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#1
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![]() 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
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![]() 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
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![]() 1 in 649,740 To be 90% sure play 584,766 hands. |
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