# Combinatorics: Number of possible 10-card hands from superdeck (10 times 52 cards)

I have the following problem from book “Introduction to Probability”, p.32

A certain casino uses 10 standard decks of cards mixed together into one big deck, which
we will call a superdeck. Thus, the superdeck has 52 · 10 = 520 cards, with 10 copies
of each card. How many different 10-card hands can be dealt from the superdeck? The
order of the cards does not matter, nor does it matter which of the original 10 decks
Hint: Bose-Einstein.

My solution:

Because the number of cards of each type in the superdeck (10) is not less than the size of the hand (10), and thus not limiting, it’s the same as sampling with replacement where the order does not matter,
so the number of possible 10-card hands would be $\binom{52+10-1}{10}$.

Is my thinking correct?

#### Solutions Collecting From Web of "Combinatorics: Number of possible 10-card hands from superdeck (10 times 52 cards)"

Your reasoning was no so evident for me at first, so I decided to dig a little and what I first found was a table with an entry that agrees with your interpretation, but then I wanted to know where that multichoose formula comes from, and after a while I found useful this explanation using the bars and stars approach.

In your case, for example, we would represent the $10$ cards of the hand as “stars” and then we would put $52-1$ “bars” between them to represent the idea of $52$ different types$^1$ of cards. The stars that are to the left of a bar are of the same type. We have then to be careful with the extreme case where all the $10$ stars are to the left of a bar, meaning that the all the $10$ cards are all of the same type. As you and @JMoravitz correctly pointed out, this is possible because the size of the hand is not greater than the multiplicity of each type in the superdeck.

Under this approach we could imagine that we have a total of $52+10-1$ bins to place the stars and the bars, and then ask in how many ways we could place the $10$ stars out of a total $52+10-1$ available bins. Now the answer is more evident to me, in

$$\binom{52+10-1}{10}$$

ways. Once you place the stars there is only one way to put the bars, since the order does not matter.

EDIT: I was wondering why Bose-Einstein would be a hint in the problem. Following the perspective proposed in this video, it seems that this problem is analogous to the Bose-Einstein statistics in that we could thought of the $10$ cards (stars) in the hand as indistinguishable particles and the $52^*$ types of cards in a deck as energy states where those particles could condensate (don’t get me wrong, I am not a quantum physicist). In any case, it’s not clear how that would be a hint to solve the problem…any hint?

$^*$ Thanks to Michael’s answer I understood that the number of energy states is $52$ and not $52+10-1$, as I originally wrote in my edit…although it still seems a rather sophisticated hint for me.

$^1$ Here the type of a card is its rank and suit.

Just to address the “Bose-Einstein” hint: in quantum physics, a set of particles are said to obey Bose-Einstein statistics when they are physically indistinguishable (so that there’s no way, even in principle, to tell them apart); and multiple particles can be in the same state. A “State” is a pretty broad concept in physics, but you can think of these states as being like energy levels in an atom. The other related concept is Fermi-Dirac statistics, in which the particles are indistinguishable and you can only have one particle in each state. The best-known particle that obeys Bose-Einstein statistics is the photon; the best-known particles that obey Fermi-Dirac statistics are the electron, proton, and neutron.

The “superdeck” problem posed is then the same as the following problem: A system of 10 particles (i.e. cards in the hand) obeying Bose-Einstein statistics are put into a system with 52 distinct one-particle states (i.e., types of cards in a standard deck.) What is the number of distinguishable states of the collective system? The answer can be found by imagining partitioning the ten particles among the 52 one-particle states, and saying that we only care about the number of particles in each one-particle state. This then becomes a standard “balls and walls” problem, with the answer of ${61 \choose 10} = 90 177 170 226$ as you found.

If we just simply follow the Bose-Einstein(sampling with replacement and order doesn’t matter), we would have $\dbinom{10+52-1}{52}$ or $\dbinom{10+52-1}{9}$
Since we are not putting 10 copies into 52 bins, it is more clear to me to think in this way $:\\$
Consider n=10 number of copies are indistinguish stars, and k=52 is distinguish bins(easier to imagine if you associate it with the idea of the degree of freedom). Then there are $10+52-1$ locations/bins for us to choose from, each location/bin contains 1 card, and we want 10 cards hand, then the solution becomes $\dbinom{10+52-1}{10}$