# 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?

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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.

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Aren't you using the notion of a deal in your thoughts? The number of points in a deal is obvious, but I'm talking about the sum of the points in all the possible 13 card combinations. Isn't that different? –  John Bucher Jan 29 '13 at 5:05
All possible $13$-card combinations are equally likely to be dealt to South, so that's not different. But in a few minutes I will write out an addendum that deals with a slightly more general problem. –  André Nicolas Jan 29 '13 at 5:20
Another elegant solution might be to show that the number of times each card appears in the total set of hands is the same for each different card, then the total points must be a multiple of the total points for the 52 individual cards which is 40. I might be able to prove that. I should have taken a course in Combinatorics. –  John Bucher Jan 29 '13 at 5:20
The probability that any particular card appears as the $i$-th card is $\frac{1}{52}$. So the probability of a count of $1$ is $\frac{4}{52}$, with the same for probability of a count of $2$, $3$, $4$. We can then compute $E(X_i)$ directly using the ordinary formula. Still need expectation of a sum. –  André Nicolas Jan 29 '13 at 7:15

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.

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Readers of this post are mis-reading the question. The question has nothing to do with 4 hand deals, as shown in the question that problem is trivial. This question is concerned with the set of unique 13 card hands that one can hold without reference to any deal. –  John Bucher Jan 30 '13 at 2:39