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? 
 A: Let $X_i$ be the number of points in the hand at position $i$. (Here $i$ goes from $1$ for South to $4$ for West). Then 
$$X_1+X_2+X_3+X_4=40,$$ 
since there are $40$ high card points in the deck.   
Take the expectation of both sides, and use the linearity of expectation. (The $X_i$ are not independent, but expectation of a sum is always the sum of the expectations.) We get
$$E(X_1+X_2+X_3+X_4)=E(X_1)+E(X_2)+E(X_3)+E(X_4)=E(40)=40.$$
By symmetry all the $E(X_i)$ are equal. So they are all equal to $\dfrac{40}{4}$.
Added: Imagine dealing out all the cards, one at a time. For $i=1$ to $52$, let random variable $Y_i$ denote the high card "value" of the $i$-th card dealt. Then 
$$X_1+X_2+\cdots+X_{52}=40.$$
Taking expectations as before, we find that 
$$E(X_1)+E(X_2)+\cdots+E(X_{52})=40.$$
By symmetry, all the $E(X_i)$ are equal. To see the symmetry, note that all orders of dealing out the cards are equally likely. 
So each $E(X_i)$ is $\frac{40}{52}$.
Now the sum of the HCP in (say) the first $k$ cards dealt is $X_1+X_2+\cdots +X_k$. The expectation of this is the sum of the expectations: It is $\frac{40}{52}k$.  The same applies to the sum of the HCP in any $k$ randomly chosen cards. Taking $k=13$ we get the earlier result.
Remark: We went into great detail, since linearity of expectation is such a useful fact. 
Finding the expected number of high card points in say the South hand becomes very unpleasant if we try to do it by first finding, for all $k$, the probability that the hand has $k$ high card points. 
And why go through all that work when the mean is intuitively obvious? The formula $E(X_1+\cdots+X_4)=E(X_1)+\cdots +E(X_4)$ shows that what is intuitively obvious is indeed true.  
A: 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.
