16
$\begingroup$

I have the following problem from Introduction to Probability (2019 2 edn) by Joseph Blitzstein, p. 32, Chapter 1.

  1. 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 the cards came from. Express your answer as a binomial coefficient. Hint: Bose-Einstein.

Is my solution below correct?

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

$\endgroup$
2
  • 1
    $\begingroup$ I wouldn't have come to this idea, but it seems completely sound. $\endgroup$
    – Peter
    Dec 6, 2014 at 21:27
  • 2
    $\begingroup$ Yes, I agree with your answer. Well done. Note, the question becomes harder in the general case if the size of the hand is larger than the number of decks used and gets into inclusion-exclusion. $\endgroup$
    – JMoravitz
    Dec 6, 2014 at 21:29

3 Answers 3

Reset to default
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I like also to imagine 52 stars. Then let's add bars as prefixes saying how many times the next card is present in the hand. Then we must fix the case of the last star which must not be followed by a bar and then let's apply S & B to get $\tbinom{51+10}{51}$ that is the same as your result $\endgroup$
    – user354674
    Aug 3, 2016 at 14:26
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer Michael. I would like to understand it better, so here some questions. What do you mean by "distinguishable states"? What part of your problem statement would allow us to imply the "sampling with replacement" approach to the problem? I don't understand well the assumption "we only care about the number of particles in each one-particle state". How is that related to your first question? $\endgroup$
    – Carlos H. Mendoza-Cardenas
    Nov 25, 2015 at 15:33
  • $\begingroup$ By "distinguishable states", I mean the number of outcomes that we are "counting" as different. In this case, we only care about the number of each type of card in the hand we end up with, and not their order. Similarly, if particles A and B obey Bose-Einstein statistics, there is no physical difference between "Particle A in state #1 and Particle B in state #2" and vice versa; so we do not count these as distinguishable outcomes. (Hopefully this answers your third question as well.) ... $\endgroup$
    – Michael Seifert
    Nov 27, 2015 at 22:10
  • 1
    $\begingroup$ ... Finally, "sampling without replacement" is equivalent to this problem only because the number of cards in the hand we're given is less than or equal to the number of each type of card in the deck. As noted by @JMoravitz in the comments on the original question, the problem would be much harder if we were given a hand of 11 cards or more, and the Bose-Einstein parallel wouldn't apply. $\endgroup$
    – Michael Seifert
    Nov 27, 2015 at 22:11
  • $\begingroup$ This is a story often told about bosons and fermions, but it's a bit of a simplification. The underlying property is that boson states are symmetric under particle exchange and fermion states are antisymmetric. If the multi-particle state is approximated as a product (for bosons) or a determinant (for fermions) of single-particle states, then bosons can occupy the same state and fermions can't. Note also that this is a quantum property that doesn't follow logically from indistinguishability; classically, even fundamentally indistinguishable particles would obey Maxwell-Boltzmann statistics. $\endgroup$
    – joriki
    Mar 31, 2016 at 15:54
0
$\begingroup$

I think your answer is correct but your thinking is half way there. Consider n=10 number of copies are indistinguish stars, and k=52 is distinguish bins. 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 $\dbinom{10+52-1}{10}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .