[callydrias]

[ QUOTE ]
We consider all possible (10 choose 2) pairs of players whose hands might coincide. For any given pair of players, the chance is:


(3^4)/(48 choose 4)

[/ QUOTE ]

It's been a while since I've taken a combinatorics class. Can you explain this part of your answer? I understand this doesn't take into account 3 or more or multiple pairs of people holding the same hand. A couple things are bugging me (which may be easily resolved once I figure this part out):

1. Does this answer taken suitedness into account? e.g Is A [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/spade.gif[/img] equal to A [img]/images/graemlins/diamond.gif[/img]2 [img]/images/graemlins/diamond.gif[/img]3 [img]/images/graemlins/diamond.gif[/img]4 [img]/images/graemlins/club.gif[/img] or not?

2. Does this answer take pairs/trips/quads in a single hand? I suspect not because the numerator term seems to describe a choice for each card which has exactly 3 options.

TIA