Cards laid out in rows

Question

A deck of 52 cards is laid out in 4 rows of 13 cards each. I need to find the probability that

1. All hearts will be in two rows
2. All hearts will be in every row

I have a few guesses about the answers, but I'm not very sure about them.

1. $$P(A_1)=\frac{C_4^2\sum_{k=1}^{12}C_{13}^k\:13!\:39!}{52!}$$
2. $$P(A_2)=\frac{\sum_{i=1}^{10}\sum_{j=1}^{11-i}\sum_{k=1}^{12-i-j}C_{13}^i\:C_{13-i}^j\:C_{13-i-j}^k\:13!\:39!}{52!}$$

Most of all I doubt factorials

Answer 1

Problem 1

Your solution is nearly correct:

• First you need to select two rows that contain hearts and you have counted this using $\binom{4}{2}$
• The number of way to order the hearts is $13!$ and you have used this as well
• The number of way to order the non - hearts is $39!$ and you have used this as well
• Among $13$ cards in the first selected row, you choose $k$ that will be hearts and you used $\binom{13}{k}$ in your expression. However, you forgot to choose $13-k$ from the other selected row.

The correct solution is then:

$$ \binom{4}{2}\times\frac{\sum_{k=1}^{12}{\binom{13}{k}\cdot\binom{13}{13-k}}\times 13!\times 39!}{52!}=\binom{4}{2}\times\frac{\left(\binom{26}{13}-2\right)\times 13!\times 39!}{52!} $$

To obtain the Right Hand Side you need to know Vandermonde (Google it). An easier way to solve this problem is to ignore the order altogether:

• Select two rows that contain hearts, $\binom{4}{2}$
• Select 13 out of 26 cards in the selected rows but eliminate possibility that all 13 hearts are in the same row, $\binom{26}{13}-2$
• The denominator shall be $\binom{52}{13}$ as we simply choose which cards in all rows are hearts

$$ \binom{4}{2}\times\frac{\binom{26}{13}-2}{\binom{52}{13}} $$

Problem 2

We can use Principle of Inclusion and Exclusion:

$$ \frac{\sum_{k=1}^{4}{\left(-1\right)^{k}\cdot\binom{4}{k}\cdot\binom{k\times 13}{13}}}{\binom{52}{13}} $$