View Full Version : likelihood of Royal Flush occuring

08-06-2002, 01:17 PM

I have a question. I hope I get the essence across without confusing everybody...

Can you calculate a "safe period" of someone getting dealt a Royal Flush pat (in 5CD)? Say you wanted to be 90% sure you get a royal flush pat in your poker career, how many hands would you have to play?

I do not know if there is an answer or if it is a ridiculously stupid question, but please humor me with your insights.


08-06-2002, 05:41 PM
Depending on how much probability you have studied some or none of this will make sense.

The number of possible hands to get dealt is C(52,5) = 52!/(5!*(52-5)!) = 2598960.

Out of theese 4 constitute royals which means that the probability of beeing dealt a royal is: 4/2598960 ~= 1.5*10^-6

Using complementary probabilities, the probability of getting dealt at least one royal in k deals is 1-(1-p)^k, where p = 4/2598960 and k is the number of tries.

Setting: 1-(1-p)^k = 0.9 yields: k = log(0.1) / log(1-p) ~= 1.5*10^6

So you would need to play approximately 1.5 million hands in order to be 90% sure to be delt a pat royal.

I might have gotten some of the terminology wrong since I have only studied probability in swedish.

Sincerely, Andreas

08-11-2002, 10:05 AM
1 in 649,740

To be 90% sure play 584,766 hands.