jjgoldy5 0 Posted November 23, 2005 Share Posted November 23, 2005 3 handed at the end of a 30+3 SNG on PacificI pick up JJ on the button and raise to 3x BBFlop comes KcJcJh to nail me with quads.I check, SB bets, BB raises, I call, SB calls.Turn comes Ac, SB goes all in, BB calls, I call.SB shows 10c / 7c, BB shows K - ragRiver comes Qc to beat my quads with a royal.Now thats the first royal I've encountered - not all too surprising at 1 in 649,740 hands.Quads, however, are 1 in 4,165.So is losing quads to a royal really (649,740)(4165) = 1 in 2,706,167,100?? Link to post Share on other sites

btlee 0 Posted November 23, 2005 Share Posted November 23, 2005 So is losing quads to a royal really (649,740)(4165) = 1 in 2,706,167,100??no Link to post Share on other sites

Cory 0 Posted November 23, 2005 Share Posted November 23, 2005 i don't think so, but i've never been much of a math guy. Link to post Share on other sites

PAYforUSC 0 Posted November 23, 2005 Share Posted November 23, 2005 since online poker is rigged, conventional math does not apply. Link to post Share on other sites

Billy 0 Posted November 23, 2005 Share Posted November 23, 2005 Wow what a tough beat. As far as the math goes, I am not to sure about that. Link to post Share on other sites

NickG 0 Posted November 23, 2005 Share Posted November 23, 2005 I don't know the exact answer to this question, but I know your basic numbers aren't right. The odds you are using are for 5-card hands (e.g. 5-card stud), not 7-card hands as in Hold-em. The chances of getting quads or a straght flush are thus MUCH higher than you estimate.In about 25,000 hands I have on poker tracker, I have ended up with quads on 0.14% of hands where the river has been dealt (regardless of whether I stayed it). Thus suggest to me that the chance of hitting quads is better than 1/1000. I know the chance of FLOPPING quads if you have a pocket pair already is 1/400. Link to post Share on other sites

Hobbes 1 Posted November 23, 2005 Share Posted November 23, 2005 The math is right, but I'm not sure it's the correct math to use... Link to post Share on other sites

project240 0 Posted November 24, 2005 Share Posted November 24, 2005 Bad Beat Forum Link to post Share on other sites

CJHunt 0 Posted November 24, 2005 Share Posted November 24, 2005 I don't care about the math. You played it horribly though. Link to post Share on other sites

joball33 0 Posted November 24, 2005 Share Posted November 24, 2005 I don't care about the math. You played it horribly though.No sw here, how would you have played it? I don't mind the slowplay since after the flop he is 99.6% to win and after the turn he's 97.6% to win. If there was a ever a time for a slow play, this is it. Just a really crappy beat. I guess he could have possibly come over the top on the flop, but why? He doesn't want to get isolate so what is the point? He want's to let the draw hit. Link to post Share on other sites

NickG 0 Posted November 24, 2005 Share Posted November 24, 2005 Yeah, I don't see any problem with the way he played it. You have to slowplay the flop, because you want people to draw to their straights and flushes and what not...especially since other people were betting for him. The one thing I don't understand is that he said he checked on the flop, and then the SB bet. If he was on the button, why was he acting first on the flop? Link to post Share on other sites

joball33 0 Posted November 24, 2005 Share Posted November 24, 2005 Yeah, I don't see any problem with the way he played it. You have to slowplay the flop, because you want people to draw to their straights and flushes and what not...especially since other people were betting for him. The one thing I don't understand is that he said he checked on the flop, and then the SB bet. If he was on the button, why was he acting first on the flop?I noticed that too, but forgot to mention it in my post. Link to post Share on other sites

MadWolf1980 0 Posted November 24, 2005 Share Posted November 24, 2005 i don´t know Pacific but on Stars u would have been at least a 2-1 dog to win the hand Link to post Share on other sites

tsalagi 0 Posted November 24, 2005 Share Posted November 24, 2005 Ask Phil Gordon. He's the math wizard. Link to post Share on other sites

anselm 0 Posted November 24, 2005 Share Posted November 24, 2005 Worst beat I've ever heard of was quad kings vs. quad aces vs. royal flush. Link to post Share on other sites

semaj550 0 Posted November 24, 2005 Share Posted November 24, 2005 Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about 285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it. Link to post Share on other sites

Mesisca 0 Posted November 24, 2005 Share Posted November 24, 2005 oh no, you are completely right in your math semaj, but i dont think that helps the question of how often does any quads lose to a royal flush... Link to post Share on other sites

bampote 0 Posted November 24, 2005 Share Posted November 24, 2005 Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about 285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it.This looks rather impressive to me, so I'm quite ready to just say that you're right. Link to post Share on other sites

semaj550 0 Posted November 24, 2005 Share Posted November 24, 2005 oh no, you are completely right in your math semaj, but i dont think that helps the question of how often does any quads lose to a royal flush...Well it does tell you how often the situation the OP encoutered will happen. I guess I'll have to go back and calculate the odds of any quads losing to any royal now. I'll return with an answer shortly.Edit: Well at 11:50 this have proved to be just too daunting a task for my tired mind. If someone hasn't calculated it by tomorrow morning I might try it again. Link to post Share on other sites

schwarz269 0 Posted November 24, 2005 Share Posted November 24, 2005 without really looking at it the number will be too high because its figured out for the exact cards and suits. in the real problem it doesnt matter if the suit is clubs heart spades or diamonds. it also doesnt matter the the other players card is as long as he has the 10 of the suit that the royal will be made with. Link to post Share on other sites

Mesisca 0 Posted November 24, 2005 Share Posted November 24, 2005 im glad we have a math fan in semaj, I am sure you will be able to work out the answer as your math skills are rather impressive. Link to post Share on other sites

semaj550 0 Posted November 24, 2005 Share Posted November 24, 2005 Well, upon further consideration the odds of quads losing to a royal depend on the quads, i.e. it is more likely that your opponent has a royal when you have quads queens than when you have quad 4's. There's just too much math involved for me to do tonight. Link to post Share on other sites

NickG 0 Posted November 24, 2005 Share Posted November 24, 2005 Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about 285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it.A few comments:With respect to the starting hands, you're calculating the specific probability that three specific players have the three specific hands you pointed out. What we are really interested in is whether ANY three players at the table have the hands we are interested in. At a 10-person table, this could occur in any of 720 combinations, so you'd have to multiply your final result by that. Although if we are giving the hero the jacks, rather than just any player, its only 72.Along the same lines, you are calculating the probably that the board will read what it did in the specific order that it did....and I'm not really sure why that's important. That particular board could occur in another 120 combinations of order, which would again be multiplied by your final result.And of course, the question was not the chance of this specific hand happening, but more generally the chance of quads losing to a royal flush. For this to happen, the pocket pair doesn't have to be jacks, the suit doesn't have to be clubs, the third hand doesn't matter, etc. The board could come in many different combinations: for instance, the royal flush could use three or four board cards, the quads could be formed from trips on the board (e.g. JhAh v. AsKs with a board of TsQsJsJcJd). So I'm sure you're number is correct for what you are calculating, but that number doesn't really have anything to do with the OP's question.BTW, in order to figure this out, you can't just multiply the chance of getting quads in Hold-em by the chance of someone else getting a royal flush. The reason is that they are not independent events. If you have quads, you know the board must be paired, and a royal flush is much less likely if the board is paired than if it is not.To answer, I would basically start backwards, by trying to figure out how many combinations of boards there are where both a royal flush and quads are possible. All such boards would follow a small number of patterns.For instance:Let's assume A, B, C, and D are distinct ranks of broadway cards, and Z is any other rank of card. s, x, and y are any suit.1.) AsBsCsZxZy, where x and y could be the same as s2.) AsBsCsCxZy, where y can also be the same as s or x3.) AsBsCsCxCy4.) AsBsBxCsCy, where y can be the same as x5.) AsBsCsDsDxAre there any possible other possible combinations of flops that could include a royal and quads? I can't think of any off the top of my head.Once the number of these combinations is determined, you have to figure out the odds that that two players were actually dealt the hands that make quads and the royal. In the case of (1) and (2), there is only one hand that can make quads and one that can make the royal.In the case of (3), one hand makes the royal, and 44 hands make quads.In the case of (4), one hand make the royal, and 2 hands make quads.In the case of (5), 44 hands make the royal, and one hand makes quads.I don't think doing the combinations of each flop would be that hard either. For instance, in the case of (1):- For the ZxZy, there are 48 combinations of hands (99 - 22 in any of six suit combination)- For the AsBsCs, there are the following combinations:AKQAKJAKTAQJAQTAJTKQJKQTKJTQJT - 10 combinations, times 4 suits for 40 combinations totalSo for the full flop, there are 48*40 = 1920 combinations that follow this pattern, but you have to multiply by 5!, or 120, to accommodate for the fact that the flop could come in any order.You would then divide this by the total number of possible flops, which equals 52! / 47!.The final number is 230400/311875200 = .00074, or 1/1350...the is the probability that the flop fits pattern #1.The chance that you have the particular hand that makes quads with this flop is 2/(48*47) = 1/1128.The chance that a specific opponent has the hand that makes the royal is 2/(46*45) = 1/1035, and the chance that any of five opponents has a this hand is about 1/115.So the chance of all these things happening simultaneous is 1 / (1350*1128*115) = 1 in 175 million.And this is only under pattern 1...to find the actual probability, you would have add the probability for patterns 2-5, and any pattern I may have missed. In any case, the chance is considerably greater than your estimate...probably somewhere in the ballpark of 1 in 50 million. Link to post Share on other sites

gkunit20 1 Posted November 24, 2005 Share Posted November 24, 2005 Ask Phil Gordon. He's the math wizard.I think you mean David Skalansky or Matt Matros Link to post Share on other sites

The Enforcer 0 Posted November 24, 2005 Share Posted November 24, 2005 Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about 285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it. A few comments: With respect to the starting hands, you're calculating the specific probability that three specific players have the three specific hands you pointed out. What we are really interested in is whether ANY three players at the table have the hands we are interested in. At a 10-person table, this could occur in any of 720 combinations, so you'd have to multiply your final result by that. Although if we are giving the hero the jacks, rather than just any player, its only 72. Along the same lines, you are calculating the probably that the board will read what it did in the specific order that it did....and I'm not really sure why that's important. That particular board could occur in another 120 combinations of order, which would again be multiplied by your final result. And of course, the question was not the chance of this specific hand happening, but more generally the chance of quads losing to a royal flush. For this to happen, the pocket pair doesn't have to be jacks, the suit doesn't have to be clubs, the third hand doesn't matter, etc. The board could come in many different combinations: for instance, the royal flush could use three or four board cards, the quads could be formed from trips on the board (e.g. JhAh v. AsKs with a board of TsQsJsJcJd). So I'm sure you're number is correct for what you are calculating, but that number doesn't really have anything to do with the OP's question. BTW, in order to figure this out, you can't just multiply the chance of getting quads in Hold-em by the chance of someone else getting a royal flush. The reason is that they are not independent events. If you have quads, you know the board must be paired, and a royal flush is much less likely if the board is paired than if it is not. To answer, I would basically start backwards, by trying to figure out how many combinations of boards there are where both a royal flush and quads are possible. All such boards would follow a small number of patterns. For instance: Let's assume A, B, C, and D are distinct ranks of broadway cards, and Z is any other rank of card. s, x, and y are any suit. 1.) AsBsCsZxZy, where x and y could be the same as s 2.) AsBsCsCxZy, where y can also be the same as s or x 3.) AsBsCsCxCy 4.) AsBsBxCsCy, where y can be the same as x 5.) AsBsCsDsDx Are there any possible other possible combinations of flops that could include a royal and quads? I can't think of any off the top of my head. Once the number of these combinations is determined, you have to figure out the odds that that two players were actually dealt the hands that make quads and the royal. In the case of (1) and (2), there is only one hand that can make quads and one that can make the royal. In the case of (3), one hand makes the royal, and 44 hands make quads. In the case of (4), one hand make the royal, and 2 hands make quads. In the case of (5), 44 hands make the royal, and one hand makes quads. I don't think doing the combinations of each flop would be that hard either. For instance, in the case of (1): - For the ZxZy, there are 48 combinations of hands (99 - 22 in any of six suit combination) - For the AsBsCs, there are the following combinations: AKQ AKJ AKT AQJ AQT AJT KQJ KQT KJT QJT - 10 combinations, times 4 suits for 40 combinations total So for the full flop, there are 48*40 = 1920 combinations that follow this pattern, but you have to multiply by 5!, or 120, to accommodate for the fact that the flop could come in any order. You would then divide this by the total number of possible flops, which equals 52! / 47!. The final number is 230400/311875200 = .00074, or 1/1350...the is the probability that the flop fits pattern #1. The chance that you have the particular hand that makes quads with this flop is 2/(48*47) = 1/1128. The chance that a specific opponent has the hand that makes the royal is 2/(46*45) = 1/1035, and the chance that any of five opponents has a this hand is about 1/115. So the chance of all these things happening simultaneous is 1 / (1350*1128*115) = 1 in 175 million. And this is only under pattern 1...to find the actual probability, you would have add the probability for patterns 2-5, and any pattern I may have missed. In any case, the chance is considerably greater than your estimate...probably somewhere in the ballpark of 1 in 50 million. :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: Link to post Share on other sites

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