King and Queen in a Deck of Cards 
In a shuffled deck of $52$ playing cards, what is the probability that at least one King and one Queen find themselves next to each other?

I read somewhere that the probability is greater than $0.6$, but can't see why. Experimentally, it certainly doesn't seem to be even greater than $0.5$. My own rude estimates are that it is indeed less than $0.5$.
 A: Hint: you need to find all the cases that atleast one pair king and queen are next to each other. May be you can subtract from total cases that no king/queen are next to each other. 
A: Continuing with Tojrah's idea, let us compute the probability that no king and queen are next to each other. 
Consider the order in which the Kings and Queens appear in the deck, without regard to suit. There are $8$ kings and queens, so there are $\binom{8}4$ possible orderings. Suppose for example the ordering was
$$
KKQKKQQQ\tag{*}
$$
These $8$ cards divide the remaining $44$ cards of  the deck into $9$ contiguous blocks. We want to count the number of ways to place these cards in these gaps so that no king is next to any queen. This is equivalent to counting nonnegative integer solutions to the equation
$$
x_1+x_2+x_3+\dots+x_9=44
$$
subject to the additional constraints
$$
x_3\ge 1,x_4\ge 1,x_6\ge 1
$$
The reason for these constraints is because gaps numbered $3,4$ and $6$ are exactly the gaps which are sandwiched by a king and queen in $(*)$. You can then count solutions to this equation using stars and bars; the number of solutions is $\binom{44+6-1}{9-1}$. Finally, multiplying by $44!$ to choose the ordering of the other cards, and $4!\times 4!$ to choose the order of the kings and queens, there are $\binom{44+6-1}{9-1}\times 44!\times 4!^2$ ways to shuffle the deck which result in no adjacent queens and kings, such that the kings and queens appear like in $(*)$. 
You will then have to repeat this same count for all of the other $\binom{8}4-1$ arrangements of kings and queens, add up the number of shuffles for each, and divide by $52!$. I would suggest using a computer to do this, but maybe there is a nice way to do it by hand.
