What is the probability of 2 named cards appearing sequentially in a randomly shuffled deck if suits are ignored? If a deck of 52 playing cards is randomly shuffled and I name two cards without specifying a suit (for example, 7 and king), what is the probability that those two cards (of any suit) will appear next to each other in the order specified somewhere in the deck?  So, if I chose 7 as my first card and king as my second card, it would be a "success" if the shuffled deck contained a 7 of any suit followed by a king of any suit.
Thanks for any help!
 A: Consider a random arrangement of the 48 cards left in the deck after the four kings have been removed. What's the probability that if I randomly place the four kings back into the deck, none of the four kings will be positioned immediately after a seven?
When placing the first king back into the deck, there are 49 possible 'slots', but four of them (the ones immediately following each of the four sevens) are ones I want to avoid, so the probability of placing the first king back in the deck so that it's not positioned immediately after a seven will be $\frac{45}{49}$. Similarly, the probability of placing the second, third and fourth kings back so they don't immediately follow a seven will be $\frac{46}{50},\frac{47}{51}$ and $\frac{48}{52}$ respectively.
Thus the probability that a randomly shuffled deck has no kings immediately following a seven is
$$\frac{45}{49} \times \frac{46}{50} \times \frac{47}{51} \times \frac{48}{52}$$
which means your probability of "success" is
$$1 - \left( \frac{45}{49} \times \frac{46}{50} \times \frac{47}{51} \times \frac{48}{52} \right) \approx 0.281$$
A: Interestingly, the correct result in @A.J.’s answer can also be obtained with this flawed argument: We can choose $4$ cards to put after the $7$s; all such choices are equally likely, so the probability that there are no kings among the chosen cards is
$$
\frac{\binom{48}4}{\binom{52}4}=\frac{48\cdot47\cdot46\cdot45}{52\cdot51\cdot50\cdot49}\;.
$$
There are two errors in this argument, and it’s interesting that they cancel. One is that a $7$ might be at the end of the deck, and then there’s no card after it. The other is that we can’t put a $7$ after itself, but the $7$s were included in the cards picked (and they must be, as we can put another $7$ after a $7$).
That raises the question whether we can formulate a correct version of this argument that derives the result by choosing cards rather than slots (as @A.J.’s answer does).
Add a $53$rd marker card and uniformly randomly arrange the $53$ cards in a circle. In choosing the card that comes (say, clockwise) after the $7$ of hearts, there are now $52$ other cards to choose from, and $48$ of them are non-kings. Then we can choose one of $51$ cards that comes after the $7$ of spades, and $47$ of them are non-kings, and so on. The probability that no king comes after a $7$ is thus
$$
\frac{48\cdot47\cdot46\cdot45}{52\cdot51\cdot50\cdot49}\;.
$$
Now open the circle into a linear deck arrangement by removing the added marker card.
So the flawed argument yields the correct result because in each choice we can replace the excess option of placing the $7$ after itself by the missing option of placing the marker card after it.
A: by P.I.E,
let the cards be $A_1, A_2, A_3, A_4, B^4, C^{44}$ where A denotes the four specific 7's, B stand for kings and C for other cards.
let
$N_0=   { 4 \choose 0}   {52!\over 44!4!} $    decks that contain at least zero AB blocks
$N_1=   { 4 \choose 1}   {51!\over 44!3!} $    decks that contain at least one AB blocks ( overcounted )
$N_2=   { 4 \choose 2}   {50!\over 44!2!} $    decks that contain at least two AB blocks ( overcounted )
$N_3=   { 4 \choose 3}   {49!\over 44!1!} $    decks that contain at least three AB blocks ( overcounted )
$N_4=   { 4 \choose 4}   {48!\over 44!0!} $    decks that contain at least four AB blocks
After using P.I.E we get
$4,669,920$ decks that contains exactly four AB's $=0.0003 $ %
$896,624,640$ decks that contains exactly three AB's $=0.07 $ %
$31,606,018,560$ decks that contains exactly two AB's $=2.5 $ %
$323,083,745,280$ decks that contains exactly one AB's $=25.55 $ %
$908,673,033,600$ decks that contains no AB's $=71.87 $ %
== COMMENT  regarding the two fast solutions above ==
they are essentially the same.
When adding K's,
first K has to be placed in one of $1 + 0 + 4 + 44$ positions
second K has to be placed in one of $1 + 1 + 4 + 44$ positions
third K has to be placed in one of $1 + 2 + 4 + 44$ positions
fourth K has to be placed in one of $1 + 3 + 4 + 44$ positions
When removing 7's
first seven is to be removed from one of $1 + 3 + 4 + 44$ positions
second seven is to be removed from one of $1 + 2 + 4 + 44$ positions
third seven is to be removed from one of $1 + 1 + 4 + 44$ positions
fourth seven is to be removed from one of $1 + 0 + 4 + 44$ positions
for exemple, first $3$ refers to 3 kings other than the last king
the second $3$ refers to three 7's other than the first 7.
$1$ refers to an extra slot wherever it is. $4$ denotes bad choices, $44$ counts "any cards" .
