Mandelbrot 0 Posted February 8, 2005 Share Posted February 8, 2005 What are the chances in a 9 handed game if you have QQ in the hole, that one or more of the other 8 players has any of the following hole cards {AK, KK, AA, AQ, KQ, or AJ}?I have an answer but I'm not 100 % sure it's correct so if anyone could post their answer and "work", I'd appreciate it greatly. :-) Link to post Share on other sites
JFarrell20 1 Posted February 8, 2005 Share Posted February 8, 2005 Slim.Your bill will be in the mail. I accept cash or money order.Good day. Link to post Share on other sites
THE ACE IS BACK 0 Posted February 8, 2005 Share Posted February 8, 2005 I would like to know the answer 2. Funny enough I had QQ last night and two other people in the same hand had KK and AA. Lovely. Link to post Share on other sites
FloppyNuts 0 Posted February 9, 2005 Share Posted February 9, 2005 I'll figure it out for $10. (Maybe we can do a Poker Stars transfer... I can give you my handle there if you're serious.) Link to post Share on other sites
Mandelbrot 0 Posted February 9, 2005 Author Share Posted February 9, 2005 Plz don't bother. I offer advice all of the time to people on this forum without attempting to charge them fees. If you know and are willing to share, thanks. If not, then please move on. Link to post Share on other sites
PA32R 0 Posted February 9, 2005 Share Posted February 9, 2005 Plz don't bother. I offer advice all of the time to people on this forum without attempting to charge them fees. If you know and are willing to share, thanks. If not, then please move on.I think it's safe to say he was just kidding about the fee..... Link to post Share on other sites
Mandelbrot 0 Posted February 9, 2005 Author Share Posted February 9, 2005 I think it's safe to say he was just kidding about the fee.....I thought so too until he mentioned about doing a xfer on PStars. Read it again. Link to post Share on other sites
FloppyNuts 0 Posted February 9, 2005 Share Posted February 9, 2005 I was semi-facetious about the fee. I'm always trying to make a buck.I'll give you my answer, but the methodology's a b!tch to lay out here. The math is quite a bit of legwork. It's actually not that slim: 20.68%, or 3.8 to 1 against. Link to post Share on other sites
akishore 0 Posted February 9, 2005 Share Posted February 9, 2005 a simple reply:(EDIT: after finishing, i see this was no simple reply. read on at your own risk...)as long as an opponent doesn't share a card with you, the probability of them getting whatever is the same.e.g. you have QQ, they will still get AA with the same frequency, same with KK. doesn't matter if they have a pocket pair or AK or any two cards, as long as neither is a Q, their probability is unnafected from your point of view (since you see two cards out of 52, and they're not going to get the other two queens).if you want to know the prob. of an opponent having the same cards as you, e.g. QQ vs QQ, the probability is simply 2/50*1/49 = approx. 1/1225, worse than a runner-runner perfect catch.if you share one card, it's a little tricker. the opponent has 6 outs from which he needs 2, and one has to be one of your cards. so let's say you have QQ and we want to see the probability of him getting AQ. if you make a tree, the equation comes out something like: (2/50*4/49)+(4/50*2/49) = 1/613 approx.if you want to see the probability of him simply sharing one card, it's just 2/50+48/50*2/49 = 1/13 approx.up to this point, though, it's all been one opponent, so for a nine-handed ring game it gets super-tricky. an example would be something like this:you have QQ, and want to know the probability of someone having sharing a queen (forget AQ or KQ, just sharing a queen). you have 8 other players for a total of 16 cards you have to account for. that becomes a pain. for just two people, the equation is:2/50 + 48/50 * (2/49 + 47/49 * (2/48 + 46/48 * (2/47))) = 15.5%. that means in a three-handed game, at LEAST one (possibly both) of your remaining queens will be in someone's hand. for a four-handed game:2/50 + 48/50 * (2/49 + 47/49 * (2/48 + 46/48 * (2/47 + 45/47 * (2/46 + 44/46 * (2/45 + 43/45 * (2/44 + 42/44 * (2/43))))))) = 29.7%. that means in a five-handed game, over 1/4 of the time one of your remaining queens (again, at least) will be in someone's hand.hell, i'll do the nine-handed calculation for you, but it's gonna be a bitch.2/50 + 48/50 * (2/49 + 47/49 * (2/48 + 46/48 * (2/47 + 45/47 * (2/46 + 44/46 * (2/45 + 43/45 * (2/44 + 42/44 * (2/43 + 41/43 * (2/42 + 40/42 * (2/41 + 39/41 * (2/40 + 38/40 * (2/39 + 37/39 * (2/38 + 36/38 * (2/37 + 35/37 * (2/36 + 34/36 * (2/35))))))))))))))) = 54.2%. so that means in a nine-handed ring game, about half the time one of your queens will be dead.but never use this in a game! this doesn't mean you should figure the outs of hitting your set to be 1/34 since you "know" the other players hold at least one queen. this is completely flawed thinking.anyway, sorry this was a super-long post. basically, i don't know how to do the prob. that your opponents hold any of the following hole cards, but i can do simpler stuff like that.aseem Link to post Share on other sites
jakoye 0 Posted February 9, 2005 Share Posted February 9, 2005 I would like to know the answer 2. Funny enough I had QQ last night and two other people in the same hand had KK and AA. Lovely.This happened to me last night as well. I think the odds are about 1 in 10,000 for this to happen? Hopefully, it won't happen to me again for a VERY long time! Link to post Share on other sites
jogsxyz 0 Posted February 9, 2005 Share Posted February 9, 2005 It's really useless information. Not worth the time to calculate. This is a game based on conditional probabilities, not unconditional. Now UTG opens for 4XBB and from a mid position Mr tight*ss raises to 20XBB. The chances of pockets queen being the best hand is close to zero. That's what this game is all about. Link to post Share on other sites
rog 0 Posted February 9, 2005 Share Posted February 9, 2005 What are the chances in a 9 handed game if you have QQ in the hole, that one or more of the other 8 players has any of the following hole cards {AK, KK, AA, AQ, KQ, or AJ}?I have an answer but I'm not 100 % sure it's correct so if anyone could post their answer and "work", I'd appreciate it greatly. :-)Before I read anyone else's math, I'll post my own.There are 1225 2-card combinations to be made from a deck of 50 cards. 1225 is 49 + 48 + 47 + 46 ...+ 2 + 1. This reduces to (n^2 - n)/2. I wont go into the reduction to formula, but the sum series comes from this: 2 card deck, you get 1 combo. Add a 3rd card, and you get 2 more combos...one each consisting of the new card and one of the old ones. So, the 50th card added brings in 49 new 2-card combos.How many are bad:KK: 6 combos: red, black, sh, sd, ch, cdAA: 6 combosAK: 16 combos (4 different kings for each of 4 Aces)AQ: 16AJ: 16KQ: 16Total 76 combos of interest.Probability that a particular player has one: 76/1225 (6.2%)P that a particular player doesn't have one: 1149/1226 (93.8%)P that N players dont have that hand: (1149/1225) ^ NP that 8 players dont have a hand (pokerstars full HE table): 59.9%P that someone has one of those hands in a full (9 handed game) against your QQ: (100 - 59.9)% or 40.1% or almost half the time.P that someone has one of your hands 6-handed: 27.4%4 handed: 17.5%Rog Link to post Share on other sites
akishore 0 Posted February 9, 2005 Share Posted February 9, 2005 How many are bad:KK: 6 combos: red, black, sh, sd, ch, cdAA: 6 combosAK: 16 combos (4 different kings for each of 4 Aces)AQ: 16AJ: 16KQ: 16Total 76 combos of interest.Probability that a particular player has one: 76/1225 (6.2%)P that a particular player doesn't have one: 1149/1226 (93.8%)P that N players dont have that hand: (1149/1225) ^ NP that 8 players dont have a hand (pokerstars full HE table): 59.9%P that someone has one of those hands in a full (9 handed game) against your QQ: (100 - 59.9)% or 40.1% or almost half the time.P that someone has one of your hands 6-handed: 27.4%4 handed: 17.5%great post, correct math, etc.just note that the OP wanted to assume he was holding QQ, so the number of AQ and KQ combos drops to 8 each, so 68 combos total.also--i might be wrong--i don't think P that N players don't have a hand is simplified to (1149/1225)^N, because the events are not independent of each other. if player one doesn't get any of the 76 or 68 combos, the number of remaining combos drops by one, which affects the next person's odds. continues down the chain of 8 players. for simplicity's sake, it'll be around the same since there are so many possible combos, just letting you know.great post, again. you make it look so easy. :-) aseem Link to post Share on other sites
rog 0 Posted February 9, 2005 Share Posted February 9, 2005 Thanks for the correction. I forgot to drop some of the combos including Q. I'm not sure about the declining odds critique. I'm going to mull that over and post if I have any thoughts on it.BTW: On another thread involving odds from flop to river, there were 2 methods posted that turned out to be equivalent. One involved the same method I used here...1 minus the probability of not making it twice. The other involved the odds of making it first time + odds of second time - odds of making it both times (subtract out double counting). You could use either method here, but applying the second method here would be mind-bogglingly difficult as the "double counting" itself with that many events involved would be extremely difficult to calculate. Rog Link to post Share on other sites
akishore 0 Posted February 9, 2005 Share Posted February 9, 2005 Thanks for the correction. I forgot to drop some of the combos including Q. I'm not sure about the declining odds critique. I'm going to mull that over and post if I have any thoughts on it.BTW: On another thread involving odds from flop to river, there were 2 methods posted that turned out to be equivalent. One involved the same method I used here...1 minus the probability of not making it twice. The other involved the odds of making it first time + odds of second time - odds of making it both times (subtract out double counting). You could use either method here, but applying the second method here would be mind-bogglingly difficult as the "double counting" itself with that many events involved would be extremely difficult to calculate. Roghmm, i think you might be right, as well, then. because another correct method is (odds of making it first time) * (odds of not making it second time if you make it first time) + (odds of not making it first time) * (odds of making it second time if you don't make it first time).e.g. you draw two cards out of a 52-card deck and want to know the chance of getting at least exactly one ace.method 1:(4/52)*(48/51) + (48/52)*(4/51) = 14.5%method 2:(4/52) + (4/51) - (4/52)*(3/51) = 15.1%method 3:1 - (48/52)*(47/51) = 15.0%so actually, they're not all the same. tell me if one of my numbers is wrong here. i know that method 1 works best in scenarios where you're interested in finding the odds that at LEAST one ace will be drawn, or at LEAST one person will have one of those specific hole cards.e.g. you have 88 and want to find out the probability that the flop will have at LEAST one of your remaining 8's.first card on the flop has a 2/50 chance of being an 8. it has a 48/50 chance of not being an 8. if it's an 8, we're done. if it's not an 8, the second card on the flop has a 2/49 chance of being an 8, and a 47/49 chance of not. if it's an 8, we're done, otherwise the third card has a 2/48 chance of being an 8, and we're done.method 1:(2/50) + (48/50) * [(2/49) + (47/49)*(2/48)] = 11.8% chance of flopping at LEAST a set, possibly quads. i know this number is correct; it's the same number on the hold'em charts at various sites.method 2:(2/50) + (2/49) + (2/48) = 12.2%. this is the same percentage that people get to using an inaccurate "rule of 6" for figuring out their chances of hitting a certain number of outs on the flop (multiply by 2, then by 3 since there are 3 cards to come).this is wrong because the 3 card probabilities are very dependent on each other, and adding probabilities only works if they're independent. the chances of the second card being an 8 is NOT (2/49), it's (2/49) GIVEN that the first card was not an 8, so you have to account for that by multiplying (2/49) by the probability that the first card was not an 8, so the real probability that the second card is an 8 is (48/50)*(2/49), which is ALMOST equal to (2/49), but not quite.if the first card WAS an 8, we don't care, but just for reference, it would be (1/50)*(1/49), an extremely small probability, so it wouldn't hurt us if we did care about double-counting and forgot to account for it. note that i didn't subtract for double-counting since we don't mind double-counting.method 3:1 - (48/50)*(47/49)*(46/48) = 11.8%. now this method is correct.so your method of 1 - (x/n) was correct, it'll give the same result as my method of (x/n) + (1-x/n)*(x/(n-1) + ...). the addition/subtraction method is inccorect, however, in these particular cases where you're interested in at LEAST one person having something and not minding if more than one person does.thanks,aseem Link to post Share on other sites
rog 0 Posted February 9, 2005 Share Posted February 9, 2005 Actually in retrospect, I think my method was flawed. For one thing, I think you do need to reduce the deck size for each successive hand. The proof is that using my method, it would be possible to take a deck containing only 4 kings and 4 aces, and deal one card each to 5 players, and have nobody holding an ace. Reductio ad absurdum.There's an even bigger hole in my math. I dont take into account that some combinations are mutually exclusive. For example, if anyone has Qx, then AQ becomes less likely, but if the first 2 people are holding Qx, then all combos with Q are dead, reducing the number of good combos. I'll have to think harder about the math of this. Maybe there's an average case to be calculated. The math of single cards is completely tractable. When combos are involved, it gets pretty hairy. I think I could write a computer simulation a lot quicker than I could work out the equations, but I'm going to bounce the problem around with some math degree friends of mine to see what they think.Rog Link to post Share on other sites
akishore 0 Posted February 9, 2005 Share Posted February 9, 2005 yeah, combinations are pretty difficult to calculate and think about. good luck, hope your math degree friends come up with a nice solution. :-) aseem Link to post Share on other sites
Mandelbrot 0 Posted February 9, 2005 Author Share Posted February 9, 2005 These are the exact same problems that I encountered when trying to figure this out. It would be a lot easier if you could assume independent hands and a 50 card deck. The best answer I could come up with was around 20%. Link to post Share on other sites
akishore 0 Posted February 9, 2005 Share Posted February 9, 2005 i have an idea. this answers the question, "if i have QQ in the hole, what is the probability that at LEAST one other person in a nine-handed game has the either AA, KK, AK, AQ, KQ, or AJ?"if you take out AJ and KJ, every hand you posted has a Q or higher. since there are 10 remaining cards that are Q or higher, there are exactly 45 possible combinations of hands:AA - 6KK - 6AK - 16AQ - 8KQ - 8AJ - 16total - 60total possible - 1225 from a 50-card deckthe probability that the first person did NOT get one of these combinations is 1165/1225. if the first person got it, you're done, otherwise, you need to calculate the probability that the second person did NOT get one of these combinations.with 48 cards in the deck, the total number of desired combinations remains 60 while the total number of possible combinations reduces to 1128. so the probability that the second person ALSO did NOT receive one of those combinations is 1068/1128.we can continue for the rest of the six people, so here is my formula.let f(n) = (n^2 - n)/2 to calculate the number of combinations remaining in a deck of n cards.p = 1 - [f(50)-60]/f(n) * [f(48)-60]/f(48) * [f(46)-60]/f(46) * ... * [f(36)-60]/f(36)this will give the probability of at least one person getting one of those combinations (AA, KK, QQ, AK, AQ, KQ).so when i calculate it out:p = 1 - (1165/1225)*(1068/1128)*(975/1035)*(886/946)*(801/861)*(730/780)*(643/703)*(570/630)p = 1 - 0.5724p = 42.76%i think this is a lot more accurate, but i'm still not completely convinced it's totally right. for example, if the first person got A2, he reduces the number of combinations from 60 to 45. however, we would still say that he didn't get one of our combinations, so there are still 60 such combinations left--this is wrong.i know this is flawed, but i think it's an improvement. any more ideas?aseemp.s. this will work with any hand you want to find out about, e.g. the chances that someone will be dealt AA. just change the number of combinations from 60 to whatever, then do the same calculations. it is still flawed, though. Link to post Share on other sites
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