For 52 cards in a line, what is the probability that the card immediately to the right of the first queen is a queen or a seven?

Question

For a standard deck of 52 cards, lay down all 52 cards in a line. What is the probability that the card immediately to the right of the first queen is a queen or a seven?

I tried 51!*28/52! but I dont think that I'm accounting for finding the first queen.

51! ways to order the items so that queen is next to another queen or seven (treat as 1 item).

28 ways to rearrange first queen with another queen or seven

52! possible outcomes.

Answer 1

Break into cases:

The first queen is followed by another queen.

First, pick the order that the queens arrive in: $4!$ choices

Next, gluing the first two queens together, pick the three positions occupied by the queens in the $51$ positions for the deck: $\binom{51}{3}$ choices

Finally, from front to back, choose how all other positions are occupied by the remaining cards: $48!$ choices, giving then a total of $4!\cdot \binom{51}{3}\cdot 48!$ in this case

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The first queen is followed by a $7$.

First, pick which queen appears first and which $7$ it was: $4^2$ choices.

Gluing the $7$ to that queen, choose the order in which the remaining queens occur: $3!$ choices.

Choose the positions occupied by the queens (remembering the $7$ is glued to the queen in the front): $\binom{51}{4}$ choices.

Finally, choose how the remaining cards are arranged in the remaining positions: $47!$ choices, for a total of $4^2\cdot 3!\cdot \binom{51}{4}\cdot 47!$ choices.

This is out of the $52!$ different options for how the deck could be arranged for a final probability of:

$$\frac{4!\cdot\binom{51}{3}\cdot 48!+4^2\cdot 3!\cdot \binom{51}{4}\cdot 47!}{52!}$$

Answer 2

Consider only the queens and $7s$, every permutation of them will necessarily have the first queen followed by a queen or $7$

If we glue the desired combo together, we shall have $7$ objects which can serve as bars creating $8$ "compartments" where the remaining $44$ cards can be put, whereas unglued, they create $9$ compartments.

Using stars and bars,

$Pr = \dfrac{\dbinom{51}{7}}{\dbinom{52}{8}}= \dfrac 2 {13}$

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It later dawned on me that "stars and bars" (of which I am perhaps inordinately fond) need not be invoked, and the issue can be looked at more simply as below:

If we glue the desired combo together, we shall have a string of $7$ to be put among the $7+44= 51$ "cards", whereas unglued, we need a string of $8$ to be put among $8+44=52$ cards. [Since only $Pr$ has been asked for, we need not permute.]

And we get exactly the same formula as before

Answer 3

Consider a related question: What is the probability that the card following the first queen is the queen of hearts?

The entire deck can be arranged in $52!$ equally likely ways. We would like to count the number of these in which the card following the first queen is the queen of hearts. To do so, suppose we remove the queen of hearts from the deck and deal out the remaining $51$ cards. This can be done in $51!$ ways. There is only one place where the queen of hearts can be inserted if it immediately follows the first queen, so there are $51!$ such arrangements. So the probability that the card following the first queen is the queen of hearts is $$\frac{51!}{52!} = \frac{1}{52}$$

The same argument applies to any of the other queens or to any of the sevens: $8$ cards in all, including the queen of hearts. So the probability that the card following the first queen is a queen or a seven is $$8 \cdot \frac{1}{52} = \frac{2}{13}$$