[aloiz]

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

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

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

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

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

aloiz