$8$ cards are drawn from a deck of cards without replacement

Question

I draw $8$ cards randomly from a shuffled deck without replacement. What is the expected value of the sum of the largest $3$ cards? The ace is given a value of 1 and the jack, queen and king are all given a value of $10$. I can generate a simulation answer but that is all. How can this be done, I am stuck?

Answer 1

For each of the possible values for the sum, $S$, of the three highest cards -- integers on $[6,30]$ -- figure out the number of combinations of eight-card hands that would give the three highest cards that sum.

For $S=30$, we can have anywhere from three to eight $10$s, so we consider those separately:

$$P(S=30, 10, 10, 10) = \frac{{16 \choose 3}{36 \choose 5} + {16 \choose 4}{36 \choose 4} + {16 \choose 5}{36 \choose 3} + {16 \choose 6}{36 \choose 2} + {16 \choose 7}{36 \choose 1} + {16 \choose 8}{36 \choose 0}}{{52 \choose 8}}.$$

For something a little more complicated, you'll need to add up more combinations. For $S=20$, for example, there are twelve possible triples for the three highest cards:

$$(7, 7, 6), \\ (8, 8, 4), (8, 7, 5), (8, 6, 6), \\ (9, 9, 2), (9, 8, 3), (9, 7, 4), (9, 6, 5), \\ (10, 8, 2), (10, 7, 3), (10, 6, 4), (10, 5, 5).$$

Let's calculate $P(S=20, 9, 8, 3).$ We draw one $9$: ${4 \choose 1}$. We draw one $8$: also ${4 \choose 1}.$ Then there needs to be at least one $3$ (but due to the constraints on the number of $2$s and $3$s, there can't be less than two!):

$$P(S=20, 9, 8, 3) = \frac{{4 \choose 1}{4 \choose 1}\left[{4 \choose 4}{4 \choose 2} + {4 \choose 3}{4 \choose 3} + {4 \choose 2}{4 \choose 4}\right]}{{52 \choose 8}}.$$

(Note that there are little subtleties like the fact that $(10, 9, 1)$ isn't possible. If you draw a $10$ and a $9$, at least two of the remaining six cards must be something higher than an ace.)

Once you have the probability of getting each sum individually, then getting the expected value is easy. Counting up the number of combinations of cards for each sum is the laborious part.

Answer 2

A little late, but here's a way to compute the answer:

We can first write the order statistic and then find the sum by linearity of expectation.

After sorting, the probability that the $k^{th}$ position is $a$ is given by (derived by considering it as a case of hypergeometric distribution): \begin{align*} \mathbb{P}\left(X_{(k)}=a\right) &= \left.\left(\sum_{j=n-k-n_a+1}^{n-k} \; \sum_{i=n-k-j+1}^{n_a} \dbinom{n_{a}}{i}\, \dbinom{(a-1)n_{a-1}}{n-i-j}\, \dbinom{52-n_a-(a-1)n_{a-1}}{j}\right) \middle/ \dbinom{52}{n} \right. \end{align*}

where:

$n$ : number of cards drawn

$n_a$ : number of cards having face value of $a$

Then, the expected value of the $k^{th}$ card is:

\begin{align*} \mathbb{E}\left(X_{(k)}\right) &= \sum_{a=1}^{10} a\, \mathbb{P}\left(X_{(k)}=a\right) \end{align*}

Hence, the required answer is: \begin{align*} \mathbb{E}\left(X_{(6)}\right)+\mathbb{E}\left(X_{(7)}\right)+\mathbb{E}\left(X_{(8)}\right) &= \boxed{\dfrac{54800638}{1929585}} \approx 28.400219736368 \end{align*}

Here's a Sage code for computation:

n=[4]*10+[16]
nn=8
sum(sum(a*sum(sum([binomial(n[a],i)*binomial(n[a-1]*(a-1),nn-i-j)*binomial(52-n[a]-(a-1)*n[a-1],j) for i in [nn-k-j+1..n[a]]]) for j in [nn-k-n[a]+1..nn-k]) for a in [1..10]) for k in [6..8])/binomial(52,8)