yall are the numbers people, can you help?

[hansdebelge]

hi there. ive posted this on general hold em but nobody knows anything there.
i was wondering if anyone could tell me how many two-card starting hands there are total. is it something like 260?
and also, how many total boards are there? (starting hand plus five community cards) am i right in thinking around 2 million. i think ive heard around there but does anybody have the exact numbers for me?
thanks a lot.

[LetYouDown]

C(52,2) = 1326 starting hands are possible. Although this includes hands like 5c 4d and 5d 4c...which are essentially the same.

C(52,5) = 2598960 total boards that are possible. Same stipulation as above.

[BruceZ]

[ QUOTE ]
i was wondering if anyone could tell me how many two-card starting hands there are total. is it something like 260?

[/ QUOTE ]

Ignoring different suits:

13 pairs
78 suited
78 non-suited non-pairs
-------------------
169


Counting different suits:

13*6 = 78 pairs
78*4 suited
78*12 non-suited non-pairs
-------------------
1326


[ QUOTE ]
and also, how many total boards are there? (starting hand plus five community cards) am i right in thinking around 2 million.

[/ QUOTE ]

7 card hands: C(52,7) = 133,784,560


1326 starting hands * C(50,5) boards

= 1326 * 2,118,760

= 2,809,475,760 hand/board combinations.

[oneeye13]

more like, what do you need this for?

[TheHammer24]

what does the C in C(52,2) mean. More precisely, can you show your mathematical equation?

[uuDevil]

[ QUOTE ]
what does the C in C(52,2) mean. More precisely, can you show your mathematical equation?

[/ QUOTE ]

The number of ways of selecting n objects from N objects when the order doesn't matter is the number of "combinations":

C(N,n)= N!/[(N-n)!n!]

where ! is the factorial operator, i.e.

N!= N*(N-1)*(N-2)*...2*1

So C(52,2)=52!/[(52-2)!(2!)]= 52!/(50!2!)=52*51/2=26*51=1326