View Full Version : Odds two stud players rolled up in same hand

10-24-2003, 06:16 PM
Saw this (http://tinyurl.com/s9i7) thread from the RGP 'golden days'. Agree/disagree with the math on this?

10-25-2003, 03:15 AM
C(8,3)*P(13,3)*4^3/C(52,3)/C(49,3)/C(46,3) +
C(8,4)*P(13,4)*4^4/C(52,3)/C(49,3)/C(46,3)/C(43,3) -
C(8,5)*P(13,5)*4^5/C(52,3)/C(49,3)/C(46,3)/C(43,3)/C(40,3) +
C(8,6)*P(13,6)*4^6/C(52,3)/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3) -
C(8,7)*P(13,7)*4^7/C(52,3)/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3)/C(34,3) +
= 1 in 5860

Some of the other posters were close, but mine is exact for the odds of 2-8 players being rolled up, by the inclusion-exclusion principle.

10-26-2003, 04:37 PM
Thanks Bruce. I'll consider that 'no chance' for playing purposes.

10-26-2003, 05:40 PM
I disagree, since for playing purposes, the interesting probability is the conditional probability that someone else is rolled up given that you are, which is much more likely.

10-26-2003, 11:12 PM
That's like the guy who always carries his own bomb aboard an airplane, since the chance of 2 people bringing a bomb are miniscule.

Once you are rolled up, it becomes over 100 times more likely that 2 players will be rolled up.

Here is the exact probability of 1 or more players being rolled up with you when you are rolled up:

C(7,2)*P(12,2)*4^2/C(49,3)/C(46,3) +
C(7,3)*P(12,3)*4^3/C(49,3)/C(46,3)/C(43,3) -
C(7,4)*P(12,4)*4^4/C(49,3)/C(46,3)/C(43,3)/C(40,3) +
C(7,5)*P(12,5)*4^5/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3) -
C(7,6)*P(12,6)*4^6/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3)/C(34,3) +

= 1 in 55.3

It isn't necessary to go through all this to get close, you can just use the first term or two:

7*12*4/C(49,3) = 1 in 54.8

10-27-2003, 06:39 PM
Thanks, that's an interesting point. I get bogged down in the math but that's why I majored in business /images/graemlins/crazy.gif

But still. I'm not going to risk playing weak/tight when rolled up because I'm afraid of a 1:55 shot. I'd rather get pounded that one time.

10-27-2003, 10:53 PM
It won't happen early in the hand, but there will come a point when you realize that he started rolled up provided you are decent and he is playing at least ok. That is, each play he makes changes the chance (increases, usu.) that he started rolled up in a Bayesian sense.