# What is the null probability of a given combination of ranks?

Following on previous posts about shuffling...

In Max Shinn's post about optimally shuffling cards, he describes looking at the "joint distribution of these frequencies compared to a null distribution." This is slightly ambiguous, but I'm going to interpret this in a similar way as last time: As the probability mass for the observed frequency of *ranks* in the hand under the null "uniform without replacement" distribution. Again, in Shinn's post, I'll assume we are drawing a hand of exactly 6 cards. How do we compute this probability mass?

Suppose we have 3 number 3 cards, 1 number 8, and 2 kings. How likely is that? The analysis goes very similar to last time. For each rank, there are exactly 4 cards of that rank, so the probability of drawing 3 number 3 cards should be \((4/52)(3/51)(2/50)\). The probability of drawing 1 number 8 is just \(4/49\). The probability of drawing 2 kings is \((4/48)(3/47)\). 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 \(6! \operatorname{P}(4, 3)\operatorname{P}(4, 2)\operatorname{P}(4, 1)/\operatorname{P}(52, 6) = \operatorname{P}(4, 3)\operatorname{P}(4, 2)\operatorname{P}(4, 1)/\operatorname{C}(52, 6)\), where \(\operatorname{P}(n, k) = n!/(n - k)!\). And as before, we can rewrite this using falling factorials as:

\[\frac{4^{\underline{3}} \cdot 4^{\underline{2}} \cdot 4^{\underline{1}}}{\operatorname{C}(52, 6)}\]

where \(\operatorname{C}(n, k)\) is a binomial coefficient.

Suppose we represent the number of cards from each suit as a vector \(\bold x = [x_1, x_2, \dots, x_{10}, x_{11}, x_{12}, x_{13}]\), with \(x_k\) giving the number of cards of rank \(k\). Then we can write the general form as:

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