I am cant figgure out a math problem i am faced with.

There is a game 5 card pot limit omaha game i play in with a royal flush jackpot.


I know that we play ruffly 42 hands per H on avg and That there are 2,3 people at showdown every hand on avg and there are on avg 7.5 people at the table at any time.


I am trying to figgure out what the odds of seeing a royal flush hit with one Hour of play.


Why i cant figgure this out:

You have to use 2 cards from your hand and 3 from the board and i cant seam to operationalize this.

Also People will not play random combinations and i have 0 idea of how to operationalize this as well.
In order to simplify it one could use random holdings here. Would be a bit of obviously but a better indicator then what i have now.


I figured that this might be the right place to ask.

Moved to probability (from SMP) as the more appropriate forum for op.

It makes the math much easier if we only consider the deal of the cards (i.e., not players' decisions). These problems are often handled assuming all players go to showdown every deal. Another way of saying the same thing is to assume that nobody ever folds at any point during the hand if there is any chance that they can make a royal flush.

I will do a straightforward derivation of the probability that a random deal of 7-player 5-card omaha results is someone making a royal flush (under the assumptions of the first paragraph). Other people may use different methods or approaches than what I am about to present, but they should all lead to the same result.

Logic: exactly three cards of a royal flush must appear among the five board cards and the other two cards that would fulfill that specific royal flush must appear in one of the seven players' five hole cards (of course, a royal flush can come in any of four suits). Think of dealing the board first. Then, when tallying the deals that yield a royal flush think of dealing that person's hole cards first (we will add a term to reflect the fact that any one of the seven players can be the one who makes the royal flush). Hopefully it will be fairly easy to translate this logic into the expression for the probability given below.

Prob = { C(4,1)*C(5,3)*C(47,2)*C(7,1)*[C(2,2)*C(45,3)]*C(42,5)*C(37,5)*C(32,5)*C(27,5)*C(22,5)*C(17,5) } / { C(52,5)*C(47,35)*C(35,5)*C(30,5)*C(25,5)*C(20,5)*C (15,5)*C(10,5)*C(5,5) }

which, after a whole bunch of simplification:

= 5 / 4,641

or approximately 1 every 928 deals.

C(X,Y) is the choose or combination function read "X choose Y" and gives the number of ways you can choose Y items out of a total of X items without replacement where the order of the choices is irrelevant. The formula for C(X,Y) is well-known and widely available on modern calculators and in many common computer environments.

The formula for a different number of players at the table (or a different number of hole cards per player) should be clear from the above.


Caution: As I always say, before using a result such as the one posted above I would await some degree of confirmation/verification from some other person or some other source.

I programmed a simulation and ran it for 100,000,000 deals of 7-player 5-card Omaha. There were 107,733 deals with a Royal Flush.

If the above 5/4641 probability is correct, we would expect 107,735 royal flushes in every 100 million deals.

So it looks like that is the correct probability.

Oh man thanks a lot this was helpfull and i am now closer to understanding what sort of number i should expect! I like the way you think and also the simulation was a good way to proof the point.

The probability that the first three cards of flop are royal flush cards is 20/52 x 4/51 x 3/50.

The probability that the next two cards are not part of that royal is 47/49 x 46/48.

Multiply those five fractions and then multiply by ten since there are ten ways the three card royal can appear on the board. That's the chances the board contains exactly three cards to a Royal.

Take that answer and multiply it by the chances that a player has the other two royal cards. That's 2/47 times 1/46 times the ten ways those two cards can be placed in the players hand.

Then simply multiply by 7 since there is no chance that more than one player can have the royal.