[LearnedfromTV]

[ QUOTE ]
pair (u)......60198
pair (s)......32274
pair (i)......11559

u = unseen rank; s = seen rank; i = identical

[/ QUOTE ]

I see what I did wrong. I had these numbers at 84480/42240/14080. 84480 is of course the number of combinations of any pair for a randomly dealt hand (out of 52C5 possibilities). 6 ways to make a pair with 4 cards available, 3 if 3, 1 if 2, and I just divided the 84480 by the appropriate factor; but the other cards elminate some of the combinations also, i.e. JJ742, the number of combinations of 77xxx (in 47C5 given JJ742 dead)isn't just changed by the missing 7, but also the missing J, J, 4, 2.

I still don't understand why the ratios aren't 6:3:1. I guess if I have JJ742, an opponent with 77 and three random cards will have two pair relatively less often than one with JJ and three random cards? Because given that he has 77, JJ77 is tougher to make than if we give him JJ and try to make JJ77. Etc.

Thanks for the post, this is great stuff.