View Full Version : JJ folded to you in the CO

08-30-2005, 06:24 PM
9-handed table. It's folded to you in the Cutoff. You have JJ.

What is the probability that at least 1 of the remaining 3 opponents yet to act has 2 overcards to your jacks? There are 50 unseen cards left, 12 of which are "overcards". I'm too lazy to do the math.

PS: Ignore the implications that the previous 5 opponents all folded junk. Assume it's perfectly plausible that as many A's, K's, and Q's got folded as did 2's, 3's, and 4's. (Example, UTG would muck A2, UTG+1 could easily muck K3, next guy may have Q4, etc..) Don't assume the previous 5 opponents mucked all low cards. Just use the 12 overs with 50 unseen cards remaining theory. In other words, pretend you are first to act at a 4-handed table! There...much easier! /images/graemlins/smile.gif

08-30-2005, 06:45 PM
Well his it's 6/25 to get one overcard and 3/49 to pair it so there's about a 1.5% chance of an opponent having an overpair. Double that for two opponents and it's 33-1 against an overpair.

08-30-2005, 08:02 PM
not overpair, just two overcards.

08-30-2005, 09:33 PM
12 overs
50 cards left
3 opponents

total hands 50c6
no overs... 38c6
1 over..... 12*38c5
2 overs.... 12c2*38c4 ...2 in same hand 20%
3 overs.... 12c3*38c3 ...2 in same hand 60%
4 overs.... 12c4*38c2
5 overs.... 12c5*38
6 overs.... 12c6

= (12c2*38c4*.2 + 12c3*38c3*.6 + 12c4*38c2 + 12c5*38 + 12c6) / 50c6 = ~.155

08-31-2005, 11:13 AM
So about 15.5%... thanks a lot.

08-31-2005, 11:39 AM
2 overs.... 12c2*38c4 ...2 in same hand 20%
3 overs.... 12c3*38c3 ...2 in same hand 60%

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Where do you get the 20% and the 60% from. I can't figure it out. TY

08-31-2005, 09:51 PM
Here's one way to look at it....

There are 3 hands... Aa Bb Cc
Put the first over into A... there are now 5 places the second over could end up, but only one of these spots (a) put both overs in the same hand... a 1 in 5 chance

Similarly, for 3 overs... put an over in A... the next over has a 4 in 5 chance of ending up in another hand... so now there will be an over in A and B... with 4 empty spots for the last over... 2 out of 4 end up in C... so the chance of all 3 overs ending up in different hands is 4/5 * 2/4 = .4
We are interested in the other times = 1 - .4 = .6