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Server Time: 11/20/2008 7:28:54 PM PACIFIC |
 Going for a Flush/7-Stud, Roy Cooke, 12. Dec 2003 10:00 | ||
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| The value of a three-suit in 7-stud is HUGELY dependent on the number of other cards of your suit that are out. For example: If you have a three-suit in an eight-handed 7-stud game and 0 of your suit is out, then you are 3.3-1 to make a flush.....If 1 of your suit is out then you are 4.1-1 to make a flush....If 2 of your suit is out 5.3-1....3/ 7.1-1...4/10-1...5/14.9-1...... When determining the value of a hand that is three-suited you need to take into account the cards in your opponents hands and how they affect your chances. Also include the value of other ways in which you may make a hand, such as high cards you may pair or straight draws you may complete. Life is Good :-) Roy Cooke | ||
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| Re: Going for a Flush/7-Stud, timmer, 12. Dec 2003 12:30 | ||
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| Roy Cooke initiating a 7 stud High thread ?!?> I cant wait timmer | ||
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| Re: Going for a Flush/7-Stud, Harold Pierce, Jr., 13. Dec 2003 07:03 | ||
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| What took him so long? Hold'em and Omaha are in fact lousy poker games! The Pierce Poker hypothesis states: The essential feature of a superior form of poker is the ability of a player to effect the future distribution of dealt cards thru his betting action. He doesn't know what cards will be in these distribution, but he does know that he can greatly increase the the chance of improving or completing his hand by eliminating as many opponents as he can as he goes to the River. Why people play Hold'em and Omaha is beyond me. MouseEars | ||
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| Re: Going for a Flush/7-Stud, Mark, 13. Dec 2003 11:39 | ||
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| on 13. Dec 2003 07:03 Harold Pierce, Jr. wrote: > What took him so long? Hold'em and Omaha are in fact lousy poker games! Mouse Ears (Harold?) Some people feel that Hold'em requires a greater amout of skill. And I have a huge problem with the following statement. (See below) > > The Pierce Poker hypothesis states: The essential feature of a superior form > of poker is the ability of a player to effect the future distribution of dealt cards > thru his betting action. I don't know of any game where a player can affect the future distribution of cards, and doing so would be cheating. (7 Stud doesn't have this ability.) While it is true that if someone folded, you may have recieved a different card (in 7 stud). But since you have no ability to choose cards, affect the cards to be dealt or know which cards are next to come, you have NO control over what card comes and therefore it doesn't matter (to your cards) whether or not someone folded. >He doesn't know what cards will be in these distribution, but he > does know that he can greatly increase the the chance of improving or completing his hand > by eliminating as many opponents as he can as > he goes to the River. This is wrong. Your odds of recieving a card do not change with the # of people in a hand. This is the same theory that some BlackJack players (wrongly) employ. Some people think that you need to play 21 as a "team" and if anyone doesn't use correct theory, it will screw up the cards for the rest of the table. This is illogical becasue no one knows which cards are next to come. It is a random distribution that you can not control. There is a probability or % that the next card dealt will be a certain card. Whether or not you play "correctly" ( or the # of people in a poker hand) has no effect on this probability. Mark | ||
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| Re: Going for a Flush/7-Stud, timmer, 13. Dec 2003 12:57 | ||
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| why,? because the best starting hand holds up more often. because you have one fewer decision rounds. Perhaps its because the open card on the end Often kills the action. | ||
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| Re: Going for a Flush/7-Stud, Harold Pierce, Jr., 13. Dec 2003 23:09 | ||
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| You should re-read carefully the Pierce Poker Hypothesis. In stud or draw games, you effect the order of distribution of dealt cards to the active players by knocking out your opponents with bets or raises and thereby increase the probability of receiving an out because there are fewer players competing for a limited pool of cards. Suppose you are in a 7 stud game and it is 6th street with 4 active players. If you bet or raise and cause two of them to fold, there are now two less positions at the table that can_siphon off_ one of your outs. Note that I have added "order of distribution of dealt cards to active players" If you knock out a player to your right, the card he would have got is now dealt to you and that card might be the out that improves or completes your hand. I can't think of any other simpler way to explain the Pierce Poker Hypothesis. Hold'em is just a fancy of version of 7 card showdown. No-limit hold'em really is not a poker game. It is Rock, Paper, Scissors, T-N-T. And the player who blinks gets blown away by T-N-T, by a player who flops a four flush and goes all in. MouseEars | ||
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| Re: Going for a Flush/7-Stud, Mikewad, 14. Dec 2003 05:26 | ||
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| yes, But maybe they will siphon off 1 of the many cards that is not 1 of your outs and the next card off will be 1 of your outs. Therefore the pierce hypothesis makes no sense! Good luck | ||
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| Re: Going for a Flush/7-Stud, Mikewad, 14. Dec 2003 05:32 | ||
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| Is there really something called the peirce hypothesis? Besides its too hard to remember all those exposed cards in stud | ||
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| Re: Going for a Flush/7-Stud, 4 POKER, 14. Dec 2003 05:36 | ||
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| on 14. Dec 2003 05:32 Mikewad wrote: > Is there really something called the peirce hypothesis? Besides its too hard to remember all those exposed > cards in stud Remembering what exposed cards have come out is one of the most important factors of the game. Many of your 3rd street decisions are based on that information - and that may alter the way you play your hand throughout. 4P- | ||
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| Re: Going for a Flush/7-Stud, Harold Pierce, Jr., 14. Dec 2003 06:45 | ||
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| Good Morning 4 POKER! Do you understand the Pierce Poker Hypothesis? Unfortunately, I left out the phrase "...order of distribution of dealt cards to the active Players..." when I first stated the PPH which is quite obvious to me. I been playing for 50 years and it an idea that has come to me slowly over a long, long time. -=-MouseEars | ||
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| Re: Going for a Flush/7-Stud, Mark, 14. Dec 2003 09:42 | ||
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| > You should re-read carefully the Pierce Poker Hypothesis. In stud or draw games, you effect the > order of distribution of dealt cards to the active players by knocking out your opponents with bets > or raises and thereby increase the probability of receiving an out because there are fewer players > competing for a limited pool of cards. Harold, The above is wrong! You DO increase your chances of winning any poker game when you get others to fold, however, you do NOT increase you chances of getting a specific card if they fold. If there are 46 unknown cards and you need one, your odds are 45:1, no matter how many opponents you have (whether 6 opponents or 1). You do not increase you odds of getting a specific card by eliminating players. By getting 2 players to fold, your odds don't magically change to 40:1 from 45:1 to get that card. >Suppose you are in a 7 stud game and it is 6th street with 4 > active players. If you bet or raise and cause two of them to fold, there are now two less positions > at the table that can_siphon off_ one of your outs. This is where your theory goes wrong. Your odds still don't change. There is one card to come and you have very specific odds, irrelevant of how many opponents there are. > > Note that I have added "order of distribution of dealt cards to active players" > If you knock out a player to your right, the card he would have got is now dealt > to you and that card might be the out that improves or completes your hand. This is very wrong. You don't know what the card is before hand, so it doesn't matter what he gets. By your own reasoning, you have the same chance that you would have gotten your 1 outter ( or 10 outer it doesn't matter) if you had NOT eliminated that player, but now you 1 outer gets dealt to the player on your left. > I can't think of any other simpler way to explain the Pierce Poker Hypothesis. Your Pierce Poker Hypothesis violates mathmetical statistic theories. Sorry Harold, you theory is wrong. Mark | ||
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| Re: Going for a Flush/7-Stud, timmer, 14. Dec 2003 12:04 | ||
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| The fact is the more cards that are exposed the better your chances of receiving a unseen remaining out. Properly sizing the odds of receiving one of your outs is a concept many players don't get right in 7 stud because the numbers are not fixed as they are in hold em they are constantly changing. determined by the numbers of players still receiving cards. Hint* 7 stud card odds start at 42 and work the way down by 2-8 on each round. hold em card odds are fixed at 50, 47 and 46 regardless of how many players seeing the pot figuring pot and especially implied odds at 7 stud in the heat of battle is infinantly more difficult. Few players get it right all the time. | ||
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| Re: Going for a Flush/7-Stud, Harold Pierce, Jr., 14. Dec 2003 12:39 | ||
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| I knew the Pierce Poker Hypothesis would stir-up the poker-playing hornets! Ouch! Ouch! Here is another example. Suppose you are in a seven card game high=low game and have on 5th street this premium four card hand: (x3h)4h5h6h. Since many of your outs are still alive, you just limp thru 5th street. The cards for 6th street are dealt out to 4 active players besides yourself. The players to your right get Ad and 7h and the two players to your left get Ah and 2h. You draw Kc. Groan! However, you still have a fair number of outs alive and being really greedy, you limp once again. The hole cards for 7th are dealt and you draw Qs, the Black Bitch. You say to yourself: This is unbelievable. How could this super great hand go bust? At the showdown the 4 remaining players turn over their hand: Player 1, royal flush in clubs; Player 2, full house 7's over 2's; Player 3, 8 high straight; and player 4, flush in hearts. All of these hands were made on 7th street. Now if you had raised on 5th street and knocked out just one player, the distribution of dealt cards on 6th street would have shifted clockwise one position, and you would have made the nuts. Usually, you raises to knock out opponents who are going your way. However, in seven card high-low, you raise to knock out "sinks" (my term) that can be dealt your outs. In this example all of the sinks were going high. Now do get the Pierce Poker Hypothesis? -=-MouseEars | ||
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| Re: Going for a Flush/7-Stud, Mark, 15. Dec 2003 09:48 | ||
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| Harold, Yes i understand the Pierce Poker Hypothesis, but I am saying it is invalid and wrong. "Now if you had raised on 5th street and knocked out just one player, the distribution > of dealt cards on 6th street would have shifted clockwise one position, and you would > have made the nuts. " The above statement is wrong, because on 5th street you don't know whether or not you will recieve one of your outs on 6th street. Yes, eliminating a player on 5th MAY make your card fall on 6th, but there is the same chance that your card will now fall to the player on your left, and had you NOT knocked out that player on 5th, you would have made the nuts on 6th Your odds of catching a card DO NOT CHANGE with the # of opponents in a hand. Here's an example. On 5th street you have 4 opponennts and are drawing to an ace high flush. There are 36 unknown cards and you have 7 outs (29 :7 or approx 4.1:1) If you raise on 5th street and eliminate 2 players, you still have 4.1:1 odds to catch a good card on 6th. The odds don't change. Now, eliminating players generally helps to win the hand, but that is the same for any poker game. Your example about eliminating someone to your right to "catch" your card is invalid. You could just as easily eliminate that player and have your card fall to the player on your left. Also, your example, if valid, shows that there is no advantage to eliminating someone on your left on 6th street. I've heard this argument before and is was wrong then. When you talk about " if i eliminated this player then that card would have fell", you are monday morning arm-chair quarter-backing. You have specific odds to catch a card and cannot change them through betting or raising. Mark | ||
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| Re: Going for a Flush/7-Stud, Harold Pierce, Jr., 15. Dec 2003 13:19 | ||
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| Actually, you have 9 outs (13-4 = 9). Or did I miss something? For the draw on 6th street, your odds against catching a heart are 36-9/9 = 3. The cards for 6th street are dealt, and your opponents all catch a heart and you don't. There are 31 cards left in the stub and you now have 6 outs. Your odds against catching a heart on 7th street are 31-6/6 = 4.2. Now suppose on 5th street you raised and two players folded. There cards for 6th street are dealt and no one receives a heart. There are 33 cards left in the stub and you still have 9 outs. Your odds against catching a heart on 7th street are now 33-9/9 = 2.7. QED, unless I really screwed up the math or missed something. You drawing odds in seven card stud (or stud and draw games) are constantly changing and depend on the number of players which can vary as the hand progresses. The tables of drawing odds in poker books in general do not take into account the number of active players in the hand, on future streets in hand or the number of outs that appear in the hand or on streets. Some poker book authors mention the assumptions used for computing drawing odds, but most authors don't. However, computing drawing odds in not complex like unified field theory, but doing this on the fly in a face-paced game is quite difficult and requires either a gift, which is given at birth, for mental math calculations or a lot of practice and hard work training one's mind to do so. Now do you understand the Pierce Poker Hypothesis? If you don't, Dr. Pierce highly recommends that you quit poker and play "Go Fish"! Sorry about that, Mark, but I just couldn't resist poking you with a really sharp poker needle. -=-MousEars | ||
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| Re: Going for a Flush/7-Stud, Mark, 15. Dec 2003 21:54 | ||
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| Harold, I still do understand your theory and still disagree with you. Bear with me but i will try to futher explain my point. In the example you pick (just to make it easy - 6 opponents, 9 live outs to a flush, 27 unknown cards on 6th street). We both agree that you have 2:1 odds to catch your card. (right?) Now you say you have better odds of completing your flush if you get rid of some of your opponents, I say it doesn't matter to completing your hand, here's why. As I said before your odds don't change with the # of opponents. Also, all your examples state that the extra players are taking away your specific outs, but this is where you go wrong. The extra players will take a random distribution of cards, some may be your outs, but most won't be. In fact, they also have 2:1 odds of catching your flush cards. But for every flush card they catch, they must also take 2 other non-flush cards (as per the odds we calculated). So you start with 2:1 odds (18:9) and, against 6 opponents, you end up with 2:1odds (14:7) when you factor in that they will recieve cards. If you raise on 6th and eliminate 3 opponents, you still end up with 2:1 odds (16:8), because as the odds state, only one opponent should end up with a flush card. So the # of opponents does NOT affect your odds. The above holds true because your opponent are subject to the same odds as you are. The problem with your theory arises from the fact that you are ASSUMING your opponents will recieve your flush cards at a higher rate than the odds dictate. Obviously, you can't predict which cards will come and which won't, beyond the odds you can calculate. I hope this explains my point. Mark P.S. We haven't even mentioned the problems with eliminating players on a draw. | ||
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| Re: Going for a Flush/7-Stud, docholdiay420, 21. Jan 2004 00:55 | ||
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| well first off I only chase a lfush if the fourth is to my suit. Now I have better then fifty percent of catching it 37/64. Second I won';t play withnumbers but here is my gt as to whyis is better to draw people out to catch that flush then keep them in. 1st this means ur betting which means bigger pot, and if they stay and still catch even bigger pot. Second lets say I got four spades, need one more. In six cards there's really good odds one of them is a spade. Now if six people are in that table for fifth street what are my chances of getting it, 1/6. And lets say there's one for eveery six. Maybe twoout of the eighteen, so if six people stay to the river I have 4/18, right for the next three rounds. But if I can everyone out except one, heads up. Now there will be six cards dealt not 18. Out of those six like I said there's a good ahcne one of them is a spade, and I'm getting to see three of them. Do the math in this case. If two of them are spades there's a real good chance I'll get it, if three are it's almost automatic, and if four or more of them are it is automatic. Speaking from experienc I have caught more flushed when I had four to it, with less people on the table then more. What if out of th six two go out for sixth stret and my card jumps over me, it has happened. Plus there's always the chance I could just win out before we ever se the river, not if everyone is playing for free. I think u are better to bet and hope people get out. Forget the math and statistics and just take a deck of cards and see for urself. sit down and say u want a spade. deal out three cards to six people, then deal out three cards to just two people. U may not gt it with six, but u almost certainly with two. | ||
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| Re: Going for a Flush/7-Stud, Mikewad, 15. Dec 2003 12:40 | ||
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| Yeah, and if my aunt had balls she d be my uncle. | ||
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