# What are the chances that 2 arbitrary ranked cards can be found adjacent to one another in a shuffled pack of cards

What are the chances that 2 arbitrary ranked cards can be found adjacent to one another in a shuffled pack of cards. Ranks being the face value of the cards (ace, two, three, etc), adjacent being that they are next to each other if the card were fanned out, the order is not important. The example is I chose Queen and Ace, I shuffle the pack, what is the chance that the pack contains a Queen next to an Ace.

My initial thoughts are that it should be a number approaching 8/13.

A first approximation is to assume the queens are far apart. There are $48$ places the aces can go and $8$ of them are unacceptable. The chance of success for the first ace is $\frac {40}{48}$, the second $\frac {39}{47}$ as one of the good places is occupied. The result of this is $\frac {40\cdot 39 \cdot 38 \cdot 37}{48 \cdot 47 \cdot 46 \cdot 45}=\frac {40!\cdot 44!}{48!\cdot 36!}=\frac {9139}{19458}\approx 47\%$

The chances are actually better than this. If a queen is at the end of the deck, there is one less failing space. If two queens are next to each other, there are two less failing spaces. If two queens are one space apart, there is one less failing space. To get the final answer, you would have to calculate the probability of each number of failing spaces from $1$ (all four queens together at one end of the deck) to $8$, the chance of success for each of these, and add them up. A simulation would probably be easier.