#1
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Royal Flush frequency - Texas Hold \'Em
Hey all.
After approx how many hands should I expect to hit a royal flush when holding both hole cards at Texas Hold em? (I hit my first this week). Taking into consideration that many sites offer some sort of bonus for hitting a royal: any thoughts on how this alters strategy? (Apologies if these are basic questions: did a search but didn't really find any answers). Thanks |
#2
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Re: Royal Flush frequency - Texas Hold \'Em
i believe the chances of getting a royal flush are in the ballpark of 650,000:1
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#3
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Re: Royal Flush frequency - Texas Hold \'Em
If you hold two cards from a royal flush, the odds of hitting it by river are C(45,2)/C(50,5). This is about 0.0467%.
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#4
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
If you hold two cards from a royal flush, the odds of hitting it by river are C(45,2)/C(50,5). This is about 0.0467%. [/ QUOTE ] Explain the functional notation of C(X,X) please. Thank you. |
#5
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Re: Royal Flush frequency - Texas Hold \'Em
C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n
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#6
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n [/ QUOTE ] further, it's the number of unique groups of k objects you can make out of a set n objects -- for example, if n = 3 and k = 2, you are counting the ways to make groups of 2 objects when you have 3 objects in front of you. Call your objects 'A', 'B', and 'C' -- we want unique groups, which means that AB and BA are the same. In this case, we can make a list: AB, AC, and BC are the only ways to group two items together, so C(3,2) should be 3. Plugging in the quoted formula, we see that 3!/(2!*1!) = 6/2 = 3. |
#7
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
If you hold two cards from a royal flush, the odds of hitting it by river are C(45,2)/C(50,5). This is about 0.0467%. [/ QUOTE ] Thanks. So when I hold both hole cards I'd expect to hit the royal by the river around once every 2141 times. Think I can work out strategy changes. [img]/images/graemlins/tongue.gif[/img] |
#8
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
[ QUOTE ] C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n [/ QUOTE ] further, it's the number of unique groups of k objects you can make out of a set n objects -- for example, if n = 3 and k = 2, you are counting the ways to make groups of 2 objects when you have 3 objects in front of you. Call your objects 'A', 'B', and 'C' -- we want unique groups, which means that AB and BA are the same. In this case, we can make a list: AB, AC, and BC are the only ways to group two items together, so C(3,2) should be 3. Plugging in the quoted formula, we see that 3!/(2!*1!) = 6/2 = 3. [/ QUOTE ] Got it! Thanks for the clarity. It's been awhile since I last took Discrete Math class. |
#9
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n [/ QUOTE ] Thanks for the re-education. |
#10
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Re: Royal Flush frequency - Texas Hold \'Em
[ QUOTE ]
[ QUOTE ] If you hold two cards from a royal flush, the odds of hitting it by river are C(45,2)/C(50,5). This is about 0.0467%. [/ QUOTE ] Thanks. So when I hold both hole cards I'd expect to hit the royal by the river around once every 2141 times. Think I can work out strategy changes. [img]/images/graemlins/tongue.gif[/img] [/ QUOTE ] However, if your hole cards are from royal and only 1 card on flop help you, your chance of getting royal with runner runner is 1/C(47,2) = 0.101% or 1 to 989 odd. I hope this helps if you want to be royal chaser. |
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