# Probability selecting three cards out of a deck

In Introduction to Probability by Blitzstein & Hwang, Chapter 2 Problem 5:

Three cards are dealt from a standard, well-shuffled deck. The first two cards are flipped over, revealing the Ace of Spades as the first card and the 8 of Clubs as the second card. Given this information, find the probability that the third card is an ace in two ways: using the definition of conditional probability, and by symmetry.

Solution:

Let A be the event that the first card is Ace of Spades, B be the event that second card is 8 of Clubs, and C be the event that third card is an Ace.

$$P(C|A,B) = \dfrac{P(A,B,C)}{P(A,B)}$$

Numerator: Having first as Ace of Spade, second as 8 of Clubs and third as an Ace, is similar to choosing three cards out of 52 cards without replacement. However, there are 3 ways for the third card to be an Ace since there are three Aces left, Ace of Hearts, Diamonds, and Clubs. $$P(A,B,C) = 3\cdot(\dfrac{1}{52})(\dfrac{1}{51})(\dfrac{1}{50})$$

Denominator: This is the same as choose two cards out of 52 without replacement. $$P(A,B) = (\dfrac{1}{52})(\dfrac{1}{51})$$

Therefore, $$P(C|A,B) = \dfrac{P(A,B,C)}{P(A,B)} = \dfrac{3\cdot(\dfrac{1}{52})(\dfrac{1}{51})(\dfrac{1}{50})}{(\dfrac{1}{52})(\dfrac{1}{51})} = \dfrac{3}{50}$$

Is this solution correct? By the way I don't get it as how to use symmetry to view this problem...

Symmetry: there are 50 cards left. Each has the same probability so the probability to get an ace is $$\tfrac{3}{50}$$ as there are 3 aces left.