��x��w�� = 2x+1 2x -x+1�k=1��x+1��k�x�w=k��x��w��= 22x => P= .52x+1 22x = .51 Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(1): A challenging Math problem Please forgive the poorness of the presentation, but I believe the idea to be correct. If jay_shark doesn't beat me to it, I will provide a more elegant proof later today. The probability of the number of heads tossed in x trials can be expressed as a function of the number of heads tossed in x-1 trials, something of the form: Let Pnx := Chance of tossing n heads in x trials. Pnx = (x-1) (Pn-1x-1 + Pnx-1) / x The summation of all n (from 0 to x) can then be written as � Pnx = 2 (x-1) � Pnx-1 / x This corresponds to the fact that the sum of any row x of Pascal's triangle will be twice ... Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(2): A challenging Math problem I am still working on an elegant solution. Phish - your answer is just wrong. As I posted earlier in this thread, the answer is 50% for any x, according to the summation of a summation. -BettinBenny Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(4): A challenging Math problem GambleAB, You have your x's offset. At x = 1 (meaning that player 1 flips one coin and player 2 flips two coins) player 2 has a 50% chance of flipping more heads than player 1. I think we both agree on this one. When player 1 flips two coins and player 2 flips three coins let's examine: Chance that player 1 flips zero heads out of two coins: .25 Chance that player 2 will flip at least 1 head out of three coins: .375 + .375 + .125 = .875 Chance that player 1 flips one head out of two coins: .5 Chance that player 2 will flip at least 2 heads out of three coins: .375 + .125 = .5 Chance that ... Posted in: Off Topic	BettinBenny Aug 31st, 2006
Re(1): A challenging Math problem P= x+1�k=1��x+1��k�.5x+1k-1�w=0��x��w�.5x�� Posted in: Off Topic	BettinBenny Aug 30th, 2006
Re(2): A challenging Math problem Some percentages when plugging in values for x: At x = 1 => P = 50% At x = 2 => P = 50% At x = 3 => P = 50% At x = 4 => P = 50% Posted in: Off Topic	BettinBenny Aug 30th, 2006
Re(2): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) For those of you who enjoy thinking about these sorts of things, I propose a question: After 221,000 hands, which of the two demonstrations is showing a smaller degree of variance' You may wish to pause and give some thought to the question before continuing (spoiler warning). The question seems trivial, but it may help some of you grasp a better understanding of what "insignificant" means to you. Quickly looking at the two data sets, some readers may be drawn to the second demonstration, since we observed a � .04% degree of variance, whereas in the first demonstration we observed a � 3% ... Posted in: Poker Discussion	BettinBenny Aug 21st, 2006
Re(1): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) Another demonstration of the proper definition of the "LONG RUN": How many hands do I have to play before the variance between the number of times I've been dealt aces and the number of times I expect it (1 in 221) is insignificant' (Note that this demonstration completely removes any SKILL vs. LUCK debate/problems, and is dealing ONLY with luck, i.e., the hole cards I am dealt.) AA is dealt 1 out of every 221 hands, or .4525% of the time. After 1000 hands, we get the following results from the binomial probability formula (k corresponds to the number of times I am actually dealt AA): At ... Posted in: Poker Discussion	BettinBenny Aug 21st, 2006
Re(4): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) Betbrett and doubledave, You are misinterpreting the demonstration. Betbrett did note that his response was written prior to recognizing the all-in preflop condition. Even so, the demonstration is not an argument for skill vs luck, but an argument for the proper definition of the term "LONG RUN." That said, both of your points are well-understood. I would like to propose some food for thought, however, relating to betbrett's "an argument can be made..." paragraph: Given a player always wins $100 with his AA winners and loses $500 with his AA losers, can you really argue that LUCK is the ... Posted in: Poker Discussion	BettinBenny Aug 20th, 2006
Re(3): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) Good points here, timzc1. For the purposes of the LONG RUN demonstration above, we are assuming both players get all their chips in preflop every time (as runner91's initial post suggested). -BettinBenny Posted in: Poker Discussion	BettinBenny Aug 19th, 2006
Re(4): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) Runner, If after 1,000 times of having AA, being 99.3% sure that it wins between 83% and 88% of the time isn't negligable enough for you, then reconfigure the number of hands it would take to get it in a range that does suit you. At ten thousand times at getting aces, I can be 99.51% sure that they will hold between 84% and 86% (� 1%) At a hundred thousand times, I can be 99.96% sure that they will hold between 84.6% and 85.4% (� .4%). At some point, the percentage chance that you're outside the true range becomes insignificant enough for you personally (at least for most poker players) - ... Posted in: Poker Discussion	BettinBenny Aug 19th, 2006
Re(1): Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) For those that don't want to read the entire post, a synopsis: After 250,000 hands or so, I can expect with 99% certainty that variance between collected odds and true odds will be within 3% --that is, if a certain hand is supposed to hold x% of the time, I can be 99% sure that after I have played this many hands, my results will demonstrate that the hand held up x � 3% of the time. For this reason, I claim that many of us have already experienced enough hands to claim that we indeed have a LONG RUN. -BettinBenny Posted in: Poker Discussion	BettinBenny Aug 19th, 2006
Defining the "Long Run" - How many hands does it take? (Re: LUCK vs. SKILL by runner91) In LUCK vs. SKILL, runner91 writes: "We love talking about 'over time I will win because I am better'. Let's get this straight. None of us will ever reach the 'long run'. Not even close. Doyle Brunson = NOT AT THE LONG RUN. Every hand ever played = CLOSE BUT NOT AT THE LONG RUN YET." I would like to refute the ideas expressed in this paragraph. The "LONG RUN" that runner91 is (poorly) defining here, is not the "LONG RUN" expressed by poker players. As far as I can tell, runner91 is defining the LONG RUN as the number of hands required to be played out so that every possible permutation of ... Posted in: Poker Discussion	BettinBenny Aug 19th, 2006
Re(5): LUCK vs SKILL by runner91 Runner, I disagree with this statement: "What we can definately prove is, 2 identical players will have different results depending on how they fall on this curve." I don't think it's definite that you can prove this. You are ignoring the fact that good players do not soley rely on the cards they are dealt. I agree that 2 identical players may be on opposite extremes of your bell curve (meaning that the first player is dealt mostly good hands and flops that match, while a second player may be dealt mostly bad hands with flops that miss), but in what part of your analysis do you refute that ... Posted in: Poker Discussion	BettinBenny Aug 19th, 2006

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