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Server Time: 11/20/2008 1:18:04 PM PACIFIC |
 Does someone have AA or KK?, Aisthesis, 11. Nov 2003 10:14 | ||
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| Here's my (possibly slightly rough) calculation of how often someone at a 9 player table is going to have AA or KK: Probability for each should be (1/13)(3/51)=1/221. So the probability for either one should be double that: 1/111. Then taking 9 times that for the probability that someone at the table has it: about 1/12. So there should be no more than a 9% worry that someone might have one of those. Right? | ||
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| Re: Does someone have AA or KK?, Mark Barnett II, 11. Nov 2003 10:16 | ||
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| hmmm no clue on the math but it looks right, but boy that means in general someone should have one or the other every 11 hands? wow that seems alot Rule #1 of Poker Circumstances alter cases Rule #2 NEVER forget rule #1 | ||
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| Re: Does someone have AA or KK?, Aisthesis, 11. Nov 2003 10:37 | ||
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| I dunno, seems about right to me. It's more than a complete round of the button for anyone at the table to get one of 'em. | ||
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| Re: Does someone have AA or KK?, TeKn0wLeD-G, 11. Nov 2003 17:06 | ||
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| you are somewhat correct. it is 110-to-1 that you will have AKo, which is 0.9% instead of 9%. AKs would be 331-to-1, which expressed as a percentage would be 0.3% | ||
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| Re: Does someone have AA or KK?, Aisthesis, 11. Nov 2003 21:17 | ||
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| Sorry, I must have phrased the question badly: It's not the chances of having AK that I'm interested in here but rather the chances of SOMEONE (at a table of 9 players) having pocket ace or king PAIR. The multiplication by 9 is what causes the unexpectedly high percentage. | ||
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| Re: Does someone have AA or KK?, PairTheBoard, 12. Nov 2003 04:57 | ||
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| Your method would give 18/221 = 8.14% or as you say about 1 chance in 12. But actually it's a little less than this because in some of those 18 times in 221 the AA's and KK's will appear on the same deal. To remove this double counting of such occurances you can approximate the Fudge Factor as the Square of Your probability of 8.14%, ie. take away about (0.0814*0.0814) = 0.66% - Although a better Fudge Factor would be (18/221 * (8/221 + 8/1225) = 0.35% which takes into account the AA's or KK's not in the deck when One of the Pairs already shows up on a deal. Anyway a better aproximation of the chance that one of your opponents will have AA or KK would be about 7.8% or closer to 1 time in about 13 deals. You might notice that's a lot more often than you ever see them, as most of them don't get called to the river. Whenever you're taking outs and adding them for the two chances to hit on the Turn and River, you should take away a fudge factor on the order of the Square of the Chance of Hitting either Street. For example: If you have 9 Flush Card outs and just add 9/48 + 9/48 = 38.3% your answer is too high. Take away (9/48)^2 = 3.5% and your answer of 34.8% is much closer to the Accurate One. The same thing can be done for quick and dirty estimates of Flush Draws with overcards, Straight Draws with overcards etc. This is kind of important for certain Brunson type plays. You may think you're a clear favorite when you're not. For example, if you are sure you have 14 good outs but it does you no good if you hit twice with them, then 14/48 + 14/48 = 58.3% looks like you're a favorite. But if you take away a Fudge Factor of (14/48 * 13/47) = 8.1% you see it's really a coin toss. | ||
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| Re: Does someone have AA or KK?, pt_Gatsby, 12. Nov 2003 09:35 | ||
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| Well... I'm not sure I agree with the numbers here... Maybe I made a mistake in my logic. Here goes: 2_c_52 - Total number of possible hands. 2_c_4 * 2 - total number of ways to be dealt AA or KK Chance of one person being dealt AA or KK: 0.5698% (or: 12/2106 times) therefore, Chance of a person not being dealt AA or KK : 99.4302 (or: 2094/2106) So, for a table not to be dealt one, you have: 1-(2094/2106)^x (where x is the number of players), Or 5.0129% of the time (or: 7.74*10^29/8.14*10^29, approx: 7.73/8.14, or should happen roughly 0.41/8.14) ---- It should happen roughly 1/19.95 times, as far as I can tell (with 9 people at a table). For ten people, it should be: 5.554%, or 1/18.01 | ||
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| Re: Does someone have AA or KK?, PairTheBoard, 12. Nov 2003 17:09 | ||
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| You have a couple of errors. First. The number of possible hands is [52 Choose 2] , which I assume you are denoting 2_C_52. The general formula for [n choose k] or if you wish k_C_n is n! / (k!)(n-k)! or as is simpler sometimes: n*(n-1)*...*(n-k+1) / k! In this case 2_C_52 = (52*51)/2 = 1326 which is different from the figure you show. Second: The number of ways AA or KK can be dealt is just the number of ways AA can be dealt plus the number of ways KK can be dealt. The number of ways AA can be dealt is 2_C_4 = (4*3)/2 = 6. So the number of ways AA or KK can be dealt is 6+6 = 12. This is not 2_C_8 if what you meant by that notation is what I assumed - you should really explain your notation and not make us guess. It's certainly not a universally known standard - but seems to agree with the actual figure you calculate. Thus the probabilty of a player being dealt AA or KK is 12/1326 = 0.904977%. Taking the reciprical of this, 1/0.00904977 gives us 1 chance in 110.5 or exactly twice the well known 1 chance in 221 or 220-1 shot of being dealt AA. That's just as it should be. You can simply add the two probabities of being dealt AA or KK BECAUSE THEY ARE MUTUALLY EXCLUSIVE EVENTS. ie. The two things never happen at the same time. This is the First Fundamental Principle for probabilty computations. To find the probability of either one event or another occurring you can add the two probabilties if they are mutually exclusive events. If they are not mutually exclusive you must reduce their sum by the probability they both occur at the same time. Third: You have a good idea to take the probability of NOT being dealt AA or KK, 1-(.00904977) = 0.99095023, and try using it to compute the probabilty of none of the 9 players being dealt AA or KK. However you miss the Second Fundamental Principle in probabilty computations. To find the probability that Both Events occur you can compute it by multiplying their individual proabilities IF THEY ARE INDEPENDENT EVENTS. In this case, you are asking that 9 seperate events ALL occur. Namely, Player 1 does not catch AA or KK, Player 2 does not catch, ... AND Player 9 does not Catch. But clearly these 9 events are not independent. The simplest way to see if they are independent is to ask yourself. Does the outcome of one or more effect the probabitly of another? In this case, if the first 8 players all DO NOT catch AA or KK you can see that the remaining deck is likely to be somewhat RICHER in Aces and Kings than if you knew nothing about what the first 8 players catch. ie. They are NOT independent events. The really correct way to make the computation you are trying to make would be to multiply the following CONDITIONAL PROBABILITIES: Probabilty Player 1 misses AA, KK Probabilty Player 2 misses GIVEN Player 1 missed. Probability Player 3 misses GIVEN Players(1,2) missed. ... Probality Player 9 misses GIVEN Players(1,2,...,8) missed. The Devil is in the Details and I'm afraid those Conditional probabilities are not at all straightforward to compute. I think the method of Aisthesis with my adjustment gives a pretty good aproximation. | ||
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| Re: Does someone have AA or KK?, Aisthesis, 12. Nov 2003 21:07 | ||
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| Yeah, glad you cleared that up as I really didn't want to get involved in dealing with the factorial calculation method. To Gatsby: The reason my shortcut should actually be completely precise on calculating the general probability of just one of these occurring is this: Taking AA, your first pocket card MUST be an A, hence 1/13. Then there are 3 cards left out of 51, either of which will make your pair. This has to yield the identical result as going through the various factorials. PairTheBoard is totally right of course in subtracting for the conjunction of the 2 events, etc. To PairtheBoard: If you haven't already, would you mind taking a look at my calculations on flopping open-ended straight draw or 4-flush given a middle suited connector (if you already have an answer posted, forgive my asking, as I haven't checked for new posts there yet)? I know I took some shortcuts on that one but hope they didn't throw the result off by more than 1% or so. | ||
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| Re: Does someone have AA or KK?, pt_Gatsby, 13. Nov 2003 08:53 | ||
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| n*(n-1)*...*(n-k+1) / k! Boy, do I feel stupid =p I'm not sure what I was doing when I was writing this guy! Yes, I meant it in the, ermm... way that makes sense. Yes... Well, 4_c_2 = 6 - which is AA, * 2, which is AA or KK possibilities (ie: to calculate any pair, you could * 13) As for the 52_c_2, yup, totally off there to. It's certainly not a universally known standard - but seems to agree with the actual figure you calculate. Inside my mind, heh. Probabilty Player 1 misses AA, KK Probabilty Player 2 misses GIVEN Player 1 missed. Probability Player 3 misses GIVEN Players(1,2) missed. ... Probality Player 9 misses GIVEN Players(1,2,...,8) missed. The Devil is in the Details and I'm afraid those Conditional probabilities are not at all straightforward to compute. I think the method of Aisthesis with my adjustment gives a pretty good aproximation. Hmmm.. so, odds are basicly: ([52_c_2]-12/[52_c_2]) * ([(50_c_2)]-12/[50_c_2]) * ... * [(32_c_2)-12]/[32_c_2] ? But, isn't there a problem in that A and K will be diminishing at an equal rate? Using the same sampling (same rate of reduction), you are in effect saying that there are still 4 copies of each, where at the 32 points, there should be only (32/52)*4 copies, or 2.6 copies, which can be paired. Wouldn't this just become {[(32_c_2)] - [2.6_c_2]} / [32_c_2] ? IOW, the chance of the 9th person being dealt AA is not 'independent' per say - the sampling, the deck itself, will diminish in both net positive outcomes and total possible outcomes. As far as I can tell, while each are independent, the rate must be kept straightline, thus the possibilities remains at the {1-([52_c_2]-12/[52_c_2])}^(x players). Its as if you took half the deck out, then delt the remaining 26 cards - under your theory, the last 4 hands would all contain AA or KK because there are 8 of them left. In truth, there would be only '2' of each - the others would be in the other half (well... theoretically). | ||
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| Re: Does someone have AA or KK?, PairTheBoard, 13. Nov 2003 10:30 | ||
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| I'm afraid you're just guessing and BSing now pt. That may work for Philosophy Essays but you can't do math that way. sorry. | ||
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| Re: Does someone have AA or KK?, pt_Gatsby, 13. Nov 2003 12:04 | ||
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| A kind way of saying I am completely wrong then, heh. My math is pretty narrow I admit, but I try to work out all the problems I can. Probably a mistake when you are still missing some principles. Maybe you could answer one question though, or just tell me I'm an idiot again :) If the chances of getting AA or KK is 12/1326, then the second guys chances would what? How would you express the deck getting richer? And out of curiousity, in terms of just theory, could you create a full permutation of, say, 52_p_4 and then count how often AA or KK are in the first two or last two positions to get the exact answer? (and therefore create like 52_p_20 to figure out ten people). Just because I read a thing about there being two cards together in a deck of cards, and it was about 20% for a particular set of 4 cards. | ||
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| Re: Does someone have AA or KK?, Aisthesis, 13. Nov 2003 12:26 | ||
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| I'm having a little trouble with your notation here, so I can't really say. For most of this stuff, which can get really complicated in some cases regardless, I'm a big fan of just trying to think it through and looking at the SIMPLEST approach. You also have to really exact in defining both assumptions and results... (on your 52_p_4 method, I'm not really sure what that number is, nor am I entirely clear on just what question it's supposed to provide the exact answer to). Anyhow, thanks very much for the very helpful (albeit complicated!) posts on all this! But I'll have to admit that I had a lot of trouble following you at times. :) | ||
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| Re: Does someone have AA or KK?, pt_Gatsby, 13. Nov 2003 13:34 | ||
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| :) Probably cause I wasn't making any sense or doing it right. Know enough to confuse people, but not enough to get it right... most of the time :) -edit- Err... 52_p_4 is all permutations of 4 card hands using the 52 cards. I'm starting to think that's the only way this will be 100% accurate now. But... I think thisone should be left to the experts now, heh. (That is, to count how often AA would be the first two or last two in those permutations). | ||
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| Re: Does someone have AA or KK?, PairTheBoard, 13. Nov 2003 13:02 | ||
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| Sorry if I was harsh pt. But you might take a closer look at my previous posts and try to understand them before slogging things out on your own. The best and completely accurate solution is given by Lee below. Definitely look at that. I don't think it can be improved on. | ||
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| Re: Does someone have AA or KK?, pt_Gatsby, 13. Nov 2003 13:30 | ||
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| NP :) I appreciate someone giving it to me straight. But your right, I probably should :) | ||
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| Re: AA or KK? The answer, Schuster, 13. Nov 2003 11:11 | ||
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| Honestly I didn't read all of the posts above, but I skimmed them. They didn't look correct, but if they are correct, they are definately too complex. The answer is pretty simple once you understand it, and you can apply it pretty generally. Before the flop, there are 52c2 or 1326 combinations of possible hands. 12 of these hands are AA or KK, so that leaves 1314 combinations of hands without AA or KK. Assuming ten players, total combinations of 10 hands dealt out preflop would be 1326c10 (which is on the order of 10^24). The total combinations of hands dealt out that do not contain AA or KK would be 1314c10. Divide the combinations where no one has AA or KK by the possible combinations and you get the answer. (1314c10) / (1326c10) = 91.282% that no one has AA or KK, or 10.47 to 1. Hope this clears things up. Lee | ||
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| Re: AA and KK?, magnus, 13. Nov 2003 12:54 | ||
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| Ok, I have a followup question to all math geniuses (which I'm not) out there . So what is the odds for a hand containing both AA and KK? I know it's probably just my selective memory, but it seams to happen a lot to me (usually on the KK side... :-) Of cause last night it happened, but this time I had QQ loosing out to both other players..... (so what would the odds be to have all 3 (QQ, KK and AA) in a hand?) -Magnus | ||
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| Re: AA and KK?, PairTheBoard, 13. Nov 2003 13:05 | ||
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| Good question magnus. A followup to your followup is, what is the probabilty my KK are up against AA when I'm facing a preflop All-In Reraise early in a tournament? lol | ||
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| Re: AA or KK? The answer, PairTheBoard, 13. Nov 2003 12:55 | ||
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| As usual Lee is right. That's the thing with these kind of problems. If you see the way to do them they're easy. But it's not always easy to see the way to do them. Aisthesis was asking for the probability of seeing the AA or KK in one of your 9 opponents' hands. Using's Lee's solution then, first compute the probability the 9 not having AA or KK: 1314C9 / 1326C9 = 92.12095% So the probabilty of none of the 9 having AA or KK is Precisely: 1- 0.9212095 = 7.8791% Thats 1 chance in 12.692 or odds of 11.692 to 1 Aisthesis method aproximated the probabilty as 8.14% and with my adjustment the Estimate was 7.8% Not too bad for estimations. | ||
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| Incorrect, felson, 13. Nov 2003 13:03 | ||
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| This is erroneous, because 1326c10 allows duplication of hands -- for example, it allows one player to have 7c6c while another has td7c. | ||
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| Re: Incorrect, PairTheBoard, 13. Nov 2003 13:09 | ||
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| You're wrong felson. The 1326C10 Formula computes the number of ways 10 hands can be drawn from 1326 Without Replacement. So there are no duplications of hands. | ||
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| Re: Incorrect, felson, 13. Nov 2003 13:26 | ||
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| There are no duplications of _hands_, but there are duplications of cards. td7c and 7c6c are both among the 1326 hands, so they can both be chosen. | ||
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| Re: Incorrect, Schuster, 13. Nov 2003 13:34 | ||
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| Crickey, he's got me... I think... I've been awake for too long I guess. My initial hunch is that this error will be averaged out over massive amount, but I can't say for sure. At least now I'm more motivated to figure it out (without computer simulation). Lee | ||
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| Re: Incorrect, PairTheBoard, 13. Nov 2003 15:21 | ||
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| omg. You are right. Damn this is tricky stuff. Isn't it interesting that using this method for a deal to 9 players produces a result so close to Aisthesis and my estimate though? | ||
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| Re: Incorrect, Aisthesis, 13. Nov 2003 16:57 | ||
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| Yeah, I think felson's right. Although in my opinion our estimates definitely have the problem solved to the degree of accuracy we're ever going to need in poker contexts, it is getting to be a mathematical challenge. I think I have an approach just for AA out of 10 hands (inspired by Lee's idea). If it's right maybe we can get a true "formula" for the AA/KK problem. Ok, here goes: There are 52!/(52-20)! ways to deal the cards among 10 hands (I'm counting in this one JdTc and TcJd as DIFFERENT hands!!!). There are going to be 4! ways of getting AA in each of the 10 hands and 50!/(50-18)! ways of dealing the remaining cards. So we would have (10)(4!)(50!)/(50-18)! AA's altogether except that we've counted some multiple times: namely every time there have been 2 sets of AA at the table. For each of the 4! AA's in each hand, there will be 18 other ways to have another AA in some other hand, hence 18(48!)/(48-16)!. ALL of these hands have been counted twice I believe. So I get: {[(10)(4!)(50!)/(50-18)!] - [(18)(48!)/(48-16)!]} / [52!/(52-20)!] If one writes this down on paper, fortunately a bunch of it cancels out, and if I got my math right, works out to 97,997/1,082,900 = 9.05% No, this is clearly wrong. I have to run for the moment, or I'd try to figure out where I went wrong. I'll check when I get back from my pool game and try getting it right if you guys haven't beat me to it. Anyhow, as complicated as this is for AA alone, I see no hope (short of looking in a computer-generated table) of getting any closer than our estimates on the AA or KK problem--and there's no doubt in my mind that we're as close as we need to get. | ||
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| Re: Incorrect, Aisthesis, 13. Nov 2003 23:29 | ||
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| Ok, I figured out my mistake on the AA problem with 10 hands: There aren't 4! but only (4)(3) ordered combinations in each hand, so we get exactly: {[(10)(4)(3)(50!)/(50-18)!] - [(18)(48!)/(48-16)!]} / (52!/32!) = 48,997/1,082,900 or 4.5% Same for KK obviously, but I think my next step is going to be to look at felson's link... :) | ||
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| Re: AA or KK? The answer, PairTheBoard, 13. Nov 2003 16:31 | ||
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| So ok, felson is right, this way counts impossible deals where cards are duplicated in different hands. I'm pretty well stumped for an elegant way to do this. Here's an idea if anyone wants to pursue it. Working on the 9 player problem as Aisthesis requested, consider the number of ways 18 cards can be dealt out, order counting. ie. 52!/(52-18)! Then consider those cases where more than 1 A or K is included in the deal. In the case where say 2 Aces - but not more than 1 King - are included in the deal do this: Take 18C2 as the number of pairs of slots those 2 Aces can appear in. . 9 of those Slot-Pairs produce AA in someone's hand. So when 2 Aces are dealt out to the 18 spots, the chances are 9/( 18C2) = 1/17 that they produce a pair in someone's hand. Do similarly for all combinations where 2, 3,4 Aces and/or 2,3,4 Kings are dealt and combine the probabilties. Aisthesis's method gave an aproximation of 8.14%. I gave first order correction to that of 0.35% for an improved aproximation of 7.79%. The 2nd order correction to this figure should be less than 0.02%. I will rest on my 7.8% estimate, guessing the correct answer is closer to 7.78% | ||
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| Re: AA or KK? The answer, Smokey27, 3. Dec 2003 14:06 | ||
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| Nice Calculation. I aprroximated the numbers by doing 9/221 per hand. I think one should not stop with KK it is best to look at it all the way down the line. This is odds higher pair dealt factoring in your hand: AA 0.00% KK 95.59% QQ 91.18% JJ 86.78% 10s 82.37% 9s 77.96% 8s 73.55% 7s 69.14% 6s 64.73% 5s 60.33% 4s 55.92% 3s 51.51% 2s 47.10% I think the key I remember is in a full ring game there is a 4.5% chance of a higher pair being dealt for each possible hand, i.e. Jacks have about a 13.5% chance that there is a better pair out there. Dominated card statistics can also be calulated but from my rough estimation if you hold qj, there is a 9/81 chance that someone holds Ak. I need to learn how to do aq to ak but it is more difficult because if you have AQ there is a lesslikelyhood of AK or KQ becoase you have one of the eight needed cards. These types of calcualtions really tighten up my play. | ||
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| Link to answers, felson, 13. Nov 2003 16:40 | ||
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| BruceZ solves some related problems here, doing a favor for a producer of the WPT. http://forumserver.twoplustwo.com/showflat.php?Cat=&Number=212797&page=&view=&sb=5&o=&vc=1 There's one where he solves the problem of someone having AA or KK at a 9-handed table if you have QQ. Clearly this answer is the same whether you have QQ or JJ or 72o. Note that the answers are approximations, correct to 3 decimal places. I believe that the answers are approximate because BruceZ concluded that an exact solution requires computing a lot of terms, which trail off quickly. | ||
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| Re: Link to answers, Schuster, 13. Nov 2003 17:09 | ||
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| After wrapping my head around it, I had a feeling that the solution would trail off quite a few terms, but using only the first 2 or 3 would give a prediction accurate within +- 0.05%. I kinda thought up a quick computer algorithm that would only take roughly 52^4 calculations, which for modern computers is no time at all. If I feel like it, I'll write it up tonight, and if not, then tomorrow. Lee | ||
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| Re: Link to answers, PairTheBoard, 13. Nov 2003 18:55 | ||
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| Thanks for the Great Link felson. All of the situations shown there are ones of considerable interest. | ||
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| Re: Link to answers, tommyhawk, 19. Nov 2003 09:39 | ||
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| klick odds on this very site. You will find the odds of someone holding AA v.s KK. On the bottom here. tommyhawk. Keep on learning | ||
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