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Are the Nuts Out There Somewhere?, Bart Mann, 7. Nov 2003 12:04 | ||
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| Here's a question for the stats gurus--what are the odds, on any given hand, that someone will have the dead nuts? Put another way--if you simulate a million hands of Texas Holdem poker, how many times will the winning hand also be the best possible hand given the structure of the board? Anyone know this? I'd like to--please reply if you do. - Bart - | ||
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Re: Are the Nuts Out There Somewhere?, FlopDaNutz, 7. Nov 2003 12:16 | ||
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| It all depends on who is dealing :) -FlopDaNutz I'm not a player I just MUCK a lot! | ||
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Re: Are the Nuts Out There Somewhere?, shorn, 7. Nov 2003 12:20 | ||
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| This cannot be calculated without making a lot of assumptions, the biggest being that all players dealt a hand stay in and don't (by accident) fold the nut hand. So, no matter how you structure your assumption, the outcome won't tell you anything statistically significant. | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 7. Nov 2003 12:34 | ||
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| I actually think this statistic can be useful. I guess we agree to disagree, but please humor me. I'll change the question slightly: "In a 10-handed HE game, what are the odds that one of the 10 hands will result in the Dead Nuts, assuming all 10 players stay in the hand until the River card is exposed." That better? | ||
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Re: Are the Nuts Out There Somewhere?, Schuster, 7. Nov 2003 12:59 | ||
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| You can't do this with simple pen and paper stat calculations as there is no easy mathematical formula to determine what the "nuts" are, but it'd be pretty easy to do by computer simulation. For every possible combination of board cards, determine the nuts, and determine the number of combinations of remaining cards that qualify. I agree with shorn though, I don't really see how this is statistically significant. Even if you hold the second nuts and want to know the chances of the nuts being out there, this statistic won't do anything since there is a huge variance in the combinations of cards. If you hold the big full, there's only 1 combo that makes someone quads, but if you hold the king high flush (with 2 of the suit in your hand), there are 5 combos that give someone the ace high flush. It gets much much worse when only 1 card is needed for the nuts. You'd do better to be able to accurately estimate the chances given a certain board while at the table than know a statistic like this. Lee | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 7. Nov 2003 13:18 | ||
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| OK--let me rephrase again. If someone out there is capable of doing this via computer simulation, I would very much appreciate it if he/she could please find the answer to the following question: "In a 10-handed Holdem game, what are the odds that one of the players will hold the Dead Nuts hand, assuming all 10 players stay in the hand through the River card?" Or to put it slightly differently . . . "If a million 10-handled holdem tables played one hand each, and every player at each table stayed in the hand through the River card, at how many tables would the winner be holding the Dead Nuts hand?" Admittedly, I wish to know the answer for no good reason--other than sheer morbid curiosity. But unfortunately I am a stats junkie. I can't turn it off. Sorry. ;-) | ||
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Re: Are the Nuts Out There Somewhere?, redsoxasu, 7. Nov 2003 14:07 | ||
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| I think you fail to realize that when everyone stays to the river, the person who has the "nuts" wins 100% of the time. If ten people stay in and the best hand was Ace high, then the "nut" is A high. If the people stay in and the best hand was a Ryal Flush, then that was also the nut. You're going to have to be more specific. On the river, the "nut" of the hand when 10 people show is always the best hand out there (even though someone may not realize it). Before the flop (Like with A-A, the probability of four of a kind, or a Royal Flush draw) it's calculatable, however by the river the absolute nut has changed at most 3 times. Maybe i dont thoroughly understand, you'd have to give a better explanation. | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 7. Nov 2003 15:06 | ||
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| I didn't start playing poker three days ago, and I'm pretty confident in my knowledge of "Poker Terms and Definitions." Therefore, it is extremely difficult--if not nearly impossible--for me to fail to realize that the person with the nuts hand wins 100% of the time. I politely ask, in the spirit of goodwill among UPF members, that you please refrain from patronizing me in the future. Per my original post, the question was framed in the context of "the makeup of the board." For example--if the board is . . . . As Ac Jd Jh 2s Without knowing what any of the 10 players are holding, the Dead Nuts hand--the best possible hand that could be made period--is four aces. In order to have the hand of four aces occur, someone at the table obviously needs to be holding AA. A second example. If the board is . . . Qs Js 10s 2d 8h The best possible hand that could be made--again, without knowing what anyone is holding--is a Royal Flush. In order to have a Royal Flush occur, someone at the table would have to be holding AsKs. With these two examples in mind (and I can certainly come up with more if necessary)--here is the question: If a million hands of 10-way Texas Holdem poker are played, and there are by definition a million winning hands, how many of those hands will have been won by a player holding the DEAD NUTS--the best hand possible given the five cards on the board? Assuming, as I mentioned previously, that all 10 players in every hand stay in the hand to the end. And again--I was asking this question for no other reason than because I was curious. I'm not looking to write a new poker book, publish a new winning strategy or turn professional and head to Vegas at any time in the near future. I know that many of you enjoy doing poker-related computer simulations, and I thought maybe one of you had either run this simulation before, or would be willing to run it in the spirit of knowledge and general curiosity. Please do not read any more into it than that. | ||
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Re: Are the Nuts Out There Somewhere?, pt_Gatsby, 7. Nov 2003 15:40 | ||
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| Well, this isn't actually all that hard - ok, it is, but I can probably summarize the 'theory' anyway. Goes something like this, anyway (from my head, without working it out...) Ways to have the nuts with 0 cards, 1 cards and 2 cards (divide the flop up - full enumerated) a)How many of those flops have the 'nuts' already? The answer is: any A high flush (therefore including straight flushes to the A - note that a straight flush is not a nut on its own) that doesn't contain a pair on the board, any 4 of a kind with an A as the 5th card - or 4 aces and a K. Any straight to the A that doesn't contain a pair or 3 suited cards. b)Then, there are the flops that have 4 cards to the nuts. This means that each person only needs to have that one card. This would happen how often? Well, to simplify it, it would happen exactly 18/47 times(or rather, x*2/47 cards, where x is the number of players - but anyway). c)Then, there are the flops that would need two cards. This ranges from a pair on a blank board (2,5,7,10,K - KK would be best) to the straight flush (AKQsxx, with J10s). The point being that you need that exact two card combination to have the nuts - which means that there is only *one* combination of those cards, of all the 2 card combinations, that would win it. This is where it gets tricky. There are multiple cards that will get you the nut straight and the nut full house. There are even 3 ways to get the nut trips on a blank board. I'm pretty sure we could work them all out (pairs, high card are irrelevant, as are flushes - all of these, there can be only one nut) All of that could, theoretically be worked out and fully enumerated. But, its not going to be me! But you are more than welcome to do it :) a) (4 of a kind + HC) + (5 suited with A - pair) + (5 card straight - pair - 3 suited cards) = 100% nuts held b) (3 of a kind - 3 or more connected suited) + (4 suited - pair - 3 or more suited connected) + (4 cards of a straight - 3 suited - pair) c) (2 of a kind - 3 or more connected suited) + (3 suited - pair - 3 or more suited connected) + (3 cards of a straight - 3 suited - pair) + (All empty flops where a pair in hand would win.) That should add up to 5_c_52. Then, a) would happen 100%, b) would happen (x*2)/47 times and c) would happen - very roughly and only for theory, ((x*2)/47)^2 times, where x is the number of players. Add them up, and there you go. To actually be accurate, you would have to full enumerate the 2_c_47, I think for c... but I'm already dizzy from this. Anyone see any errors, flaws or... maybe I'm just completely out of it. Seems to me I'm missing something. Probably come to me right after I post. | ||
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Re: Are the Nuts Out There Somewhere?, pt_Gatsby, 7. Nov 2003 15:44 | ||
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| Yup, already noticing errors. To enumerate a) properly, the royal flushes have to be seperated from the flushes... so, it would be: 5 suited connected to the A and 5 suited unconnected by at least 2 gaps of 3 containing the A. (ie: 23 78 A). By definition, excludes pairs, so they aren't relevant. | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 10. Nov 2003 06:33 | ||
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| Is your degree in Math, Economics or Finance? | ||
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Re: Are the Nuts Out There Somewhere?, pt_Gatsby, 10. Nov 2003 08:26 | ||
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| None of the above... this is a hobby! As an aside, another correction: All flops where 3 of a kind of a value < a pair means you need a pair of the pair (ermmm... yah), so that falls under needing 2 exact cards rather than just 1. | ||
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Re: Are the Nuts Out There Somewhere?, Mark Barnett II, 10. Nov 2003 11:12 | ||
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| hey at least there is one other person on this planet strange and bizarre like me. these are questions i randomly come up with at times that 99.9% of the time will be totally useless but then again who knows what lead newton to come up with his law of gravity besides an apple falling on his head. Rule #1 of Poker Circumstances alter cases Rule #2 NEVER forget rule #1 | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 10. Nov 2003 11:55 | ||
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| Mark: Thanks for the kind words. Sometimes the people on this board can really make you feel like an idiot for asking something. It seems like every time I post a question like this, the first three responses are something to the effect of "Bart, WHY would you want to know THIS?" or "That's not practical or useful in a game." Be that as it may, inquiring minds stil want to know. Like the formula for figuring out the odds of someone having a bigger pocket pair than you (220 / # of higher pairs / # of other players in the pot). Sure, maybe it's not practical in a game situation, but when my pocket Kings get cracked heads up by pocket Aces, at least I can say to myself "Well, it was a 220 to one shot." Makes me feel at least a little better. | ||
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Re: Are the Nuts Out There Somewhere?, Mark Barnett II, 11. Nov 2003 10:15 | ||
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| worlds biggest believer in there are no bad questions except those unasked (i may be tired of answering the same one for the 107th time but i will never imply someone shouldnt ask) Rule #1 of Poker Circumstances alter cases Rule #2 NEVER forget rule #1 | ||
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Re: Are the Nuts Out There Somewhere?, Aisthesis, 15. Nov 2003 22:43 | ||
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| I don't have a problem with the question at all--if it really were even remotely easy to answer. Working it out longhand would require a LOT of time and might be completely undoable. I'm guessing it would be on the order of how many years devoted full-time to this problem, but I could be erring on the high or the low side depending on how the various algorithms turn out in various scenarios. A computer simulation seems doable, but unless it's actually already been done, I would think that would involve a very heavy amount of case-by-case programming. I certainly wouldn't know how to do it. This being said the answer is: It's rare. | ||
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Re: Are the Nuts Out There Somewhere?, Phish, 23. Nov 2003 21:18 | ||
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| on 7. Nov 2003 12:04 Bart Mann wrote: > Here's a question for the stats gurus--what are the odds, on any given hand, > that someone will have the dead nuts? Put another way--if you simulate a > million hands of Texas Holdem poker, how many times will the winning hand also > be the best possible hand given the structure of the board? Anyone know this? > I'd like to--please reply if you do. > > - Bart - I'm more interested in how this information can possibly be of any use to anyone at the poker table. | ||
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Re: Are the Nuts Out There Somewhere?, Bart Mann, 24. Nov 2003 19:35 | ||
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| It's a question based only on general curiosity. Nothing else. Guess I didn't realize this thread had a subtitle of "But Only Stuff that is Useful at the Table." My mistake. | ||
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