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:
( (4 C 1)(3 C 1) 50!) / 52! = 1/221
[Math] Probability ace on top and ace on bottom of shuffled deck
probability
Best Answer
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 $\frac{4}{52}\cdot \frac{3}{51} = \frac{1}{221}$