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What is the null probability of a given combination of suits?

In Max Shinn's post about optimally shuffling cards, he describes looking at the "joint distribution of these frequencies compared to a null distribution." What does that mean? Two possible interpretations:

  1. We compute a single suit-related number from the hand. We use the "null distribution" ("uniform without replacement") to compute the probability of seeing a number this extreme or more extreme. That's the null hypothesis significance testing (NHST) interpretation.
  2. We compute the probability mass for the observed frequency of suits in the hand under the null "uniform without replacement" distribution. That's the probability mass interpretation.

I am going to assume Shinn means (2) for practical reasons. If it is (1) then I can't think of an easy way to both (a) quantify the suit distribution in just one number, such that it is easy to (b) compute the probability. (I'd be happy to find out there's an easy way...) So, let's assume he means (2). As in Shinn's post, I'll assume we are drawing a hand of exactly 6 cards. How do we compute (2)?

Suppose we have 3 hearts, 2 clubs, and 1 diamond. How likely is that? There are 13 cards of any given suit, so the probability of drawing 3 hearts should be \((13/52)(12/51)(11/50)\), and likewise the probability of drawing 2 clubs should be \((13/49)(12/48)\) and the probability of drawing 1 diamond is \((13/47)\). We don't actually care what order we draw the six cards in, so I think we need to multiply by \(6! = 720\). So, the probability is \(6! \operatorname{P}(13, 3)\operatorname{P}(13, 2)\operatorname{P}(13, 1)/\operatorname{P}(52, 6) = \operatorname{P}(13, 3)\operatorname{P}(13, 2)\operatorname{P}(13, 1)/\operatorname{C}(52, 6)\), where \(\operatorname{P}(n, k) = n!/(n - k)!\). We can also write this using falling factorials as:

\[\frac{13^{\underbar{3}} \cdot 13^{\underbar{2}} \cdot 13^{\underbar{1}}}{\operatorname{C}(52, 6)}\]

Suppose we represent the number of cards from each suit as a vector \(\bold x = [x_1, x_2, x_3, x_4] = [s, d, c, h]\) giving respectively the number of spades, diamonds, clubs, and hearts drawn. Then we can write the general form as:

\[P(X = x) = \frac{\prod_{j=1}^4 13^{\underline{x_j}}}{\operatorname{C}(52, 6)}\]