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.