Odds of a Royal Flush in Omaha
I usually just lurk around and read, but I can't find the answer to this anywhere, so I'm hoping someone here spent more time actually attending a statistics class than I did.
After getting my first royal flush playing holdem tonight I started trying to figure out what the odds were for a royal flush in Omaha. I thought I had it until I realized I had to eliminate all the times that I don't have exactly two of the five royal flush cards in my hand. How would I account for this when figuring out the probability?
1 Reply
I'll do it the long way, 3 steps. This happens one of every 10,829 hands.
There's a permutations calculator here: http://www.mathsisfun.com/combinatorics/...
1. Calculate Numerator: 42,807,600
There's four straight flushes (one in each suit): 4
There's 4 choose 2 different hands: 6
There's 5 choose 3 different boards: 10
There's 47 choose 4 different hands: 178,365.
Multiply these together: 4 * 6 * 10 * 178,365
2. Calculate Denominator: 463,563,500,400
There are 52 c
I know this is a sort of old and seemingly settled topic, but I worry that your permutation calculation doesn't take into consideration the hands where you have MORE THAN TWO hole cards to a royal in your hand which invalidates the hand entirely towards chances of getting a royal. Does this factor in here?
