Quick question

[SA125]

Sorry if asked a million times. What are the odds you'll have the same hand as another player in a full (10) ring game?

Zero.

[BruceZ]

[ QUOTE ]
Sorry if asked a million times. What are the odds you'll have the same hand as another player in a full (10) ring game?

[/ QUOTE ]

P(pair)*P(matching pair) + P(XYs)*P(other XYs) + P(XYo)*P(other XYo)

= (1/17)*P(matching pair) + (4/17)*P(other XYs) + (12/17)*P(other XYo)

From the inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=):

1/17*[9*1/C(50,2)] +

4/17*[9*3/C(50,2) -
C(9,2)*3*2/C(50,2)/C(48,2) +
C(9,3)*3*2*1/C(50,2)/C(48,2)/C(46,2)] +

12/17*[9*7/C(50,2) -
C(9,2)*(4*3 + 2*2 + 1*2)/C(50,2)/C(48,2) +
C(9,3)*(4*3*1 + 2*2*1 + 1*2*1)/C(50,2)/C(48,2)/C(46,2)]

=~ 4.1%.

[SA125]

n/m

Yeah thanks a lot for the help.
Like Iv'e said before in another post, I'm only 17 and my math knowledge goes up to only a first semester of AP Calculus AB, so I really have to learn most of this stuff. But thanks for helping me, I'll get right to reading your post (we learn something new every day /images/graemlins/smile.gif!)
From now on I won't psot any math unless I know it's right, so I don't make myself look like an idiot.