[BruceZ]

This is what we need. I only found one thing that needs to be recomputed, and one comment.

[ QUOTE ]
Need to assume that xx does not contain an A,2, and in the last two cases a 3 for the following to be correct.

getting A2xx =
4 * 4 * C(44,2)/C(52,4) =~ .056

anyone at a 10 person table getting A2xx (need inclusion exclusion for this)=
10*4*4*C(44,2)/C(52,4)
- C(10,2) *4*4*C(44,2) * 3*3*C(42,2)/C(52,4)/C(48,4)
+ C(10,3) *4*4*C(44,2) * 3*3*C(42,2) * 2*2*C(40,2)/C(52,4)/C(48,4)/C(44,4)
- C(10,4) * 16 * C(44,2) * 9 * C(42,2) * 4 * C(40,2) * C(38,2)/C(52,4)/C(48,4)/C(44,4)/C(40,4) =~ .467

[/ QUOTE ]

OK


[ QUOTE ]
A specific person having A2xx given that I have A2xx
Your calculation is correct

[/ QUOTE ]

His denominator needs to be C(48,4) instead of C(52,4).


[ QUOTE ]
Someone else having A2xx given that I have A2xx (again inclusion/exclusion)=
9*3*3*C(42,2)/C(52,4)
- C(9,2) *3*3*C(42,2) * 2*2*C(40,2)/ C(48,4)/C(44,4)
+ C(9,3) *9*C(42,2) * 4*C(40,2) * C(38,2)/ C(48,4)/C(44,4)/C(40,4) =~ .226

[/ QUOTE ]

The denominator in the first term should be C(48,4). That would change this to ~.327 if everything else is right.


[ QUOTE ]
A specific person having A2xx given that I have A3xx:
Your's looks right

[/ QUOTE ]

His denominator needs to be C(48,4) instead of C(52,4).


[ QUOTE ]
Someone else having A2xx given that I have A3xx =
9*4*3*C(41,2)/C(48,4)
- C(9,2) *4*3*C(41,2) * 3*2*C(39,2)/C(48,4)/C(44,4)
+ C(9,3) *4*3*C(41,2) * 3*2*C(39,2) * 2*C(37,2) / C(48,4)/C(44,4)/C(40,4) =~ .398

[/ QUOTE ]

You said this would not allow a 3, but the C(41,2) in each term allows 3s. 52 - 4 aces - 4 deuces - 1 three - 2 x = 41. To exclude 3s we would make it C(38,2). Allowing 3s may be preferable though.