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thestevo31
08-18-2005, 01:18 AM
What are the odds of 2 players to have the same pocket pair?

ie. both with pocket 7s?

weevil
08-18-2005, 02:53 AM
[ QUOTE ]
What are the odds of 2 players to have the same pocket pair?

ie. both with pocket 7s?

[/ QUOTE ]

I couldn't answer that, but when the other person holding pocket sevens is a girl you have a crush on, and you both have a straight at the end, and you're trying to softplay her, because c'mon, she's a girl, and you don't want her to not like you do you, and then you both turn them over. Well that's how poker is played lady.

BruceZ
08-18-2005, 03:58 AM
[ QUOTE ]
What are the odds of 2 players to have the same pocket pair?

ie. both with pocket 7s?

[/ QUOTE ]

For 10 players and any duplicate pocket pair (not a specific pair) by the inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=&amp;Number=417383&amp;page=&amp;v iew=&amp;sb=5&amp;o=&amp;fpart=):

C(10,2)*13*6/C(52,2)/C(50,2) -

C(10,2)*C(8,2)/2*13*6*12*6/C(52,2)/C(50,2)/C(48,2)/C(46,2) +

C(10,2)*C(8,2)*C(6,2)/3!*13*6*12*6*11*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2) -

C(10,2)*C(8,2)*C(6,2)*C(4,2)/4!*13*6*12*6*11*6*10*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2) +

C(10,2)*C(8,2)*C(6,2)*C(4,2)*C(2,2)/5!*13*6*12*6*11*6*10*6*9*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2)/C(36,2)/C(34,2)

=~ 462-to-1.

thestevo31
08-18-2005, 05:29 PM
Thanks Bruce

hukilai
08-18-2005, 10:49 PM
Bruce,

In this case using inc/exc is a kind of an overkill, I think...

Here is my solution:
We know prob for two people to have the same pair:
3/51*2/50*1/49, we multiply it by number of ways we can choose two people from ten (C(10,2)) and this is it:

3/51*2/50/49*C(10,2) = 0.00216086435, which turns into 461.78:1

Any flaws here?

BruceZ
08-19-2005, 02:45 PM
[ QUOTE ]
Bruce,

In this case using inc/exc is a kind of an overkill, I think...

Here is my solution:
We know prob for two people to have the same pair:
3/51*2/50*1/49, we multiply it by number of ways we can choose two people from ten (C(10,2)) and this is it:

3/51*2/50/49*C(10,2) = 0.00216086435, which turns into 461.78:1

Any flaws here?

[/ QUOTE ]

That's the same as my first term. It's an approximation since it over counts the times that more than 1 pair of players have the same pair, but the other terms are so small that they can be ignored in this case. I just showed them because it was easy to do.