Would someone please explain the answer below:
What is the probability that a randomly shuffled deck of $52$ cards has an ace as the top card and an ace as the bottom card?
ANSWER: $$\frac{\binom{4}{1}\times\binom{3}{1}\times50!}{52!}= \frac{1}{221}$$
On
As alluded to in the comments above, instead of approaching via direct counting techniques, it may be more intuitive to approach via conditional probability arguments.
$Pr(\text{First card is an ace}\wedge \text{Last card is an ace}) = Pr(\text{First is an ace})\cdot Pr(\text{Last is an ace}|\text{First is an ace})$
Now, notice that we may choose the first card first, and the last card second. Given that the first card is an ace, every other spot is equally likely to contain an ace.
The probability that an ace occupies the first spot is $\frac{4}{52}$ as there are four aces available in the deck of $52$ cards total.
Given that the first card is an ace, the probability that the last card is an ace is then $\frac{3}{51}$ as there are three remaining aces (does not matter which ace the first was) out of $51$ remaining cards.
The combined probability is then $\frac{4}{52}\cdot \frac{3}{51} = \frac{1}{221}$
$4~C~1=$ combination of $4$ taken $1$ times ---> the number of cases to choose the ace from the top
$3~C~1=$ combination of $3$ taken $1$ times----> the number of cases to choose the bottom ace (because one ace has been used on top)
$50!=$ this number represent how many chooses you have to arrange the cards between the two aces. $50!$ = arrangement of $50$ taken $50$ times.
$52!=$ the total possibilities to arrange your deck = permutation of $52$
$((4~C~1) (3~C~1) 50!)=$ number of favorable cases $52!$ = number of total cases
Probability is then: $\dfrac{(4~C~1) (3~C~1) 50!}{52!}$