Royal Flush frequency - Texas Hold 'Em

[Audi]

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

[FishHooks]

i believe the chances of getting a royal flush are in the ballpark of 650,000:1

[gamble4pro]

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%.

[NeoGeo]

[ 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.

[gamble4pro]

C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n

[elitegimp]

[ 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.

[Audi]

[ 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. /images/graemlins/tongue.gif

[NeoGeo]

[ 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.

[NeoGeo]

[ QUOTE ]
C(n,k)= n!/k!(n-k)!, where n! = 1*2*...*n

[/ QUOTE ]

Thanks for the re-education.

[NeoGeo]

[ 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. /images/graemlins/tongue.gif

[/ 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.

[NeoGeo]

[ 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 ]

I understand the logic of C(45,2), but I am not sure of why C(50,5). Can you show me the logic how it derives to royal count. It is quite intereting. Thank you.