I show the probability that any one player gets a royal is 42,807,600 /463,563,500,400 = 0.000092. The probability of not having a royal would be 1-0.000092=0.999908. An estimate of the probability there are no royals in 9 hands is 0.999908^9 = 0.999169. So the probability of at least one royal in 9 hands is 1-0.999169=0.000831.

Before somebody mentions it, this assumes that each hand is dealt from a separate deck. The probabilities are correlated when they all come from the same deck, but I think that effect is very minor.

Quote:WizardFor the benefit of others, in Omaha the player gets four hole cards, and there are five community cards. The player must use exactly two of his hole cards and three community cards.

I show the probability that any one player gets a royal is 42,807,600 /463,563,500,400 = 0.000092. The probability of not having a royal would be 1-0.000092=0.999908. An estimate of the probability there are no royals in 9 hands is 0.999908^9 = 0.999169. So the probability of at least one royal in 9 hands is 1-0.999169=0.000831.

Before somebody mentions it, this assumes that each hand is dealt from a separate deck. The probabilities are correlated when they all come from the same deck, but I think that effect is very minor.

Of course, this assumes that every player will play every hand all the way to the river, which is never the case. Using a fairly generous assumption that five out of nine players see the flop, and that two such players, on the average, play all the way to the river (many pots are resolved before then), you would have to multiply the figure above by ( (.55)(.4) ) to get a real-world estimation.

Quote:mkl654321Of course, this assumes that every player will play every hand all the way to the river, which is never the case. Using a fairly generous assumption that five out of nine players see the flop, and that two such players, on the average, play all the way to the river (many pots are resolved before then), you would have to multiply the figure above by ( (.55)(.4) ) to get a real-world estimation.

I'm not sure which "figure above" you are referring to. The way I would make that adjustment is to estimate the number out of 9 players who will stay in the whole way. Let's say that is 4 players. Then instead of the ^9 in my math above, do a ^4. I've never played a hand of Omaha in my life, so that 4 is just as example. To take it to another level, one should assume that with a possible royal, it is more likely the player will stay in than with random cards.

Quote:WizardI'm not sure which "figure above" you are referring to. The way I would make that adjustment is to estimate the number out of 9 players who will stay in the whole way. Let's say that is 4 players. Then instead of the ^9 in my math above, do a ^4. I've never played a hand of Omaha in my life, so that 4 is just as example. To take it to another level, one should assume that with a possible royal, it is more likely the player will stay in than with random cards.

The final figure: 0.000831, or whatever it was; I can't go back to that screen without losing this post.

If you figure that each hand, nine "possible royals" are dealt, then you lose four of them as players fold preflop. (I know that most dealt hands don't contain possible royals at all, since they would have to have two suited cards ten or higher, but let's ignore that for the moment.) The play of the hand eliminates three out of those five before the river (and not every possible royal would play that far; if the flop only contains one helping card, then the player might still fold even though he would have gotten a runner-runner royal if he had stayed).

That is why I used the .55*.4 calculation. 5/9 of the players will play their hands, and 2/5 of those will play their hands all the way to the river.

It would be interesting to know how many Omaha hands contain "possible royals". Such hands would have to contain exactly two cards 10-A of a given suit, and could not contain a third such card . A hand could contain two such pairs of cards in different suits, of course. You could then assume, just for grins, that any such hand would play to the river, given that the flop would necessarily contain at least one helping card. That might give the best estimation of the "chances for a royal".

The reason I think this might be different from your previous result is that the majority of Omaha hands aren't even in the running for a royal from the start, so perhaps the best way to look at it might be from (percentage of Omaha hands that can make a royal)*(chance of completing the royal with five additional cards).

Quote:mkl654321The final figure: 0.000831, or whatever it was; I can't go back to that screen without losing this post.

If you figure that each hand, nine "possible royals" are dealt, then you lose four of them as players fold preflop. (I know that most dealt hands don't contain possible royals at all, since they would have to have two suited cards ten or higher, but let's ignore that for the moment.) The play of the hand eliminates three out of those five before the river (and not every possible royal would play that far; if the flop only contains one helping card, then the player might still fold even though he would have gotten a runner-runner royal if he had stayed).

That is why I used the .55*.4 calculation. 5/9 of the players will play their hands, and 2/5 of those will play their hands all the way to the river.

It would be interesting to know how many Omaha hands contain "possible royals". Such hands would have to contain exactly two cards 10-A of a given suit, and could not contain a third such card . A hand could contain two such pairs of cards in different suits, of course. You could then assume, just for grins, that any such hand would play to the river, given that the flop would necessarily contain at least one helping card. That might give the best estimation of the "chances for a royal".

The reason I think this might be different from your previous result is that the majority of Omaha hands aren't even in the running for a royal from the start, so perhaps the best way to look at it might be from (percentage of Omaha hands that can make a royal)*(chance of completing the royal with five additional cards).

so basically, once every 1200 dealt hands at a full Omaha table.

Quote:mkl654321That is why I used the .55*.4 calculation. 5/9 of the players will play their hands, and 2/5 of those will play their hands all the way to the river.

That is not the correct way to do it. Suppose you ask what is the probability that if 10 people toss a coin 5 times, that at least one of them will get 5 tails? That answer is 1-(1-(1/2)^5)^10 = 1-(31/32)^10 = 1-.72798 = .27202.

No suppose half of them drop out before finishing. I think by your logic you would say the probability drops to .5*.27202 = .13601.

How it in fact changes to 1-(1-(1/2)^5)^5 = 1-(31/32)^5 = 1-.853215 = .146785.

Close, but no cigar.

Agreed that not all 10 people can get a royal. I admitted it was an estimate.

Quote:WizardThat is not the correct way to do it. Suppose you ask what is the probability that if 10 people toss a coin 5 times, that at least one of them will get 5 tails? That answer is 1-(1-(1/2)^5)^10 = 1-(31/32)^10 = 1-.72798 = .27202.

No suppose half of them drop out before finishing. I think by your logic you would say the probability drops to .5*.27202 = .13601.

How it in fact changes to 1-(1-(1/2)^5)^5 = 1-(31/32)^5 = 1-.853215 = .146785.

Close, but no cigar.

Agreed that not all 10 people can get a royal. I admitted it was an estimate.

If you only need one occurence of an event to fulfill a desired condition, aren't the probabilities additive? So if a nine-handed game has X probability of one player winding up with a royal, doesn't that mean that any single player has X/9 probability of hitting a royal--and if five people at least see the flop, then the chances of the table hitting a royal are 5*(X/9)?

Quote:mkl654321If you only need one occurence of an event to fulfill a desired condition, aren't the probabilities additive? So if a nine-handed game has X probability of one player winding up with a royal, doesn't that mean that any single player has X/9 probability of hitting a royal--and if five people at least see the flop, then the chances of the table hitting a royal are 5*(X/9)?

Probabilities are only additive if the deseried condition must happen withing so many trials. For example drawing the ace of spades from a single deck of cards. The probability of getting it in 10 cards is 10x that of one card.

Quote:WizardProbabilities are only additive if the deseried condition must happen withing so many trials. For example drawing the ace of spades from a single deck of cards. The probability of getting it in 10 cards is 10x that of one card.

The main point of the question was to determine when the royal flush jackpot at UB becomes a good bet.

The jackpot tables rake a portion of any pot over 10 BB, up to 1 BB.

65% of the jackpot goes to the following distribution:

50% to the RF holder

25% to the rest of the players at the table

25% to the rest of the players at the same stakes.

25% of the remainder feeds the next jackpot, 10% to UB.