An elegant proof for Average HCP in game in Bridge. The average points for any of the $\binom{52}{13}$ hands in the game of bridge is exactly $10$.  This can be computed fairly easily as there are only $16$ cards of interest.  An Ace counts $4$, King $3$, Queen $2$ and Jack $1$, all other cards are zero.  The average points for any of $5.6\times 10^{28}$ deals is obviously also $10$ since every deal has an exact average of $10$ HCP as there are always $40$ HCP in the deck and always $4$ hands dealt.  For some reason I can't reason out why the average hand is also $10$, the only thing I could think of was that the $635,013,559,600$ hands can be grouped into groups of $4$ where each group is a valid deal, (no card appears twice in a deal) and no hand appears in more than one of the $\frac{\binom{52}{13}}{4}$ deals. So my actual question is: How can this possibility be proved?  Or is there a better easier proof?
Note, simply adding all the points up yields $\binom{52}{13} \cdot 10$, but doesn't answer my question of why is that so.  Is there some elegant reason why? 

One elegant solution appears when one realizes that for the set of combinations C(a,b) each element of a will repeat exactly C(a,b) * b/a.  In which case, each card of the 52 card deck will appear C(52,13) * 13/52 times.  Or 1/4 of the time.  Therefore, The entire deck will repeat this number of times.  We know that the deck contains 40 HCP.  So the average HCP will
be (C(52,13) * 1/4 * 40) / C(52,13) = 10 HCP.
This solution is more elegant than actually computing the total points contained in the full set of combinations.
