JJ folded to you in the CO

MickeyHoldem · #1 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

#2 08-31-2005, 11:13 AM

So about 15.5%... thanks a lot.

KJL · #3 08-31-2005, 11:39 AM

[ QUOTE ]
2 overs.... 12c2*38c4 ...2 in same hand 20%
3 overs.... 12c3*38c3 ...2 in same hand 60%

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

MickeyHoldem · #4 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