[gaming_mouse]

What is the chance that there are two people at the table in a full ring game who have the same hand?

We consider all possible (10 choose 2) pairs of players whose hands might coincide. For any given pair of players, the chance is:

(3^4)/(48 choose 4)

Our first order approximation (using inclusion-exclusion) is therefore:

(10 choose 2)*(3^4)/(48 choose 4)=0.0187326549

This is an upper-bound on the true probability. To refine our appoximation, we now consider the 2nd term.

Finding the second term (which we must subtract off) is more difficult, as there are ((10 choose 2) choose 2) = 990 pairs of pairs of players. We can break these pairs of pairs down into two types:

1. Those that overlap (eg, {1,3} & {1,7})
2. those that do not (eg, {1,3} & {2,4})

By multinomial, there are (10!/(2!*2!*6!))/2=630 non-overlapping pairs of pairs. There are therefore 360 pairs of pairs of type 1 (overlapping).

Note that an overlapping pair of a pairs corresponds to 3 people all having the same hand, which we calculate as:

((3^4)/(48 choose 4))*((2^4)/(44 choose 4))

Call this P1.

Whereas the chance of a non-overlapping pair of pairs is simply:

((3^4)/(48 choose 4))^2

Call this P2

The second term of inclusion-exclusion is: P1*360 + P2*630, which works out to:

0.000126835808

Refining our approximation now (which is guaranteed to be as good as the above number)

0.0187326549-0.000126835808=0.0186058191

About 1 in 54