45 replies to this topic

[Mesisca]

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.

[semaj550]

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.

[NickG]

semaj550 said:

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.

[gkunit20]

tsalagi said:

Ask Phil Gordon. He's the math wizard.

I think you mean David Skalansky or Matt Matros

[The Enforcer]

semaj550 said:

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.

Quote

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.

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[gkunit20]

semaj550 said:

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.

Quote

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.

I see Skalansky has posted :shock: :shock: Now if he could post in english, it would be very helpful.

[NickG]

I just realized what I posted contains a mistake...on the odds of having the quad and flush hands, where I use 2/(48*47), it should be 2/(47*46), and where I use 2/(46*45), it should be 2/(45*44)...changes the final result to 1 in 160 million, not 1 in 175 million.

[milestodavid]

NickG said:

semaj550 said:

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.

oh i get it now

[Jeepster80125]

gkunit20 said:

Now if he could post in english, it would be very helpful.

He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".

[pwoblo]

I guess you are assuming two players. 5 community cards. 2 hole cards.Of the five community, 3 must be broadway and of one suit. Total combinations? C(5,3)*4 = 40. [the 4 was for each suit].Several ways to go after this on the board:a.pairs below tens(99-22). Combinations? 8*[C(4,2)] = 48b.pair of broadway cards including the royal card. 2*[C(3,1)] = 6.c.pairing one of the already existing broadways, and any non royal card: 3*3*38=342.d. pairing royal card, and another royal card: 3*3*2=18.e.double pairing the royal cards: 9*6= 54.f.putting a set of one of the royal cards on board= 3*3=9.hole cards:a. royal holes * quad holes = 1 * 1 = 1b. royal * quad = c[44,1]*1 = 44.c. royal * quad = 1*1=1d. 44*1=144e. 1*2=2f. 1 * 44 = 44.combos of [letter]:a. 1*48=48b.6*44=264c. 342*1=342d.18*44 = 792e. 54*2=108f.9*44=396sum total = 1950total combos of community cards = C(52,5) = 2,598,960total combos of hole cards = C(47,2) * C(45,2) = 1070190probability of quads and royal = (combos of 3 royals)*(sum of each letter combo)/combos of community cards/ combos of hole cards= (40)*(1950)/(2,598,960)/(1070190) = 2.8044E-8 = 2.8044 out of 10^8 = = one out of 35.66 millionSo for 2 players, 2 hole cards, 5 board cards, it's one out of 35.66 mil. Is everything copamalecetic with the calculations? I think this is right...

[pwoblo]

I guess you are assuming two players. 5 community cards. 2 hole cards.Of the five community, 3 must be broadway and of one suit. Total combinations? C(5,3)*4 = 40. [the 4 was for each suit].Several ways to go after this on the board:a.pairs below tens(99-22). Combinations? 8*[C(4,2)] = 48b.pair of broadway cards including the royal card. 2*[C(3,1)] = 6.c.pairing one of the already existing broadways, and any non royal card: 3*3*38=342.d. pairing royal card, and another royal card: 3*3*2=18.e.double pairing the royal cards: 9*6= 54.f.putting a set of one of the royal cards on board= 3*3=9.hole cards:a. royal holes * quad holes = 1 * 1 = 1b. royal * quad = c[44,1]*1 = 44.c. royal * quad = 1*1=1d. 44*1=144e. 1*2=2f. 1 * 44 = 44.combos of [letter]:a. 1*48=48b.6*44=264c. 342*1=342d.18*44 = 792e. 54*2=108f.9*44=396sum total = 1950total combos of community cards = C(52,5) = 2,598,960total combos of hole cards = C(47,2) * C(45,2) = 1070190probability of quads and royal = (combos of 3 royals)*(sum of each letter combo)/combos of community cards/ combos of hole cards= (40)*(1950)/(2,598,960)/(1070190) = 2.8044E-8 = 2.8044 out of 10^8 = = one out of 35.66 millionSo for 2 players, 2 hole cards, 5 board cards, it's one out of 35.66 mil. Is everything copamalecetic with the calculations? I think this is right...

[jjgoldy5]

when this happened, I laughed hysterically.... then quit poker for a week. I'd say I handled it pretty well!At the time I figured it to be 1-24million though I dont have my work anymore, but Im thinking its higher than that.Maybe it would be easier to find the odds of pocket pair drawn to quads dying to a straight flush, then the odds of a straight flush being a royal flush?

[pwoblo]

Jeepster80125 said:

gkunit20 said:

Now if he could post in english, it would be very helpful.

He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".

Hah... let the truth be told.

[Fluffdog87]

im a poker math guy.the odds are low.its as irrelevant as possible.

[troutsmart]

I'm in your elite company OP. I too had quads taken down by a royal, though I turned my quads vs. flopping them. And I was holding a non-paired pocket (QJ) vs. your pocket (JJ). The odds are the stuff that could give MIT grads a headache. What I found extraordinary in my case was the fact that this hand occured when we were playing 3-handed at the end of a tournament. The hilarious part was the reaction of the 5 people left in the room. When my opponent went all-in on the river, which gave him the royal, I simply said in a rather dry manner as I called, "I have the queen." He flipped over his K J on the board of Q Q T Q A . I was stunned, as was everybody else, but not less than 10 seconds later we were playing the next hand, and most of the commotion was over. Everybody was talking about the royal, but nobody seemed to care about the quads losing. I still find it one of the quirky moments in my life that seems surreal.

[Fluffdog87]

I think everyone has had this happen to them.so its much lower how many hands have you ever played.its not even 1 in 50 million.think about itits not even 1 in a million.its not even 1 in 100k.we have all seen it happen to others and ourselves, so the number is so far off its just hilarious.my friend beat quads with a runner runner royal flush like 3-4 weeks ago. :cry:

[Dixie Wrecked]

PAYforUSC said:

since online poker is rigged, conventional math does not apply.

LMAO did you math freaks factor in PayforUSC's "online poker is rigged" constant. If you did, I bet the odds would be a lot less from 1 in eleventy brazillian to something like 1 in 5.

[Greg Kelley]

Debug error. Double posted.

[Greg Kelley]

Edit: double post

[Greg Kelley]

Jeepster80125 said:

gkunit20 said:

Now if he could post in english, it would be very helpful.

He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".

Mr Jeepster Sir,Why is it that your posts seem to turn every topic into an intelligence contest? What is your deal? I cruise through the forum infrequently, yet I always seem to see a post from you flaming away at someone. The guy was making a light hearted comment... and , as usual, it somehow ends up being about how "smart" you are, and how stupid someone else is.Try happiness. You might like it. Yeah, I know... I'm an idiot too...At least I'm not an insecure prick.