4 Responses

1. m203

   February 25th, 2010 at 1:27 pm

   1

   It would be 1st A 2nd B and 3rd C
   regardless of chip count they were all-in so chips became mute. Since A had KKQQ10 they won
   Since B had QQ1010A(higher kicker wins the hand of B,C)
   Since C had QQ10105(lower kicker lesser hand loses to B)
   If A had not had “cowboys” B would have won correct? thus B would have beat C. This being said C would have beaten A without the “cowboys” they would have taken the second side pot and B taking the main pot leaving A with third side pot or what his opponents could not match.
   Pot 1(everyone matching B’s all-in)150
   Pot 2(A matching C’s all-in)100
   Pot 3(A being the only one automaticly wins this pot)100 of course there would only be a third pot if A went all-in first.
   A(without K’s) 100 chips
   B(without A having K’s) 150 chips
   C(without A having K’s) 100 chips

2. True That

   February 25th, 2010 at 2:02 pm

   2

   I would have to say person C. But, it depends on if person B bets or checks out their hand. If person B bets first and doesn’t check, then he will win. It is all about betting at the right time…

3. Will R

   February 25th, 2010 at 3:55 pm

   3

   “C” more chips

4. shortncu

   February 25th, 2010 at 5:07 pm

   4

   B