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Server Time: 2/13/2012 2:20:55 PM PACIFIC |
 A-K, Kingstalker, 19. Dec 2003 23:17 | ||
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| If your dealt A-K in a 9 handed game or any hand with a ace in it, what are the odds that somebody has A-A? | ||
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| Re: A-K, Angel, 19. Dec 2003 23:33 | ||
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| 1 chance in approximately 51 1/2 | ||
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| Re: A-K, SmallFeesh, 20. Dec 2003 00:40 | ||
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| 9 player game, 21.9-1 10 player game, 19.7-1 possibility that no one else has an ace but you in 10 handed, 25.3% on that note i have a story! ita a 10-20 holdem game at the Mirage, i get A-K, one guy raises, i re-raise, and we see a flop... flop comes A K 3 rainbow, i chek, he bets, i raise, he re-raises, i re-raise, and he comes back over the top! im like, ok... weve got the same hand... or hes got A Q.. but hes been losing and has shown some crazy draws, so a blank on the turn and i come out betting... and he raises! so now it hits me... does this guy have AA or KK? this guy who has had alot of junk... now have the monster?? hes paid everyone off but me... now im paying him off!?! needless to say i back way off and check call on the river.. sure enough! pocket rockets! VERY expensive hand for me! after that he picks up his chips, not even half of what he's lost, and gets out of dodge! leaves me with a bunch of rocks sympathizing with me and a depleted stack! lol.... my luck huh? | ||
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| Re: A-K, Angel, 20. Dec 2003 13:27 | ||
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| on 20. Dec 2003 00:40 SmallFeesh wrote: > 9 player game, 21.9-1 I'm fighting the flu right now so perhaps it's my brain that's congested but could you tell me how you did the math? For what it's worth - this was how I arrived at my conclusion of 51.5:1 (3/50)*(2/49) = x 1 - x = y y^8 = z 1 - z = 0.019 or 1.9% or 51.5:1 against | ||
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| Re: A-K, iceman5, 20. Dec 2003 13:30 | ||
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| I dont know the odds, but it happened to me today in a SNG. Luckily, I was the one with AA. | ||
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| Re: A-K, stdioh, 26. Dec 2003 20:37 | ||
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| First off, the chance of an opponent having AA heads up when you hold one ace is obviously 3/48*2/47 which is 1/376. Now you also have to work out the odds of him getting a single ace. Then you break it into cases. If he got one ace then the next player has 2/48*1/47 and if he didn't get an ace then that player has 3/48*2/47. The chance of him having gotten a single ace is 3/48*45/47+45/48*3/47 which is equal to about 3/25. So the chance that the second opponent has AA is about 3/25*2/48*1/47+22/25*3/48*2/47 which is about 1:409 Now you say, "Assuming that neither player got AA what is the chance that one of the got exactly one ace and what is the chance that each of them got exactly one ace?" Then you work out the next guy's chances for one ace used and no aces used again, with the reduced set of cards. Obviously if there are 2 aces used, then there is a 0% chance of AA appearing in a subsequent hand. Thus you work it through for as many players as you want. It would be time consuming to do the entire calculation, but you can see from here that each subsequent chance decreases and we are starting with 1/376 and then 1/409 ... so an absolute upper bound to the number at a 9 handed table would be 1/376 + 7/409 which would be an absolute highest liklihood of 1/27.1 ... however with the decreasing series I would estimate this number to be more like 1/59.2 If anybody is really curious they can carry out the entire arithmetic of it. | ||
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| Re: A-K, TrippH, 29. Dec 2003 18:16 | ||
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| wow, I think I see what you're trying to do, but it makes it so much more difficult ... break it into cases, but for the number of aces that are in hands besides yours: it's 9-handed, you have Ax ... so what are the odds someone holds 2 of the remaining 3 aces in his hand? ... there obviously have to be either 2 or 3 aces among the other 16 cards dealt out of what is now a 50-card population for anyone to have AA, so figure the odds of there being exactly 2 aces dealt, then the odds that they are both in one hand, and add it to the odds that all 3 aces were dealt and that 2 are in the same hand: exactly 2 aces were dealt: ((3/50)(2/49) x (47/48)(46/47......35/36)(34/35))(16!/14!2!) (probability of (2 random cards among the 16 being aces x no other dealt card being the other aces)(the number of ways this arrangement can occur) = 51/245 ... if 2 aces are dealt, the probability they are in the same hand -- 1/15 (pick the ace of the higher suit; 1 of the other 15 cards must be the other ace for there to be AA) ... (51/245)(1/15) = 51/3675 = roughly .013878 exactly 3 aces were dealt: ((3/50)(2/49)(1/48))((16!/13!3!)) (probability of 3 random cards among the 16 being aces)(number of ways this arrangement can occur) = 1/35 ... if 3 aces are dealt, the probability 2 are in the same hand -- like above, pick A of highest suit -- probability one of the other aces is with it = 2/15 ... if not, then pick A of 2nd highest suit -- probability the last ace is with it = 1/13 ... 2/15+ 1/13 = 41/195 ... (1/35)(41/195) = 41/6825 = roughly .006007 and .013878 + .006007 = .019885 = roughly 1/50.3 | ||
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| Re: A-K, stdioh, 31. Dec 2003 09:39 | ||
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| Very good! Your math is correct and much simpler. It has been too long since I finished my math degree and I think that the working world has taught me to overengineer a little :) | ||
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| Re: A-K, TrippH, 31. Dec 2003 18:45 | ||
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| ha, I have an English degree, that's probably why my math remains pure ... but I was a math contest dork in high school, so probability is like my second language ... really only comes in handy in poker ... it's a good thing I discovered poker or I may have forgotten it all | ||
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| Re: A-K, SmallFeesh, 22. Dec 2003 01:42 | ||
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| math! haha... im no good at math! i have to memorize the stuff! i got it from Ken Warrens "winners guide to texas holdem poker"... i went back and looked it up... and i was right... he might be wrong.... i doubt it... if i have the will power ill look it up else where... but i myself am suffering from a TERRIBLE hang over... so i dont feel like doin to much of anything! | ||
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| Re: A-K, TrippH, 22. Dec 2003 15:23 | ||
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| nah, the numbers you're quoting are the odds that you hold AK and an opponent holds EITHER AA or KK ... the table is for odds that your hand is an underdog ... the question is slightly different -- you hold an ace (not necessarily AK but it really doesn't matter in the question), and he's asking what are the odds that someone holds AA ... Angel was close enough but simplified the math somewhat ... if I didn't screw up the math, the answer for a 9-handed game is about 1 in 50.3 | ||
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| Re: A-K, SmallFeesh, 23. Dec 2003 05:50 | ||
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| ah yes... i looked AGAIN... and your right... lol | ||
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| Re: A-K, Aisthesis, 30. Dec 2003 09:30 | ||
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| I get an answer very close to that of Angel's. Here's my thinking: There are 50!/34! ways to deal the remaining 8 hands (ordered). Then, in any of the 8 positions, there will be 6*(48!)/(34!) deals on which the hand in question is AA. Since, in this case, AA in any hand excludes AA in any other hand (since we know we ourselves have AK), there are hence 8*6*(48!)/(34!) deals on which someone else has AA. Probability is thus: [48*(48!)*(34!)]/[(34!)*(50!)] = 24/1225 = 1.96% = 1/51.04 | ||
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| Re: A-K, stdioh, 31. Dec 2003 09:41 | ||
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| You neglect one key piece of information. The ways to deal AA have to take into account that one of the aces has to belong to AK ... you can't just use that fact for the mutex. | ||
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| Re: A-K, Aisthesis, 5. Jan 2004 18:47 | ||
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| Not sure exactly what you mean there. Basically what I was doing was pulling an AK out of the deck, then dealing the other 8 hands randomly with the remaining cards. | ||
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