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#1
08-06-2002, 01:17 PM
 Guest Posts: n/a
likelihood of Royal Flush occuring

Hi.

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.

Stephan
#2
08-06-2002, 05:41 PM
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Re: likelihood of Royal Flush occuring

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

#3
08-11-2002, 10:05 AM
 Guest Posts: n/a
Re: likelihood of Royal Flush occuring

1 in 649,740

To be 90% sure play 584,766 hands.

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